State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 1

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 2

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 3

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 4

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 5

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 6

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

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Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 8

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 9

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 10

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 11

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 12

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 13

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 14

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 15

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 16

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 17

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 18

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 19

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 20

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 21

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

eSolutions Manual - Powered by Cognero Page 22

Study Guide and Review - Chapter 4

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.  

State whether each sentence is true or false . If false , replace the underlined term to make a true sentence.

1. The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse.

SOLUTION:  The sine of an acute angle in a right triangle is the ratio of the lengths of its opposite leg to the hypotenuse. The cosine is the ratio of the lengths of its adjacent leg to the hypotenuse. The tangent is the ratio of the lengths of its opposite leg to the adjacent leg.

2. The secant ratio is the reciprocal of the sine ratio.

SOLUTION:  The secant ratio is the reciprocal of the cosine ratio, while the cosecant ratio is the reciprocal of the sine ratio. The cotangent is the reciprocal of the tangent ratio.

3. An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object below the line.

SOLUTION:  An angle of elevation is the angle formed by a horizontal line and an observer’s line of sight to an object above the line, while an angle of depression isthe angle formed by a horizontal line and an observer’s line of sight to an object below the line.

4. The radian measure of an angle is equal to the ratio of the length of its intercepted arc to the radius.

SOLUTION:  The measure θ in radians of a central angle is equal to the ratio of the length of the intercepted arc s to the radius r of the circle. A central angle has a measure of 1 radian if it intercepts an arc with the same length as the radius of the circle.  

5. The rate at which an object moves along a circular path is called its linear speed.

SOLUTION:  The rate at which an object moves along a circular path is called its linear speed. The rate at which the object rotates about a fixed point is its angular speed.

6. 0 , , and  are examples of reference angles.

SOLUTION:  

0 , , and  are examples of quadrantal angles. 

Quadrantal angles lie on one of the coordinate axes. These angles are not used as reference angles.

7. The period of the graph of y = 4 sin 3x is 4.

SOLUTION:  The amplitude of the graph of y = 4 sin 3x is 4. The

period is . The frequency is .

8. For f (x) = cos bx, as b increases, the frequency decreases.

SOLUTION:  For f (x) = cos bx, as b increases, the frequency increases while the period decreases. The frequencyand period are reciprocals.

9. The range of the arcsine function is [0, ].

SOLUTION:  

The range of the arccosine function is [0, ], while

the range of the arcsine function is .

10. The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles.

SOLUTION:  The Law of Sines can be used to determine unknown side lengths or angle measures of some triangles. Such occasions are when AAS, ASA, and SSA are known. Sometimes, when SSS and SAS areknown, the Law of Cosines will be needed as well.

Find the exact values of the six trigonometric functions of θ.

11. 

SOLUTION:  

The length of the side opposite  is 36, the length of the side adjacent to  is 15, and the length of the hypotenuse is 39.

12. 

SOLUTION:  

The length of the side opposite θ is 9, the length of

the side adjacent to  is 40, and the length of the hypotenuse is 41.

Find the value of x. Round to the nearest tenth,if necessary.

13. 

SOLUTION:  An acute angle measure and the length of the hypotenuse are given, so the sine function can be

used to find the length of the side opposite .

14. 

SOLUTION:  An acute angle measure and the length of a leg is given, so the cosine function can be used to find the length of the hypotenuse.

Find the measure of angle θ. Round to the nearest degree, if necessary.

15. 

SOLUTION:  

Because the length of the side opposite θ and the hypotenuse are given, use the sine function.

16. 

SOLUTION:  Because the lengths of the sides opposite and

adjacent to θ  are given, use the tangent function.

Write each degree measure in radians as a multiple of π and each radian measure in degrees.

17. 135

SOLUTION:  To convert a degree measure to radians, multiply by

18. 450

SOLUTION:  To convert a degree measure to radians, multiply by

19. 

SOLUTION:  To convert a radian measure to degrees, multiply by

20. 

SOLUTION:  To convert a radian measure to degrees, multiply by

Identify all angles coterminal with the given angle. Then find and draw one positive and one negative angle coterminal with the given angle.

21. 342

SOLUTION:  

All angles measuring are coterminal

with a 342  angle.    Sample answer: Let n = 1 and −1.  

22. 

SOLUTION:  

All angles measuring are coterminal with a

 radian angle. 

  Sample answer: Let n = 1 and −1.  

 

Find the area of each sector.

23. 

SOLUTION:  

The measure of the sector’s central angle  is 31° and the radius is 14 inches. Convert the central anglemeasure to radians.

Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 53.0 square inches.

24. 

SOLUTION:  

The measure of the sector’s central angle   is 

 and the radius is 10 meters.

  Use the central angle and the radius to find the area of the sector.

Therefore, the area of the sector is about 246.1 square meters.

Sketch each angle. Then find its reference angle.

25. 240

SOLUTION:  The terminal side of 240º lies in Quadrant III.

Therefore, its reference angle is  ' = 240 – 180or 60º.

26. 75

SOLUTION:  

The terminal side of 75 lies in Quadrant I.

Therefore, its reference angle is  ' = 75 – 0 or

75 .

27. 

SOLUTION:  

The terminal side of  lies in Quadrant III. 

Therefore, its reference angle is  ' = .

28. 

SOLUTION:  

The terminal side of  lies in Quadrant II. 

Therefore, its reference angle is  ' =

.

 

Find the exact values of the five remaining trigonometric functions of θ.

29. cos  =  , where sin  > 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ and tan  are positive,  must lie in Quadrant I. This means that both x and y are positive.  

Because cos  =   or  , x =2 and r = 5. Find y .

 

 

Use x = 2, y = , and r = 5 to write the five remaining trigonometric ratios.

30.  where sin  > 0 and cos   < 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is positive and cos  is negative,  must lie in Quadrant II. This means that x is

negative and y is positive.  

Because tan  =   or – , use the point (–4, 3) to

find r.

 

  Use x = –4, y = 3, and r = 5 to write the five remaining trigonometric ratios.

31.  where cos  > 0 and cot  < 0

SOLUTION:  

where cos  > 0 and cot   < 0

To find the other function values, find the

coordinates of a point on the terminal side of .

Because cos θ is positive and cot θ is negative,

 must lie in Quadrant IV. This means that x is positive and y is negative.  

Because sin  =   or – , y = –5 and r = 13.

Find x.

 

  Use x = 12, y = –5, and r = 13 to write the five remaining trigonometric ratios.

32. cot θ = , where sin  < 0 and tan   > 0

SOLUTION:  To find the other function values, find the

coordinates of a point on the terminal side of .

Because sin θ is negative and tan  is positive,  must lie in Quadrant III. This means that both x

and y are negative.  

Because cot  =   or  , use the point (–2, –3) to

find r.

 

 

Use x = –2, y = –3, and r =  to write the five remaining trigonometric ratios.

Find the exact value of each expression.  If undefined, write undefined.

33. sin 180

SOLUTION:  

Because the terminal side of   lies on the negative x-axis, the reference angle  ' is 180°.

34. cot

SOLUTION:  

Because the terminal side of  lies in Quadrant IV, 

the reference angle θ' of    is  .

In quadrant IV cot θ is negative.  

35. sec 450

SOLUTION:  

Because the terminal side of  lies on the positive y-

axis, the reference angle  ' is  .

 

36. 

SOLUTION:  

Because the terminal side of θ lies in Quadrant III,

the reference angle θ' of    is  .  

 

Describe how the graphs of f (x)and g (x) are related. Then find the amplitude and period of g (x) , and sketch at least one period of both functions on the same coordinate axes.

37. f (x) = sin x; g(x) = 5 sin x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded

vertically. The amplitude of g(x) is |5| or 5, and the

period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one

period, 2π, on the interval [0, 2 ]. Then use the amplitude of g(x) to find corresponding points on its graph.  

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x g(x) = 5 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

38. f (x) = cos x; g(x) = cos 2x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

horizontally. The amplitude of g(x) is |1| or 1, and the

period is or .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                         

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x g(x) = cos 2x

Max (0, 1) (0, 1)

x-int

Min (π, –1)

x-int

Max (2π, 1) (π, 1)

39. f (x) = sin x; g(x) =  sin x

SOLUTION:  The graph of g(x) is the graph of f (x) compressed

vertically. The amplitude of g(x) is  or  , and

the period is or 2 .

  Create a table listing the coordinates of the x-intercepts and extrema for f (x) = sin x for one period, 2π, on the interval [0, 2π]. Then use the amplitude of g(x) to find corresponding points on its graph.  

    Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = sin x

g(x) =

 sin x

x-int (0, 0) (0, 0)

Max

x-int (π, 0) (π, 0)

Min

x-int (2π, 0) (2π, 0)

40. f (x) = cos x; g(x) = −cos x

SOLUTION:  The graph of g(x) is the graph of f (x) expanded horizontally and reflected in the x-axis. The

amplitude of g(x) is |–1| or 1, and the period is or

6 .   Create a table listing the coordinates of the x-intercepts and extrema for f (x) = cos x for one

period, 2π, on the interval [0, 2 ]. Then use g(x) to find corresponding points on its graph.  

                           

  Sketch the curve through the indicated points for each function. Then repeat the pattern to complete asecond period.

Functions f (x) = cos x

g(x) = –cos

x

Max (0, 1) (0, −1)

x-int

Min (π, –1) (3π, 1)

x-int

Max (2π, 1) (6π, −1)

State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then graph two periods of the function.

41. y = 2 cos (x – )

SOLUTION:  

In this function, a = 2, b = 1, c =   , and d = 0. Because d = 0, there is no vertical shift.  

  Graph y = 2 cos x shifted units to the right.

42. y = −sin 2x + 1

SOLUTION:  

In this function, a = , b = 2, c =  0, and d = 1. 

  Graph y = – sin 2x shifted 1 unit up.

43. 

SOLUTION:  

In this function, a = , b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y =  cos x shifted units to the left.

44. 

SOLUTION:  

In this function, a = 3, b = 1, c =   , and d = 0.

Because d = 0, there is no vertical shift.  

 

Graph y = 3 sin x shifted units to the left.

Locate the vertical asymptotes, and sketch the graph of each function.

45. y = 3 tan x

SOLUTION:  The graph of y = 3 tan x is the graph of y = tan x

expanded vertically. The period is  or  . Find

the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = 3 tan x for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x y = 3 tan x

Vertical Asymptote

Intermediate Point x-int   (0, 0) (0, 0)

Intermediate Point

Vertical Asymptote

46. 

SOLUTION:  

The graph of  is the graph of y =

tan x compressed vertically and translated  units 

right. The period is  or  . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [0, ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = tan x

Vertical Asymptote x = 0

Intermediate Point x-int   (0, 0)

Intermediate Point

Vertical Asymptote x = π

47. 

SOLUTION:  

The graph of  is the graph of y = cot

x shifted units to the left. The period is  or  .

Find the location of two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π

48. y = −cot (x – )

SOLUTION:  

The graph of  is the graph of y = cot

x reflected in the x-axis and shifted π units to the

left. The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [ , 2 ].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = cot x

Vertical Asymptote x = 0

Intermediate Point x-int  

Intermediate Point

Vertical Asymptote x = π x = 2π

49. 

SOLUTION:  

The graph of  is the graph of y = sec x

expanded vertically and expanded horizontally. The

period is  or 4 . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on [−π, 3π].

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

Vertical Asymptote

Intermediate Point

(0, 1) (0, 1)

x-int   x = π

Intermediate Point

Vertical

Asymptote x = 3π

50. y = –csc (2x)

SOLUTION:  

The graph of  is the graph of y = csc x

compressed horizontally and reflected in the x-axis.

The period is  or  . Find the location of two

consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = csc x

y = –csc 2x

Vertical Asymptote x = −π

Intermediate Point x-int   x = 0 x = 0

Intermediate Point

Vertical Asymptote x = π

51. y = sec (x – )

SOLUTION:  

The graph of y = sec (x − ) is the graph of y = sec

x translated π units to the right. The period is  or 

2 . Find the location of two vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for y = sec (x − π) for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function   y = sec x

y = sec (x − π)

Vertical Asymptote

Intermediate Point

(0, 1)

x-int  

Intermediate Point

Vertical Asymptote

52. 

SOLUTION:  

The graph of  is the graph of y =

csc x compressed vertically and translated  units to

the left. The period is or 2 . Find the location of

two consecutive vertical asymptotes.

     and     

Create a table listing the coordinates of key points

for for one period on .

 

  Sketch the curve through the indicated key points forthe function. Then repeat the pattern to sketch at least one more cycle to the left and right of the first curve.

Function  

y = csc x

Vertical Asymptote Intermediate

Point x-int   x = 0

Intermediate Point

Vertical Asymptote x = π

Find the exact value of each expression, if it exists.

53. sin−1 (−1)

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of –1.

 

When t = , sin t = –1. Therefore, sin – 1

–1 =

.

54. 

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

 

When t = , cos t = . Therefore, cos– 1

=

.

55. 

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, tan–1

 =  .

56. arcsin 0

SOLUTION:  Find a point on the unit circle on the interval

 with a y-coordinate of 0.

When t = 0, sin t = 0. Therefore, arcsin  0 = 0.

57. arctan −1

SOLUTION:  Find a point on the unit circle on the interval

 such that  =  

When t = , tan t = . Therefore, arctan

 =  .

58. arccos

SOLUTION:  Find a point on the unit circle on the interval

 with a x-coordinate of .

When t = , cos t = . Therefore, arccos =

.

59. 

SOLUTION:  

The inverse property applies, because  lies on 

the interval [–1, 1]. Therefore, =

.

60. 

SOLUTION:  

The inverse property applies, because cos –3π lies on the interval [–1, 1]. Therefore,

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measurements to the nearest degree.

61. a = 11, b = 6, A = 22

SOLUTION:  Draw a diagram of a triangle with the given dimensions.  

  Notice that A is acute and a > b because 11 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C  180 –

(22 + 12 ) or about 146 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B  12 , C  146 , and c  16.4.

62. a = 9, b = 10, A = 42

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a < b because 9 < 10. Therefore, two solutions may exist. Find h.

9 > 6.69, so two solutions exist. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(42 + 48 ) or about 90 . Apply the Law of Sines to find c.

  When,  then C  180 – (42 + 132 ) orabout 6º. Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  48 , C  90 , and c  13.5 or B = 132 , C =

6 , and c = 1.4.

63. a = 20, b = 10, A = 78

SOLUTION:  Draw a diagram of a triangle with the given dimensions.

  Notice that A is acute and a > b because 20 > 10. Therefore, one solution exists. Apply the Law of Sines to find B.

  Because two angles are now known, C  180 –

(78 + 29 ) or about 73 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B  29 , C  73 , and c  19.5.

64. a = 2, b = 9, A = 88

SOLUTION:  

Notice that A is acute and a < b because 2 < 9. Find h.

Because a < b and a < h, no triangle can be formed

with sides a = 2, b = 9, and A = 88 . Therefore, this problem has no solution.

Solve each triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree.

65. a = 13, b = 12, c = 8

SOLUTION:  Use the Law of Cosines to find an angle measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, A  78 , B  65 , and C  37 .

66. a = 4, b = 5, C = 96

SOLUTION:  Use the Law of Cosines to find the missing side measure.

Use the Law of Sines to find a missing angle measure.

Find the measure of the remaining angle.

Therefore, c = 6.7, A = 36 , and B = 48 .

67. CONSTRUCTION  A construction company is installing a three-foot-high wheelchair ramp onto a landing outside of an office. The angle of the ramp must be 4°. a. What is the length of the ramp? b. What is the slope of the ramp?

SOLUTION:  a. We are given the angle and the height opposite theangle. Draw a diagram.

  We need to find the length of the ramp, so we can use the tangent function.

  b. The slope is equal to the rise over the run. One coordinate is (0, 0) and another coordinate is (43, 3).

68. NATURE  For a photography project, Maria is photographing deer from a tree stand. From her sight30 feet above the ground, she spots two deer in a straight line, as shown below. How much farther away is the second deer than the first?

SOLUTION:  Draw a diagram of the situation.

  The 40° and 60° angles shown are equal to the angles in the original picture because they are alternate interior angles.   We can use trigonometry to find x and y and then

find y − x, which is the distance between the deer.

 

69. FIGURE SKATING  An Olympic ice skater performs a routine in which she jumps in the air for 2.4 seconds while spinning 3 full revolutions. a. Find the angular speed of the figure skater. b. Express the angular speed of the figure skater in degrees per minute.

SOLUTION:  a. The rate at which an object rotates about a fixed point is its angular speed. The angular speed is given

by . Each rotation requires 2 radians, so θ =

6 .

The angular speed is 2.5 radians per second.     b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

eSolutions Manual - Powered by Cognero Page 23

Study Guide and Review - Chapter 4

  b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

  b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

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Study Guide and Review - Chapter 4

  b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

  b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

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Study Guide and Review - Chapter 4

  b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

  b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

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  b. Convert the seconds to minutes then convert the radians to degrees.

The angular speed is about 27,000 degrees per minute.

70. TIMEPIECES  The length of the minute hand of a pocket watch is 1.5 inches. What is the area swept by the minute hand in 40 minutes? 

SOLUTION:  After 40 minutes, the minute hand has traveled

 of a revolution. Multiply by 360 to determine

the central angle θ.

The area of a sector is determined by A = r2

,

where r is the radius and  is the measure of the central angle in radians. We need to convert the degrees to radians.

Substitute the known values of r and to determine A.

The area is about 4.71 square inches.

71. WORLD’S FAIR  The first Ferris wheel had a diameter of 250 feet and took 10 minutes to complete one full revolution. a. How many degrees would the Ferris wheel rotate in 100 seconds? b. How far has a person traveled if he or she has been on the Ferris wheel for 7 minutes? c. How long would it take for a person to travel 200 feet?

SOLUTION:  a. One complete revolution is done in 10 minutes, so divide 100 seconds by 10 minutes to determine the fraction of a revolution that has been completed in 100 seconds.

One sixth of a revolution is equal to  or 60 

degrees.   b. If someone has been on it for 7 minutes, then they

have made  of a revolution. The wheel has a 

diameter of 250 feet, so the circumference is 250

feet. After 7 minutes,  of the circumference has 

been traveled.

They have traveled about 550 feet.   c. It takes 10 minutes to travel the circumference, or

250 feet. Use a proportion to solve for t.

It would take about 2.5 minutes to travel 200 feet.

72. AIR CONDITIONING  An air-conditioning unit turns on and off to maintain the desired temperature.On one summer day, the air conditioner turns on at

8:30 A.M. when the temperature is 80 Fahrenheit and turns off at 8:55 A.M. when the temperature is 74°. a. Find the amplitude and period if you were going touse a trigonometric function to model this change in temperature, assuming that the temperature cycle will continue. b. Is it appropriate to model this situation with a trigonometric function? Explain your reasoning.

SOLUTION:  a. The range of temperatures is 6 degrees. Half of this range is 3, so the amplitude is 3. The temperature takes 25 minutes to go down and we are assuming that it will take 25 minutes to go back up, so one full period is 50 minutes.   b. Sample answer: The function appears to not be periodic because it might take more or less time for

the temperature to warm back up to 80 . The temperature will increase at a different rate depending on the time of day. For example, it will gethotter much quicker between 10 in the morning and 2 in the afternoon.

73. TIDES  In Lewis Bay, the low tide is recorded as 2 feet at 4:30 A.M., and the high tide is recorded as 5.5 feet at 10:45 A.M. a. Find the period for the trigonometric model. b. At what time will the next high tide occur?

SOLUTION:  a. The period is from one low tide to the next low tide. From low tide to high tide, the time elapsed is 6 hours and 15 minutes. Therefore, the period is twice that, or 12 hours and 30 minutes.   b. The last high tide occurred at 10:45 in the morning, so the next one will occur 12 hours and 30 minutes from then, or at 11:15 in the evening.

74. MUSIC  When plucked, a bass string is displaced 1.5 inches, and its damping factor is 1.9. It produces a note with a frequency of 90 cycles per second. Determine the amount of time it takes the string’s

motion to be dampened so that −0.1 < y   < 0.1.

SOLUTION:  First, write a trigonometric function that models that motion of the string. The maximum displacement of

the string occurs when t = 0, so y = ke–ct

cos t can be used to model the motion of the string because the graph of y = cos t has a y-intercept other than 0.   The maximum displacement occurs when the string is plucked 1.5 inches. The total displacement is the maximum displacement M minus the minimum displacement m, so k = M – m = 1.5 in.   You can use the value of the frequency to find .

  Write a function using the values of k , , and c.

y = 1.5e–1.9t

t is one model that describes the motion of the string.  

Use a graphing calculator to graph y = 1.5e–1.

t, y = –0.1, and y = 0.1. Select the intersectfeature under the TRACE menu to find the point

where the graphs of y = 1.5e–1.9t

t and

y = –0.1 or y = 0.1 first intersect within the interval –0.1 ≤ y  ≤ 0.1.

From the graph, you can see that it takes approximately 1.41 seconds for the graph of y =

1.5e–1.9t

t to oscillate within the interval –0.1 ≤ y  ≤ 0.1.

75. PAINTING  A painter is using a 15-foot ladder to paint the side of a house. If the angle the ladder

makes with the ground is less than 65 , it will slide out from under him. What is the greatest distance that the bottom of the ladder can be from the side of the house and still be safe for the painter?

SOLUTION:  The smaller the angle, the larger the distance from the ladder to the house. Therefore, to find the greatest safe distance from the house, we can use the smallest safe angle. The minimum angle that we can use is 65 degrees, so find the distance from the ladder to the side of the house at this minimum angle.Use the cosine function.  

At the minimum angle, the ladder is about 6.3 feet away from the side of the house.

76. NAVIGATION  A boat is 20 nautical miles from a port at a bearing 30° north of east. The captain sees a second boat and reports to the port that his boat is 15 nautical miles from the second boat, which is located due east of the port. Can port personnel be sure of the second boat’s position? Justify your answer.

SOLUTION:  Draw a diagram of the situation.  

  From the diagram, there are two possible positions where the second boat can be located. This follows the ambiguous case of the Law of Sines. 10 < 15 < 20, so there are two possible triangles with the given dimensions.

77. GEOMETRY  Consider quadrilateral ABCD.

a. Find C. b. Find the area of ABCD.

SOLUTION:  a. Use SAS to find the length of BD.

Use SSS to find C.  

  b. Use SAS to find the area of each triangle formed by BD.

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