Chapter 4 Trig Other Book

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C H A P T E R 4

Trigonometry

Section 4.1 Radian and Degree Measure . . . . . . . . . . . . . . . . 335

Section 4.2 Trigonometric Functions: The Unit Circle . . . . . . . . . 344

Section 4.3 Right Triangle Trigonometry . . . . . . . . . . . . . . . . 350

Section 4.4 Trigonometric Functions of Any Angle . . . . . . . . . . 360

Section 4.5 Graphs of Sine and Cosine Functions . . . . . . . . . . . 373

Section 4.6 Graphs of Other Trigonometric Functions . . . . . . . . . 386

Section 4.7 Inverse Trigonometric Functions . . . . . . . . . . . . . . 397

Section 4.8 Applications and Models . . . . . . . . . . . . . . . . . . 410

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432

Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436



335

C H A P T E R 4

Trigonometry

Section 4.1 Radian and Degree Measure

You should know the following basic facts about angles, their measurement, and their applications.

■ Types of Angles:

(a) Acute: Measure between 0 and 90.

(b) Right: Measure 90.

(c) Obtuse: Measure between 90 and 180 .

(d) Straight: Measure 180.

■ and are complementary if They are supplementary if

■ Two angles in standard position that have the same terminal side are called coterminal angles.

■ To convert degrees to radians, use radians.

■ To convert radians to degrees, use 1 radian

■ one minute of 1.

■ one second of 1.

■ The length of a circular arc is where is measured in radians.

■ Linear speed

■ Angular speed t srt

arc length

time

s

t

s r

160 of 1 136001

1601

180 .

1 180

180. 90.

Vocabulary Check

1. Trigonometry 2. angle

3. coterminal 4. radian

5. acute; obtuse 6. complementary; supplementary

7. degree 8. linear

9. angular 10. A 12r 2

1.

The angle shown is approximately

2 radians.

2.


5.5 radians.

3.


radians.3

4.


radians.4

5.


1 radian.

6.


6.5 radians.



336 Chapter 4 Trigonometry

7. (a) Since lies in Quadrant I.

(b) Since lies in Quadrant III. <7

5<

3

2 ;

7

5

0 <

5<

2 ;

5

9. (a) Since lies in Quadrant IV.

(b) Since lies in Quadrant III. < 2 <

2; 2

2 <

12 < 0 ;

12

8. (a) Since lies in Quadrant III.

(b) Since lies in Quadrant III. <9

8 <

3

2;

9

8

<11

8 <

3

2;

11

8

10. (a) Since lies in Quadrant IV.

(b) Since lies in Quadrant II.11

9

3

2 <

11

9 < ,

2< 1 < 0; 1

11. (a) Since lies in Quadrant III.

(b) Since lies in Quadrant II.

2< 2.25 < ; 2.25

< 3.5 <

3

2 ; 3.5 12. (a) Since lies in Quadrant IV.

(b) Since 4.25 lies in Quadrant II.3

2 < 4.25 < ;

3

2< 6.02 < 2 ; 6.02

13. (a)

(b)

−2

3

π

x

y

2

3

5

4

π

x

y

5

4 14. (a)

(b)

x

y

5

2

π

5

2

−7

4

π

y

x

7

4 15. (a)

(b)

−3

x

y

3

11

6

π

x

y

11

6

16. (a) 4

4

x

y (b)

7π

x

y7



Section 4.1 Radian and Degree Measure 337

26.


120.

27.


60º.

28.


330.

17. (a) Coterminal angles for

(b) Coterminal angles for

5

6 2

7

6

5

6 2

17

6

5

6

6 2

11

6

6 2

13

6

6 18. (a)

(b)

11

6 2

23

6

11

6 2

6

7

6 2

5

6

7

6 2

19

619. (a) Coterminal angles for


12 2

23

12

12 2

25

12

12

2

3 2

4

3

2

3 2

8

3

2

3

20. (a)

(b)

2

15 2

32

15

2

15 2

28

15

9

4

4 7

4

9

4 2

4 21. (a) Complement:

Supplement:

(b) Complement: Not possible, is greater than

Supplement: 3

4

4

2.

3

4

3

2

3

2

3

6

22. (a) Complement:

Supplement:


Supplement: 11

12

12

2.

11

12

12

11

12

2

12

5

12 23. (a) Complement:

Supplement:

(b) Complement: Not possible, 2 is greater than

Supplement: 2 1.14

2.

1 2.14

2 1 0.57

24. (a) Complement: Not possible, 3 is greater than

Supplement:

(b) Complement:

Supplement: 1.5 1.64

2 1.5 0.07

3 0.14

2. 25.

The angle shown is approximately .210º




31. (a) Since lies in Quadrant II.

(b) Since lies in Quadrant IV.270 < 285 < 360, 285

90 < 130 < 180, 130 32. (a) Since lies in Quadrant I.

(b) Since lies in

Quadrant III.

180 < 257 30 < 270, 257 30

0 < 8.3 < 90, 8.3

33. (a) Since

lies in Quadrant III.

(b) Since lies in

Quadrant I.

360 < 336 < 270, 336

180 < 132 50 < 90, 132 50 34. (a) Since lies in

Quadrant II.

(b) Since lies in Quadrant IV.90 < 3.4 < 0, 3.4

270 < 260 < 180, 260

35. (a)

(b)

150°

x

y

150

x

30°

y

30 36. (a)

(b)

−120

°

x

y

120

−270°

x

y

270

37. (a)

(b)

480°

x

y

480

405°

x

y

405

29.

The angle shown is approximately .165

30.

The angle shown is approximately 10.

38. (a)

(b)

−600°

x

y

600

−750°

x

y

750




54. (a)

(b) 34 15

34 15 180

408

11

6

11

6 180

330 55.

2.007 radians

115 115

180 56.

1.525 radians

87.4 87.4

180

57. 216.35 216.35

180 3.776 radians

59. 532 532

180 9.285 radians

58. radians48.27 48.27

180 0.842

60. 345 345

180 6.021 radians

39. (a) Coterminal angles for 45

45 360 405

45 360 315


36 360 324

36 360 396

36

40. (a)

(b)

420 360 60

420 720 300

120 360 240

120 360 480 41. (a) Coterminal angles for


180 360 540

180 360 180

180

240 360 120

240 360 600

240

42. (a)

(b)

230 360 130

230 360 590

420 360 60

420 720 300 43. (a) Complement: 90 18 72

Supplement:

(b) Complement: Not possible, 115 is greater than 90 .

Supplement: 1180 115 65

180 18 162

44. (a) Complement:

Supplement:

(b) Complement:


90 64 26

180 3 177

90 3 87 45. (a) Complement:

Supplement:



90.150

180 79 101

90 79 11

46. (a) Complement: Not possible, is greater than

Supplement:



90.170

180 130 50

90.130 47. (a)

(b) 150 150

180 5

6

30 30

180

6

48. (a)

(b) 120 120

180 2

3

315 315

180 7

4 49. (a)

(b) 240 240

180 4

3

20 20

180

9 50. (a)

(b) 144 144

180 4

5

270 270

180 3

2

51. (a)

(b)7

6

7

6 180

210

3

2

3

2 180

270 52. (a)

(b)

9

9180

20

7

12

7

12180

105 53. (a)

(b) 11

30

11

30 180

66

7

3

7

3 180

420




61. radian0.83 0.83

180 0.014 62. 0.54 0.54

180 0.009 radians

63.

7

7180

25.714 64.

5

11

5

11180

81.818 65.15

8

15

8 180

337.500

66.13

2

13

2 180

1170.000 67.

756.000

4.2 4.2 180

68. 4.8 4.8 180

864.000

69. 2 2180

114.592 70. 0.57 0.57180

32.659

71. (a)

(b) 128 30 128 3060 128.5

54 45 54 4560 54.75

73. (a)

(b) 330 25 330 25

3600 330.007

85 18 30 85 1860 303600 85.308

75. (a)

(b) 145.8 145 0.860 145 48

240.6 240 0.660 240 36

77. (a)

(b)

3 34 48

3 34 0.860

3.58 3 0.5860

2.5 2 30

72. (a)

(b) 2 0.2 2.2212 2 1260

245.167 245 0.16724510 245 1060

74. (a)

(b)

408.272

408 0.2667 0.0056

408 16 20 408 1660 20

3600

135 0.01 135.01

135 36 135 363600

76. (a)

(b) 0 27

0 27 0.45 0 0.4560

345 7 12

345 7 0.260

345.12 345 0.1260

78. (a)

(b)

0 47 11.4 0 47 11.4

0 47 0.1960

0.7865 0 0.786560

0 21 18 0 21 18

0 21 0.360

0.355 0 0.35560

79.

65 radians

6 5

s r 81.

327 4

47 radians

32 7

s r 80.

2910 radians

29 10

s r

82.

Because the angle represented is

clockwise, this angle is radians.45

60

75

4

5 radians

60 75

s r 83.

6

27

2

9 radians

6 27

s r 84. feet, feet

s

r

8

14

4

7 radians

s 8r 14




100. sr 24

5 4.8 radians 4.8180

275

91.

8.38 square inches

A 1

242 3

8

3 square inches

A 1

2r 2 92.

56.55 mm2

18 mm2

A 1

2r 2

1

2122 4

r 12 mm,

4 93.

12.27 square feet

A 1

22.52225

180

A 1

2r 2

94.

square miles 5.6421.56

12 A

1

21.42330

180 r 1.4 miles, 330 95.

miless r 40000.14782 591.3

8.46972 0.14782 radian

41 15 50 32 47 39

96.

s r 40000.1715 686.2 miles

0.1715 radian

47 37 18 37 47 36 9 49 42

r 4000 miles97.

s

r

450

6378 0.071 radian 4.04

98. kilometers

The difference in latitude is about radians 3.59.0.062716

0.062716

400

6378

400 6378

s r

r 3189 99. s

r

2.5

6

25

60

5

12 radian

85.

25

14.5

50

29 radians

25 14.5

s r 87. in radians

47.12 inches

s 15180

180 15 inches

s r , 86. kilometers,

kilometers

s

r

160

80 2 radians

s 160

r 80

88. feet,

9.42 feet

s r 9 3 3 feet

60

3r 9 89. in radians

s 31 3 meters

s r , 90. centimeters,

15.71 centimeters

s r 20 4 5 centimeters

4r 20

101. (a) 65 miles per hour feet per minute

The circumference of the tire is feet.

The number of revolutions per minute is

revolutions per minuter 5720

2.5 728.3

C 2.5

655280

60 5720 (b) The angular speed is

Angular speed radians per minute4576 radians

1 minute 4576

5720

2.5 2 4576 radians

t .




102. Linear velocity for either pulley: inches per minute

(a) Angular speed of motor pulley: radians per minute

Angular speed of the saw arbor: radians per minute

(b) Revolutions per minute of the saw arbor: revolutions per minute

1700

2 850

v

r

3400

2 1700

v

r

3400

1 3400

17002 3400

103. (a)

(b)

9869.84 feet per minute

31412

3 feet per minute

Linear speed 7.25

2 in. 1 ft

12 in.52002

1 minute

32,672.56 radians per minute

10,400 radians per minute

Angular speed 52002 radians

1 minute104. (a)

(b)

628.32 feet per minute

Linear speed 2525.13274 feet per minute

r

t 200 feet per minute

r 25 ft

25.13 radians per minute

8 radians per minute

4 rpm 42 radians per minute

105. (a)

Interval:

(b)

Interval: 2400 , 6000 centimeters per minute

62002 ≤ Linear speed ≤ 65002 centimeters per minute

400 , 1000 radians per minute

2002 ≤ Angular speed ≤ 5002 radians per minute

106.

A 1

2125

180 252 112 175 549.8 square inches

r 25 14 11

R 25

25125°

14r

A 1

2 R2 r 2 107.

1496.62 square meters

476.39 square meters

12352140 180140°

35

A 1

2r 2

108. (a) Arc length of larger sprocket in feet:

Therefore, the chain moves feet, as does the smaller rear sprocket. Thus, the angle of the

smaller sprocket is

and the arc length of the tire in feet is:

—CONTINUED—

14 feet

3 seconds

3600 seconds

1 hour

1 mile

5280 feet 10 miles per hour

Speed s

t

14 3

1 second

14

3 feet per second

s 4 14

12 14

3 feet

s r

s

r

2 3 feet

212 feet

4

r 2 inches 212 feet. 2 3

s 1

32

2

3 feet

s r




115. The arc length is increasing. In order for the angle to

remain constant as the radius r increases, the arc length

s must increase in proportion to r, as can be seen from

the formula s r .

116. The area of a circle is The circumference of a circle is

For a sector, Thus, for a sector. A

r r

2

1

2 r

2

C

s

r .

Cr

2 A

C 2 A

r

C 2 Ar 2 r

C 2 r . A r 2 ⇒ A

r 2.

108. —CONTINUED—

(b) Since the arc length of the tire is feet and the cyclist is pedaling at a rate of one revolution per second,

we have:

Distance

(c)

(d) The functions are both linear.

14

3 feet per second 1 mile

5280 feett seconds 7

7920 t miles

Distance Rate Time

14

3

feet

revolutions1 mile

5280 feetn revolutions 7

7920n miles

14 3

109. False. An angle measure of radians corresponds to

two complete revolutions from the initial to the terminal

side of an angle.

4

111. False. The terminal side of lies on the negative x -axis.1260

113. Increases, since the linear speed is proportional to the radius.

110. True. If and are coterminal angles, then

where n is an integer. The difference

between is n360 2 n. and

n360

112. (a) An angle is in standard position if its vertex is at the origin and its initial side is on the positive x -axis.

(b) A negative angle is generated by a clockwise rotation of the terminal side.

(c) Two angles in standard position with the same terminal sides are coterminal.

(d) An obtuse angle measures between 90° and 180°.

114. 1 radian

so one radian is much larger than one degree.

180

57.3,

117.4

4 2

4

4 2 2

2

4 2

8

2

2 118.

5 5

2 10

5

2 510

5

2 12

5

2 2 2

2

5 2

4

119. 22 62 4 36 40 4 10 2 10 120.

208 16 13 4 13

172 92 289 81




Section 4.2 Trigonometric Functions: The Unit Circle

■ You should know the definition of the trigonometric functions in terms of the unit circle. Let t be a real number and

the point on the unit circle corresponding to t .

■ The cosine and secant functions are even.

■ The other four trigonometric functions are odd.

■ Be able to evaluate the trigonometric functions with a calculator.

cott cot t tant tan t

csct csc t sint sin t

sect sec t cost cos t

cot t x

y , y 0tan t y

x , x 0

sec t 1

x , x 0cos t x

csc t 1

y, y 0sin t y

x , y

Vocabulary Check1. unit circle 2. periodic

3. period 4. odd; even

121.

Graph of shifted

to the right by two units

y x 5

432−2

3

2

1

−1

−

2

−3

x

y y = x 5

y = ( x − 2)5

f x x 25 122.

Vertical shift four units

downward

4

2

−2

−6

32−2−3 1

y = x 5

y = x

5

−

4

x

y f x x 5 4

123.

Graph of reflected

in x -axis and shifted

upward by two units

y x 5

321−2−3

6

5

4

3

1

−1

−2

−3

x

y

y = x 5

y = 2 − x 5

f x 2 x 5 124.

Reflection in the x -axis

and a horizontal shift

three units to the left

3

2

1

−1

−2

−3

21−3−4−5 x

y

y = x 5

y = −( x + 3)5 f x x 35



Section 4.2 Trigonometric Functions: The Unit Circle 345

7. corresponds to 3

2,

1

2.t 7

6 8. t

5

4, x , y 2

2,

2

2

10. t 5

3, x , y 1

2,

3

2 9. corresponds to 1

2,

3

2 .t 4

3

11. corresponds to 0,1.t 3 2

13. corresponds to

sin

cos

tan t y

x 1

t x 2

2

t y 2

2

2

2, 2

2 .t

4

15. corresponds to

sin

cos

tan t y

x

1

3

3

3

t x 3

2

t y 1

2

3

2,

1

2.t

6

12. t , x , y 1, 0

14.

tan

3

32

12 3

cos

3

1

2

sin

3

3

2

t

3, x , y 1

2, 3

2

16.

tan

4 22

22 1

cos

4 2

2

sin

4 2

2

t

4, x , y 2

2,

2

2

1.

sin csc

cos sec

tan cot x

y

8

15

y

x

15

8

1

x

17

8 x

8

17

1

y

17

15 y

15

17

x 8

17, y

15

17

3.

sin csc

cos sec

tan cot x

y

12

5

y

x

5

12

1

x

13

12 x

12

13

1

y

13

5 y

5

13

x 12

13, y

5

13

5. corresponds to 2

2, 2

2 .t

4

2.

cot x

y

12

5tan

y

x

5

12

sec 1

x

13

12cos x

12

13

csc 1

y

13

5sin y

5

13

x 12

13, y

5

13

4.

cot x

y

4

3tan

y

x

3

4

sec 1

x

5

4cos x

4

5

csc 1

y

5

3sin y

3

5

x 4

5, y

3

5

6. t

3, x y 1

2, 3

2




19. corresponds to

sin

cos

tan t y

x

1

3

3

3

t x 3

2

t y 1

2

3

2,

1

2.t 11

620.

tan5

3

32

12 3

cos5

3

1

2

sin5

3

3

2

t 5

3, x , y 1

2,

3

2

21. corresponds to

sin

cos

tan is undefined.t y

x

t x 0

t y 1

0, 1.t 3

222.

tan2 0

1 0

cos2 1

sin2 0

t 2 , x , y 1, 0

23. corresponds to

sin csc

cos sec

tan cot t x

y 1t

y

x 1

t 1

x 2t x

2

2

t 1

y 2t y

2

2

2

2, 2

2 .t 3

424.

cot5

6

1

tan t 3tan

5

6

12

32

3

3

sec5

6

1

cos t

2 3

3cos

5

6

3

2

csc5

6

1

sin t 2sin

5

6

1

2

t 5

6, x , y 3

2,

1

2

25. corresponds to

sin csc

cos sec is undefined.

tan is undefined. cot t x

y 0t

y

x

t 1

x t x 0

t 1

y

1t y 1

0,1.t

226.

is undefined.

is undefined. cot3

2

0

1 0tan

3

2

sec3

2cos

3

2 0

csc3

2

1

sin t

1sin3

2

1

t 3

2, x , y 0,1

17. corresponds to

sin

cos

tan t y

x 1

t x 2

2

t y 2

2

2

2, 2

2 .t 7

4 18.

tan4

3 3212

3

cos4

3 1

2

sin4

3 3

2

t 4

3, x , y 1

2, 3

2




29. sin 5 sin 0 31. cos8

3 cos

2

3

1

2

33. cos15

2 cos

2 0

35. sin9

4 sin7

4 2

237.

(a)

(b) csct csc t 3

sint sin t 1

3

sin t 1

3

39.

(a)

(b) sect 1

cost 5

cos t cost 1

5

cost 1

5

30. cos 5 cos 1

32. sin9

4 sin

4

2

2 34. sin

19

6 sin

7

6

1

2

36. cos8

3 cos4

3

1

2

38.

(a)

(b) csc t 1

sin t

8

3

sin t sint 3

8

sint 3

840.

(a)

(b) sect sec t 1

cos t

4

3

cost cos t 3

4

cos t 3

4

41.

(a)

(b) sint sin t 4

5

sin t sin t 4

5

sin t 4

543. sin

4 0.7071

45. csc 1.3 1

sin 1.3 1.0378

42.

(a)

(b) cost cos t 4

5

cos t cos t 4

5

cos t 4

5

44. tan

3 1.7321 46. cot 1

1

tan 1 0.6421

47. cos1.7 0.1288 49. csc 0.8 1

sin 0.8 1.3940

51. sec 22.8 1

cos 22.8 1.4486

53. (a)

(b) cos 2 x 0.4

sin 5 y 1

48. cos2.5 0.8011

50. sec 1.8 1

0051.8 4.4014 52. sin0.9 0.7833

54. (a)

(b) cos 2.5 x 0.8

sin 0.75 y 0.7

27. corresponds to

sin csc

cos sec

tan cot t x

y 3

3t

y

x 3

t 1

x 2t x

1

2

t 1

y

2 3

3t y

3

2

1

2,

3

2 .t 4

328.

cot7

4

1

tan t 1tan

7

4

22

22 1

sec7

4

1

cos t 2cos

7

4

2

2

csc7

4

1

sin t 2sin

7

4

2

2

t 7

4, x , y 2

2,

2

2




57.

(a)

(b) From the table feature of a graphing utility we see that seconds.

(c) As t increases, the displacement oscillates but decreases in amplitude.

y 0 when t 5

yt 1

4 et

cos 6t

t 0 1

y 0.25 0.0138 0.08830.02490.1501

34

12

14

58.

(a)

(b)

(c) y1

2 1

4 cos 3 0.2475 feet

y1

4 1

4 cos

3

2 0.0177 feet

y0 1

4 cos 0 0.2500 feet

yt 1

4 cos 6t 59. False. means the function is odd, not

that the sine of a negative angle is a negative number.

For example:

Even though the angle is negative, the sine value is positive.

sin

3

2 sin

3

2 1 1.

sint sin t

60. True. since the period of tan is .tan a tana 6 61. (a) The points have y-axis symmetry.

(b) since they have the same y-value.

(c) since the x -values have the

opposite signs.

cos t 1 cos t

1

sin t 1 sin t

1

62.

So and are even.

So and are odd.

So and are odd.cot tan

cot x

y cot

cot x

y

tan y

x tan

tan y

x

csc sin

csc 1

y csc

x

y

θ

θ −

( x , y)

( x , − y)

csc 1

y

sin y sin

sin y

cos sec

sec 1

x sec

cos x cos 63.

f 1 x 2

3 x

2

3

2

3 x 1

2

3 x

2

3 y

2 x 3 y 2

x 1

23 y 2

y 1

23 x 2

f x 1

23 x 2

55. (a)

(b)

t 1.82 or 4.46

cos t 0.25

t 0.25 or 2.89

sin t 0.25 56. (a)

(b)

t 0.7 or t 5.6

cos t 0.75

t 4.0 or t 5.4

sin t 0.75




64.

f 1 x 3 4 x 1

y 3 4 x 1

4 x 1 y3

x 1 1

4 y3

x 1

4 y3 1

y 1

4 x 3 1

f x 1

4 x 3 1

66.

f 1 x 22 x 1

x 1

y 22 x 1

x 1

y x 1 4 x 2

xy y 4 x 2

xy 4 x y 2

x y 4 y 2

x y 2

y 4

y x 2

x 4

f x x 2

x 465.

f 1 x x 2 4, x ≥ 0

± x 2 4 y

x 2 y2 4

x y2 4

y x 2 4

f x x 2 4, x ≥ 2

67.

Intercept:

Vertical asymptote:

Horizontal asymptote: y 2

x 3

0, 0

y

x −2−4−6 2 4 6 8 10−2

−4

−6

−8

4

6

8

2

f x 2 x

x

3

x 0 1 2 4 5 6

y 0 8 5 4411

2

1

68.

Horizontal asymptote:

Vertical asymptote:

Intercept: 0, 0

x 3, x 2

x 0

y

x −4 4 6 8

−2

−4

−6

−8

2

4

6

8

2−2

f x 5 x

x 2

x 6

5 x

x 3 x 2, x 3, 2

x 0 1 3 5

y 025

24

5

2

5

4

5

2

10

3

5

4

246

69.

Intercepts:

Vertical asymptote:

Horizontal asymptote:

Hole in the graph at 2,7

8

y 1

2

x 2

5, 0, 0,5

4

y

x −1−5−6 1 2−1

−2

−3

−4

1

2

3

4

−2

f x x 2 3 x 10

2 x 2 8

x 5 x 2

2 x 2 x 2 x 5

2 x 2, x 2

x 0 1 3

y 0 2 14

5

5

41

1

4

1345




70.

Vertical asymptote:

Slant asymptote:

intercept:

intercept: 5.86, 0 x -

0,1

8 y-

y 1

2 x

7

4

x

5 ± 89

4 ; x

1.11, x 3.61

y

x −4 46 82−2−6

−2

x 5 ± 52 428

22

2 x 2 5 x 8 0

f x x 3 6 x 2 x 1

2 x 2 5 x 8

1

2 x

7

4

15 x 4

42 x 2 5 x 8

Section 4.3 Right Triangle Trigonometry

■You should know the right triangle definition of trigonometric functions.

(a) (b) (c)

(d) (e) (f)

■ You should know the following identities.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j) (k)

■ You should know that two acute angles are complementary if and that cofunctions of complemen-

tary angles are equal.

■ You should know the trigonometric function values of 30 , 45 , and 60 , or be able to construct triangles from which you

can determine them.

90, and

1 cot2 csc2 1 tan2 sec2

sin2

cos2

1cot cos

sin tan sin

cos

cot 1

tan tan

1

cot sec

1

cos

cos 1

sec csc

1

sin sin

1

csc

cot adj

oppsec

hyp

adjcsc

hyp

opp

tan opp

adjcos

adj

hypsin

opp

hyp

0 2 3 4 7

9 5 1294

32

18

15532

175

154 y

13234 x

Vocabulary Check

1. (i) (v) (ii) (iv) (iii) (vi)

(iv) (iii) (v) (i) (vi) (ii)

2. opposite; adjacent; hypotenuse

3. elevation; depression

opposite

adjacent tan

opposite

hypotenuse sin

adjacent

hypotenuse cos

hypotenuseopposite

csc adjacentopposite

cot hypotenuseadjacent

sec



Section 4.3 Right Triangle Trigonometry 351

5.

tan opp

adj

1

2

2

2

4

cos adj

hyp

2 2

3

sin opp

hyp

1

3θ

31

adj 32 12 8 2 2

cot adj

opp

2 2

sec hyp

adj

3

2 2

3 2

4

csc hyp

opp 3

tan opp

adj

2

4 2

1

2 2

2

4

cos adj

hyp

4 2

6

2 2

3

sin opp

hyp

2

6

1

3θ

62

adj 62 22 32 4 2

cot adj

opp

4 2

2 2 2

sec hyp

adj

6

4 2

3

2 2

3 2

4

csc hyp

opp

6

2 3

The function values are the same since the triangles are similar and the corresponding sides are proportional.

1.

tan opp

adj

6

8

3

4

cos adj

hyp

8

10

4

5

sin opp

hyp

6

10

3

5

6

8

θ

hyp 62 82 36 64 100 10

cot adj

opp

8

6

4

3

sec hyp

adj

10

8

5

4

csc hyp

opp

10

6

5

3

2.

513

b

θ sin csc

cos sec

tan cot adj

opp

12

5

opp

adj

5

12

hyp

adj

13

12

adj

hyp

12

13

hyp

opp

13

5

opp

hyp

5

13

adj 132 52 169 25 12

3.

tan opp

adj

9

40

cos adj

hyp

40

41

sin opp

hyp

9

41θ 9

41

adj 412 92 1681 81 1600 40

cot adj

opp

40

9

sec hyp

adj

41

40

csc hyp

opp

41

9

4.

4

4

θ

sin csc

cos sec

tan cot adj

opp

4

4 1

opp

adj

4

4 1

hypadj

4 24 2 adj

hyp 4

4 2 1 2

22

hyp

opp

4 2

4 2

opp

hyp

4

4 2

1

2 2

2

hyp 42 42 32 4 2




6.

cot adj

opp

15

8tan

opp

adj

8

15

sec hyp

adj

17

15cos

adj

hyp

15

17

csc hypopp

178

sin opphyp

8

17

hyp 152 82 289 17

15

8

θ

cot adj

opp

7.5

4

15

8tan

opp

adj

4

7.5

8

15

sec hyp

adj

172

7.5

17

15cos

adj

hyp

7.5

172

15

17

csc hyp

opp

172

4

17

8

sin opp

hyp

4

172

8

17

hyp 7.52 42 17

2

7.5

4

θ

The function values are the same because the triangles are similar, and corresponding sides are proportional.

7.

tan opp

adj

3

4

cos adj

hyp

4

5

sin opp

hyp

3

5

θ

4

5

opp 52 42 3

cot adj

opp

4

3

sec hyp

adj

5

4

csc hyp

opp

5

3

The function values are the same since the triangles are similar and the corresponding sides are proportional.

tan

opp

adj

0.75

1

3

4

cos adj

hyp

1

1.25

4

5

sin opp

hyp

0.75

1.25

3

5θ

1

1.25

opp 1.252 12 0.75

cot adj

opp 1

0.75

4

3

sec hyp

adj

1.25

1

5

4

csc hyp

opp

1.25

0.75

5

3

8.

cot adj

opp

2

1 2tan

opp

adj

1

2

sec hyp

adj 5

2cos

adj

hyp

2

5

2 5

5

csc hyp

opp 5

1 5sin

opp

hyp

1

5 5

5

hyp 12 22 5

1

2

θ

cot 6

3 2tan

3

6

1

2

sec 3 5

6 5

2cos

6

3 5

2

5

2 5

5

csc 3 5

3 5sin

3

3 5

1

5 5

5

hyp 32 62 3 5

3

6

θ

The function values are the same because the triangles are similar, and corresponding sides are proportional.




13. Given:

cot adj

opp

1

3

sec hyp

adj 10

csc hyp

opp

10

3

cos adj

hyp 10

10

sin opp

hyp

3 10

10

hyp 10

32 12 hyp2

3

1

10

θ

tan 3 3

1

opp

adj14. Given:

cot adj

opp

1

35 35

35

csc hyp

opp

6

35

6 35

35

tan opp

adj

35

1 35

cos adj

hyp

1

6

sin opp

hyp 35

6

opp 62 12 35

6

1

θ

35

sec 6

1

hyp

adj

9. Given:

cot adj

opp 7

3

sec hyp

adj

4 7

7

csc hyp

opp

4

3

tan opp

adj

3 7

7

cos adj

hyp 7

4

adj 7

32 adj2 42

θ

7

43

sin 3

4

opp

hyp

11. Given:

cot adj

opp 3

3

csc hyp

opp

2 3

3

tan opp

adj 3

cos adj

hyp

1

2

sin opp

hyp 3

2

opp 3

1

23

θ

opp2

12

22

sec 2 2

1

hyp

adj

10. Given:

cot adj

opp

5

2 6

5 6

12

sec hyp

adj

7

5

csc hyp

opp

7

2 6

7 6

12

tan

opp

adj

2 6

5

sin opp

hyp

2 6

7

opp 72 52 24 2 6

5

72 6

θ

cos 5

7

adj

hyp

12. Given:

sec hyp

adj

26

5

csc hyp

opp 26

1 26

tan opp

adj

1

5

cos adj

hyp

5

26

5 26

26

sin opp

hyp

1

26 26

26

hyp 52 12 26

261

5

θ cot 5

1

adj

opp




15. Given:

sec hyp

adj 13

3

csc hyp

opp 13

2

tan opp

adj

2

3

cos adj

hyp

3

13

3 13

13

sin opp

hyp

2

13

2 13

13

hyp 13

22 32 hyp2 2

3

13

θ

cot 3

2

adj

opp16. Given:

cot adj

opp 273

4

sec hyp

adj

17

273

17 273

273

tan opp

adj

4

273

4 273

273

cos

adj

hyp

273

17

sin opp

hyp

4

17

adj 172 42 273

417

273

θ

csc 17

4

hyp

opp

17.

sin 30 opp

hyp

1

2

30 30 180

6 radian

30°

60°

1

2

3

18.

cos 45 adj

hyp

1

2 2

2

45 45 180

4 radian

2

1

1

45°

19.

tan

3

opp

adj

3

1 3

3

3180

60

1

23

π

3

π

6

20.

sec

4

hyp

adj

2

1 2

4

4180

45

2

1

1

π

4

21.

60

3 radian

cot 3

3

1

3

adj

opp

1

3

30°

60°

22.

45 45 180

4 radian

csc 2 hyp

opp

2

1

1

45°

23.

cos 6 adj

hyp 3

2

6

6180

30

12

3

π

3

π

6

24.

sin 4 opp

hyp 1 2

22

4

4180

45

2

1

1

π

4

25.

4 radian

45 45 180

cot 1 1

1

adj

opp2

1

1

45°

45°

26.

6 radian

30 30 180

tan 3

3

1

3

opp

adj

30°

60°1

2

3




31.

(a)

(b)

(c)

(d) sin90 cos 1

3

cot cos sin

1

3

2 2

3

1

2 2 2

4

sin 2 2

3

sin2 8

9

sin2 1

32

1

sin2 cos2 1

sec 1cos

3

cos 1

3 32.

(a)

(b)

(c)

(d)

1 125

2625

265

1 1

52

csc 1 cot2

tan90º cot 1

tan

1

5

1

1 52

1

26 26

26

cos 1

sec

1

1 tan2

cot 1

tan 1

5

tan 5

33. tan cot tan 1

tan 1 34. cos sec cos 1

cos 1

27.

(a)

(b)

(c)

(d) cot 60 cos 60

sin 60

1

3 3

3

cos 30 sin 60 32

sin 30 cos 60 1

2

tan 60 sin 60

cos 60 3

sin 60 3

2, cos 60

1

2 28.

(a)

(b)

(c)

(d) cot 30 1

tan 30

3

3

3 3

3 3

cos 30 sin 30

tan 30

1

2

3

3

3

2 3 3

2

cot 60 tan90 60 tan 30 3

3

csc 30 1

sin 30 2

sin 30 1

2, tan 30

3

3

29.

(a)

(b)

(c)

(d) sec90 csc 13

2

tan sin

cos

2 13

13

3 13

13

2

3

cos

1

sec

3

13

3 13

13

sin 1

csc

2

13

2 13

13

csc 13

2, sec

13

3 30.

(a)

(b)

(c)

(d) sin tan cos 2 6 1

5 2 6

5

cot90º tan 2 6

cot 1

tan

1

2 6 6

12

cos 1

sec

1

5

sec 5, tan 2 6




35. tan cos sin

cos cos sin 36. cot sin cos

sin sin cos

37.

sin

2

sin2 cos2 cos2

1 cos 1 cos 1 cos2 38. 1 sin 1 sin 1 sin2 cos2

39.

1

1 tan2 tan2

sec tan sec tan sec2 tan2

41.

csc sec

1

sin

1

cos

1

sin cos

sin

cos

cos

sin

sin2 cos2

sin cos

40.

2 sin2 1

sin2 1 sin2

sin2 cos2 sin2 1 sin2

42.

1 cot2 csc2

1 cot

1

cot

tan cot

tan

tan

tan

cot

tan

43. (a)

(b)

Note: cos 80 sin90 80 sin 10

cos 80 0.1736

sin 10 0.1736 44. (a)

(b) cot 66.5 1

tan 66.5 0.4348

tan 23.5 0.4348

45. (a)

(b) csc 16.35

1

sin 16.35 3.5523

sin 16.35 0.2815 46. (a)

(b) sin 73 56 sin73 56

60

0.9609

cos 16 18 cos16 18

60 0.9598

47. (a)

(b) csc 48 7 1

sin 48 760 1.3432

sec 42 12 sec 42.2 1

cos 42.2 1.3499

49. (a)

(b) tan 11 15 tan 11.25 0.1989

cot 11 15

1

tan 11.25

5.0273

51. (a)

(b) tan 44 28 16 tan 44.4711 0.9817

csc 32 40 3 1

sin 32.6675 1.8527

48. (a)

(b)

1.0036

sec 4 50 15 1

cos 4 50 15

0.9964

cos 4 50 15 cos4 50

60

15

3600

50. (a)

(b) cos 56 8 10 cos56 8

60

10

3600 0.5572

sec 56 8 10

sec56

8

60

10

3600

1.7946

52. (a)

(b) cot9

5 30 32 0.0699

sec9

5 20 32 2.6695




62.

r 20

sin 45

20

22 20 2

sin 45 20

r

63.

Height of the building:

Distance between friends:

323.34 meters

cos 82 45

y ⇒ y

45

cos 82

123 45 tan 82 443.2 meters

x 45 tan 82

45 m

82°

x y

tan 82 x

45

53. (a)

(b) csc 2 ⇒ 30

6

sin 1

2 ⇒ 30

655. (a)

(b) cot 1 ⇒ 45

4

sec 2 ⇒ 60

354. (a)

(b) tan 1 ⇒ 45º

4

cos 2

2 ⇒ 45

4

59.

x 30 3

1 3 30 x

tan 30 30

x

30°

30

x

60.

y 18 sin 60 18

3

2 9 3

sin 60 y

18

61.

x 32

3

32 3

3

3 x 32

3 32

x

tan 60 32

x

60°

32

x

64. (a) (b)

(c)

h 270 feet

2135 h

tan 6

3

h

135

h

3

6

132

Not drawn to scale

65.

0

6

sin 1500

3000

1

2

θ

1500 ft3000 ft

66.

w 100 tan 54 137.6 feet

tan 54 w

100

tan opp

adj

56. (a)

(b) cos 1

2 ⇒ 60

3

tan 3 ⇒ 60

3

57. (a)

(b) sin 2

2 ⇒ 45

4

csc

2 3

3 ⇒ 60

3

58. (a)

tan 3

3 3 ⇒ 60

3

cot 3

3 (b)

cos 1

2 2

2 ⇒ 45

4

sec 2




67.

(a)

x 145sin 23

371.1 feet

sin 23 145

x

150 ft

5 ft23°

x

(b)

y 145tan 23

341.6 feet

tan 23 145

y(c) Moving down the line:

feet per second

Dropping vertically:

feet per second145

6 24.17

145sin 23

6 61.85

68. Let the height of the mountain.

Let the horizontal distance from where the angle of

elevation is sighted to the point at that level directly below

the mountain peak.

Then tan

Substitute into the expression for tan

The mountain is about 1.3 miles high.

1.2953 h

13 tan 9 tan 3.5tan 9 tan 3.5

h

13 tan 9 tan 3.5 htan 9 tan 3.5

h tan 3.5 13 tan 9 tan 3.5 h tan 9

tan 3.5 h tan 9

h 13 tan 9

tan 3.5 h

h

tan 9 13

3.5. x h

tan 9

tan 9 h

x ⇒ x

h

tan 9

3.5 h

x 13 and tan 9

h

x .

9 x

h 69.

x 2, y

2 28, 28 3

x 2 cos 60°56 1

256 28

cos 60° x

2

56

y2 sin 60°56 3

2 56 28 3

sin 60 y

2

56

60°

56

( , ) x y2 2

x 1, y

1 28 3, 28

x 1 cos 3056

3

2 56 28 3

cos 30 x

1

56

y1 sin 3056 1

256 28

sin 30

y1

56

30°

56

( , ) x y1 1

70.

d 5 2 x 5 215 tan 3 6.57 centimeters

x 15 tan 3

tan 3 x

15




72.

tan 20 y

x 0.36

cos 20 x

10

0.94

sin 20 y

10 0.34

x 9.4, y 3.4

csc 20 10

y 2.92

sec 20 10

x

1.06

cot 20 x

y 2.75

73. True,

csc x 1

sin x ⇒ sin 60 csc 60 sin 60 1

sin 60 1

75. False, 2

2 2

2 2 1

77. False,

sin 2 0.0349

sin 60

sin 30

cos 30

sin 30 cot 30 1.7321;

74. True, because sec90 csc .sec 30 csc 60

76. True, because

cot2 csc2 1.

cot2 csc2 1

1 cot2 csc2

cot2 10 csc2 10 1

78. False,

tan25 tan 5tan 5 0.0077

tan52 tan 25 0.4663

tan52 tan25. 79. This is true because the corresponding sides of similar

triangles are proportional.

80. Yes. Given can be found from the identity 1 tan2 sec2 .tan , sec

81. (a) (b) In the interval

(c) As approaches 0, approaches .sin

0, 0.5, > sin .0.1 0.2 0.3 0.4 0.5

0.0998 0.1987 0.2955 0.3894 0.4794sin

71. (a)

(b)

(c)

(d) The side of the triangle labeled

h will become shorter.

h 20 sin 85 19.9 meters

sin 85 h20

h20

85°

(e)Angle, Height (in meters)

80 19.7

70 18.8

60 17.3

50 15.3

40 12.9

30 10.0

20 6.8

10 3.5

(f) The height of the balloon

decreases as decreases.

20 h

θ

82. (a)

—CONTINUED—

sin 0 0.3090 0.5878 0.8090 0.9511 1

cos 1 0.9511 0.8090 0.5878 0.3090 0

90725436180





Section 4.4 Trigonometric Functions of Any Angle 361

Vocabulary Check

1. 2. 3.

4. 5. 6.

7. reference

cot x

r cos

r

x

tan y

x csc sin

y

r

1. (a)

r 16 9 5

x , y 4, 3 (b)

r 64 225 17

x , y 8, 15

tan y

x

3

4

cos x

r

4

5

sin y

r

3

5

cot x

y

4

3

sec r

x

5

4

csc r

y

5

3

tan y

x

15

8

cos x

r

8

17

sin y

r

15

17

cot x

y

8

15

sec r

x

17

8

csc r

y

17

15

2. (a)

cot x

y12

5

12

5

sec r

x

13

12

13

12

csc r

y

13

5

13

5

tan y

x 5

12

5

12

cos x

r

12

13

sin y

r

5

13

r 122 52 13

x 12, y 5 (b)

cot x

y1

1 1

sec r

x 2

1 2

csc r

y

2

1

2

tan y

x

1

1 1

cos x

r 1

2

2

2

sin y

r

1

2 2

2

r 12 12 2

x 1, y 1

3. (a)

r 3 1 2

x , y 3, 1 (b)

r 16 1 17

x , y 4, 1

tan y

x

3

3

cos x r 3

2

sin y

r

1

2

cot x

y 3

sec r

x

2 33

csc r

y 2

tan y

x

1

4

cos x r 4 17

17

sin y

r 17

17

cot x

y 4

sec r x 17

4

csc r

y 17




4. (a)

cot x

y

3

1 3

sec r

x 10

3

csc r

y 10

1 10

tan y

x

1

3

cos x

r

3

10

3 10

10

sin y

r

1

10 10

10

r 32 12 10

x 3, y 1 (b)

cot x

y

4

4 1

sec r

x

4 2

4 2

csc r

y

4 2

4 2

tan y

x 4

4 1

cos x

r

4

4 2 2

2

sin y

r 4

4 2

2

2

r 42 42 4 2

x 4, y 4

5.

cot x

y

7

24

sec r

x

25

7

csc r

y

25

24

tan y

x

24

7

cos x

r

7

25

sin y

r

24

25

r 49 576 25

x , y 7, 24 6.

cot x

y

8

15

sec r

x

17

8

csc r

y

17

15

tan y

x

15

8

cos x

r

8

17

sin y

r

15

17

r 82 152 17

x 8, y 15 7.

cot x

y

2

5

sec r

x

29

2

csc r

y 29

5

tan y

x

5

2

cos x

r

2 29

29

sin y

r

5 29

29

r 16 100 2 29

x , y 4, 10

8.

cot x

y5

2

5

2

sec r

x 29

5

29

5

csc r

y 29

2

29

2

tan y

x 2

5

2

5

cos x

r 5

29

5 29

29

sin y

r 2

29

2 29

29

r 52 22 29

x 5, y 2 9.

cot x

y

35

68 0.5

sec r

x

5849

35 2.2

csc r

y 5849

68 1.1

tan y

x

68

35 1.9

cos x

r

35 5849

5849 0.5

sin y

r

68 5849

5849 0.9

r 12.25 46.24 5849

10

x , y 3.5, 6.8




18.

cot

8

15tan

y

x 15

8

15

8

sec 17

8cos

x

r

8

17

csc 17

15sin

y

r 15

17

15

17

tan < 0 ⇒ y 15

cos x

r

8

17 ⇒ y 15

15.

tan y

x

3

4

cos x

r

4

5

sin y

r

3

5

in Quadrant II ⇒ x 4

sin y

r

3

5 ⇒ x 2 25 9 16

10.

cot x

y

72

314

14

31 0.5tan

y

x 314

72

31

14 2.2

sec r x 11574

72 1157

14 2.4cos

x r

72 11574

14 1157

1157 0.4

csc r

y 11574

314

1157

31 1.1sin

y

r

314

11574

31 1157

1157 0.9

r 7

22

31

4 2

1157

4

x 31

2

7

2, y 7

3

4

31

4

11.

sin < 0 and cos < 0 ⇒ lies in Quadrant III.

cos < 0 ⇒ lies in Quadrant II or in Quadrant III.

sin < 0 ⇒ lies in Quadrant III or in Quadrant IV.

13.

sin > 0 and tan < 0 ⇒ lies in Quadrant II.

tan < 0 ⇒ lies in Quadrant II or in Quadrant IV.

sin > 0 ⇒ lies in Quadrant I or in Quadrant II.

12. and

and

Quadrant I

x

r > 0

y

r > 0

cos > 0sin > 0

14. and

and

Quadrant IV

x

y< 0

r

x > 0

cot < 0sec > 0

cot x

y

4

3

sec r

x

5

4

csc r

y

5

3

16.

in Quadrant III

tan y

x

3

4

cos x

r

4

5

sin y

r

3

5

⇒ y 3

cos x

r 4

5 ⇒ y2 25 16 9

cot 4

3

sec 5

4

csc 5

3

17.

is in Quadrant IV

tan y

x

15

8

cos

x

r

8

17

sin y

r

15

17

x 8, y 15, r 17

y < 0 and x > 0.

⇒sin < 0 and tan < 0 ⇒

tan y

x 15

8

cot x

y

8

15

sec

r

x

17

8

csc r

y

17

15




19.

cot x

y 3tan

y

x

1

3

sec r

x 10

3cos

x

r

3 10

10

csc r

y 10sin

y

r

10

10

x 3, y 1, r 10

cos > 0 ⇒ is in Quadrant IV ⇒ x is positive;

cot x

y

3

1

3

120.

cot 15tan y

x

15

15

sec

4 15

15cos

x

r

15

4

csc 4sin y

r

1

4

cot < 0 ⇒ x 15

csc r

y

4

1 ⇒ x 15

21.

cot x

y

3

3tan

y

x 3

sec

r

x 2cos

x

r

1

2

csc r

y

2 3

3sin

y

r 3

2

sin > 0 ⇒ is in Quadrant II ⇒ y 3

sec r

x

2

1 ⇒ y2 4 1 3 22.

is undefined

is undefinedcot x

ytan

y

x 0

sec r x 1cos x

r r

r 1

csc r

ysin 0

y 0, x r

sec 1 ⇒ 2 n

sin 0 ⇒ 0 n

23.

cot is undefinedtan 0

sec 1cos 1

csc is undefinedsin 0

cot is undefined,

2 ≤ ≤

3

2 ⇒ y 0 ⇒ 24. tan is undefined

is undefined.

is undefined. cot x

y

0

y 0tan

y

x

sec

r

x cos

x

r

0

r

0

csc r

y 1sin

y

r r

r 1

≤ ≤ 2 ⇒ 3

2, x 0, y r

⇒ n

2

25. To find a point on the terminal side of use any point on

the line that lies in Quadrant II. is one

such point.

cot 1

sec 2

csc 2

tan 1

cos

1

2

2

2

sin 1

2 2

2

x 1, y 1, r 2

1, 1 y x

26. Let

Quadrant III

cot x

y

x

13 x 3

sec r

x

10 x 3

x

10

3

csc r

y

10 x 3

13 x 10

tan y

x

13 x

x

1

3

cos x r x

10 x 3 3 10

10

sin y

r

13 x

10 x 3

10

10

r x 2 1

9 x 2

10 x

3

x , 1

3 x ,

x > 0.




27. use any point on

the line Quadrant III. is one

such point.

cot 1

2

1

2

sec 5

1 5

csc 5

2

5

2

tan 2

1 2

cos 1

5

5

5

sin 2

5

2 5

5

x 1, y 2, r 5

1, 2 y 2 x that lies in

To find a point on the terminal side of , 28. Let

Quadrant IV

tan 3

4tan

y

x

43 x

x

4

3

sec 5

3cos

x

r

x

53 x

3

5

csc 5

4sin

y

r

43 x

53 x

4

5

r

x

2 16

9 x 2

5

3 x

x , 4

3 x ,

4 x 3 y 0⇒ y 4

3 x

x > 0.

29.

sin y

r 0

x , y 1, 0, r 1 30.

since corresponds to 0, 1.3

2

csc3

2

r

y

1

1 1 31.

sec undefined3

2

r

x

1

0 ⇒

x , y 0, 1, r 1

32.

since corresponds to 1, 0.3

2

sec r

x

1

1 1 33.

sin

2

y

r 1

x , y 0, 1, r 1 34. (undefined)

since corresponds to 1, 0.

cot x

y1

0

35.

undefinedcsc r

y

1

0 ⇒

x , y 1, 0, r 1 36.

since corresponds to 0, 1.

2

cot

2 x

y 0

1 0

37.

′θ

203°

x

y

203 180 23

203 38.

′θ

309°

x

y

360 309 51

309



45. Quadrant III

tan 225 tan 45 1

cos 225 cos 45 2

2

sin 225 sin 45 2

2

225, 360 225 45,

47. is coterminal with

Quadrant I

tan 750 tan 30 3

3

cos 750 cos 30 3

2

sin 750 sin 30 1

2

30,

30. 750

46. Quadrant IV

sin

cos

tan 300 tan 60 3

300 cos 60 1

2

300 sin 60 3

2

300, 360 300 60,


Quadrant IV

sin

cos

tan405 tan 45 1

405 cos 45 2

2

405 sin 45 2

2

360

315

45,

315. 405


39.

′θ

−245°

x

y

180 115 65

360 245 115 coterminal angle

245 40. is coterminal with

′θ

−145°

x

y

215 180 35

215. 145

41.

2

3

3

′θ

2

3

π

x

y

2

342.

2 7

4

4

′θ

7

4

π

x

y 7

4

43.

3.5

′θ

3.5

x

y 3.5 44. is coterminal

with

2 5

3

3

5

3.

′θ

x

y

11

3

π

11

3





Quadrant III

tan150 tan 30 3

3

cos150 cos 30 3

2

sin150 sin 30 1

2

210 180 30,

210. 150 50. .

sin

cos

tan840 tan 60 3

840 cos 60 1

2

840 sin 60 3

2

240 180 60, Quadrant III

840 is coterminal with 240

51. Quadrant III

tan4

3 tan

3 3

cos4

3 cos

3

1

2

sin4

3 sin

3

3

2

3,

4

3, 52. Quadrant I

sin

cos

tan

4 1

4 2

2

4 2

2

4,

4, 53. Quadrant IV

tan

6 tan

6

3

3

cos

6 cos

6 3

2

sin

6 sin

6

1

2

6,

6,


sin

cos

tan is undefined.

2 tan3

2

2 cos3

2 0

2 sin3

2 1

3

2.


Quadrant II

tan11

4 tan

4 1

cos11

4 cos

4

2

2

sin11

4 sin

4 2

2

3

4

4,

3

4 .

11

4 56. is coterminal with

Quadrant III

sin

cos

tan10

3 tan

3 3

10

3 cos

3

1

2

10

3 sin

3

3

2

4

3

3,

4

3.

10

3


tan3

2 tan

2 which is undefined.

cos3

2 cos

2 0

sin3

2 sin

2 1

2.

2,

3

2 58. .

tan

25

4 tan

4 1

cos25

4 cos

4 2

2

sin25

4 sin

4 2

2

2 7

4

4 in Quadrant IV.

25

4 is coterminal with

7

4



74. cot 1.35 1

tan 1.35 0.2245

77. sin0.65 0.6052 78. sec 0.29 1

cos 0.29 1.0436

79. cot11

8 1

tan11

8 0.4142 80. csc15

14 1

sin15

14 4.4940


59.

in Quadrant IV.

cos 4

5

cos > 0

cos2 16

25

cos2 1 9

25

cos2 1 3

52

cos2 1 sin2

sin2 cos2 1

sin 3

560. cot

sin 1

csc

1

10 10

10

csc 1

sin

10 csc

csc > 0 in Quadrant II.

10 csc2

1 32 csc2

1 cot2 csc2

3 61.

in Quadrant III.

sec 13

2

sec < 0

sec2 13

4

sec2

1

9

4

sec2 1 3

22

sec2 1 tan2

tan 3

2

62.

cot 3

cot < 0 in Quadrant IV.

cot2 3

cot2 22 1

cot2 csc2 1

1 cot2 csc2

csc 2 63.

sec 1

58

8

5

cos

1

sec ⇒ sec

1

cos

cos 5

8 64.

.

tan 65

4

tan > 0 in Quadrant III

tan2 65

16

tan2 9

42

1

tan2 sec2 1

1 tan2 sec2

sec 9

4

65. sin 10 0.1736 66. sec 225

1

cos 225 1.4142 67. cos110 0.3420

68. csc330 1

sin330 2.0000 69. tan 304 1.4826 70. cot 178

1

tan 178 28.6363

71. sec 72 1

cos 72 3.2361 72. tan188 0.1405 73. tan 4.5 4.6373

75. tan

9 0.3640 76. tan

9 0.3640




81. (a) reference angle is 30 or is in Quadrant I or Quadrant II.

Values in degrees: 30 , 150

Values in radians:

(b) reference angle is 30 or is in Quadrant III or Quadrant IV.


Values in radians:7

6,

11

6

6and sin

1

2 ⇒

6,

5

6

6and sin

1

2 ⇒

82. (a) cos reference angle is 45 or and is in Quadrant I or IV.


Values in radians:

(b) cos reference angle is 45 or and is in Quadrant II or III.


Values in radians:3

4,

5

4

4

2

2 ⇒

4,

7

4

4

2

2 ⇒

83. (a) reference angle is 60 or and is in Quadrant I or Quadrant II.


Values in radians:

(b) reference angle is 45 or and is in Quadrant II or Quadrant IV.


Values in radians:3

4,

7

4

4

cot 1 ⇒

3,

2

3

3csc

2 3

3 ⇒

84. (a) sec reference angle is 60 or and is in

Quadrant I or IV.


Values in radians:

(b) sec reference angle is 60 or and is

in Quadrant II or III.


Values in radians:2

3,

4

3

3 2 ⇒

3

,5

3

3 2 ⇒ 85. (a) reference angle is 45 or and is in

Quadrant I or Quadrant III.


Values in radians:

(b) reference angle is 30 or and

is in Quadrant II or Quadrant IV.


Values in radians:5

6,

11

6

6cot 3 ⇒

4,

5

4

4tan 1 ⇒




86. (a) sin reference angle is 60 or and is

in Quadrant I or II.


Values in radians:

(b) sin reference angle is 60 or and

is in Quadrant III or IV.


Values in radians:4

3,

5

3

3

32

⇒

3,

2

3

3

3

2 ⇒ 87. (a) New York City:

Fairbanks:

(b)

(c) The periods are about the same for both models,

approximately 12 months.

F 36.641 sin0.502t 1.831 25.610

N 22.099 sin0.522t 2.219 55.008

88.

(a) For February 2006,

(b) For February 2007,

(c) For June 2006,

(d) For June 2007,

S 23.1 0.44218 4.3 cos 18

6 26,756 units

t 18.

S 23.1 0.4426 4.3 cos 6

6 21,452 units

t 6.

S 23.1 0.44214 4.3 cos 14

6 31,438 units

t 14.

S 23.1 0.4422 4.3 cos 2

6 26,134 units

t 2.

S 23.1 0.442t 4.3 cos t

689.

(a)

(b)

(c) y1

2 2 cos 3 1.98 centimeters

y1

4 2 cos3

2 0.14 centimeter

y0 2 cos 0 2 centimeters

yt 2 cos 6t

Month New York City Fairbanks

February

March

May

June

August

September

November 6.546.8

41.768.6

55.675.5

59.572.5

48.663.4

13.941.6

1.434.6

90.

(a)

centimeters

(b)

centimeters

(c)

centimeters y12 2e12 cos6 1

2 1.2

t 12

y14 2e14 cos6 1

4 0.11

t 14

y0 2e0 cos 0 2

t 0

yt 2et cos 6t 91.

ampere I 0.7 5e1.4 sin 0.7 0.79

I 5e2t sin t




92. sin

(a)

miles

(b)

miles

(c)

milesd 6

sin 120 6.9

120

d 6

sin 90º

6

1 6

90

d 6

sin 30

6

12 12

30

6

d ⇒ d

6

sin 93. False. In each of the four quadrants, the sign of the secant

function and the cosine function will be the same since

they are reciprocals of each other.

94. False. For example, if and but is not the reference angle.

The reference angle would be For in Quadrant II, For in Quadrant III,

For in Quadrant IV, 360 .

180. 180 . 45.

360n 1350 ≤ 135 ≤ 360, 225,n 1

95. As increases from to x decreases from 12 cm to 0 cm and y increases from 0 cm to 12 cm.

Therefore, increases from 0 to 1 and decreases from 1 to 0. Thus,

increases without bound, and when the tangent is undefined. 90tan y

x

cos x

12sin

y

12

90,0

96. Determine the trigonometric function of the reference angle and prefix the appropriate sign.

97.

x -intercepts:

y-intercept:

No asymptotes

Domain: All real numbers x

0, 4

−2−6−8 2 4 6 8−2

−4

−8

2

4

6

8

x (1, 0)(−4, 0)

(0, −4)

4, 0, 1, 0

y x 2 3 x 4 x 4 x 1 98.

x -intercepts:

y-intercepts:

No asymptotes


0, 0

−1−2−3 1 2 3 4 5−1

−2

−3

−4

1

2

y

x (0, 0) , 05

2( (

0, 0, 52, 0

y 2 x 2 5 x x 2 x 5

99.

x -intercept:

y-intercept:

No asymptotes


0, 8

2, 0

−6 −4−8 2 4 6 8

−4

10

12

x

(0, 8)

(−2, 0)

f x x 3 8 100.

x -intercepts:

y-intercepts:

No asymptotes


0, 3

−2−3−4 2 3 4

−3

−4

1

2

3

4

x (−1, 0)

(0, −3)

(1, 0)

1, 0, 1, 0

x 2 3 x 1 x 1

g x x 4 2 x 2 3 x 2 3 x 2 1





Section 4.5 Graphs of Sine and Cosine Functions 373

Section 4.5 Graphs of Sine and Cosine Functions

■ You should be able to graph

■ Amplitude:

■Period:

■ Shift: Solve

■ Key increments: (period)1

4

bx c 0 and bx c 2 .

2

b

a

y a sinbx c and y a cosbx c. Assume b > 0.

Vocabulary Check

1. cycle 2. amplitude

3. 4. phase shift

5. vertical shift

2

b

105.

Domain: All real numbers except

x -intercepts:

Vertical asymptote: x 0

±1, 0

x 0

−12 −9 −6 −3 3 6 9 12

6

9

12

(−1, 0) (1, 0) x

y y ln x 4

106.

To find the x -intercept, let

x -intercepts:

To find the y-intercept, let

y-intercepts:

The vertical asymptote is the horizontal asymptote of translated two units to the left.

Vertical asymptote:

Domain: All real numbers such that x > 2 x

x 2

y log10 x

0, 0.301

y log10

x 2 log10

2 0.301

x 0:

1, 0

0 log10

x 2 ⇒ 10 x 2 ⇒ x 1

y 0:3

2

−1

−2

−3

321−1−3

(−1, 0) x

y y log10

x 2




7.

Period:

Amplitude: 2 2

2

1 2

y 2 sin x 8.

Period:

Amplitude: a 1 1

2

b

2

23 3

y cos2 x

3 9.

Period:

Amplitude: 3 3

2

10

5

y 3 sin 10 x

10.

Period:

Amplitude: a 1

3

2

b

2

8

4

y 1

3 sin 8 x 11.

Period:

Amplitude: 1

2 1

2

2

23 3

y 1

2 cos

2 x

3 12.

Period:

Amplitude: a 5

2

2

b

2

14 8

y 5

2 cos

x

4

13.

Amplitude: 1

4 1

4

Period:2

2 1

y 1

4 sin 2 x 14.

Period:

Amplitude: a 2

3

2

b

2

10 20

y 2

3 cos

x

10 15.

The graph of g is a horizontal shift

to the right units of the graph of

f a phase shift.

g x sin x

f x sin x

16.

g is a horizontal shift of f units

to the left.

f x cos x , g x cos x 17.

The graph of g is a reflection in

the x -axis of the graph of f.

g x cos 2 x

f x cos 2 x 18.

g is a reflection of f about the

y-axis.

f x sin 3 x , g x sin3 x

19.

The period of f is twice that of g.

g x cos 2 x

f x cos x 20.

The period of g is one-third the

period of f .

f x sin x , g x sin 3 x 21.

The graph of g is a vertical shift

three units upward of the graph of f.

f x 3 sin 2 x

f x sin 2 x

1.

Period:

Amplitude: 3 3

2

2

y 3 sin 2 x 2.

Period:

Amplitude: a 2

2

b

2

3

y 2 cos 3 x 3.

Period:

Amplitude: 5

2 5

2

2

12 4

y 5

2 cos

x

2

4.

Period:

Amplitude: a 3 3

2

b

2

13 6

y 3 sin x

35.

Period:

Amplitude: 1

2 1

2

2

3 6

y 1

2 sin

x

3 6.

Period:

Amplitude: a 3

2

2

b

2

2 4

y 3

2 cos

x

2




28.

Period:

Amplitude: 1

Symmetry: origin

Key points: Intercept Maximum Intercept Minimum Intercept

Since the graph of is the graph of but stretched horizontally by a factor of 3.

Generate key points for the graph of by multiplying the coordinate of each key point of by 3. f x x -g x

f x ,g x g x sin x 3 f x

3,

2 , 03

2, 1 , 0 2

, 10, 0

2

b

2

1 2

− 2

2

π 6

f g

x

y f x sin x

29.

Period:

Amplitude: 1

Symmetry: axis

Key points: Maximum Intercept Minimum Intercept Maximum

Since the graph of is the graph of but translated upward by one unit.

Generate key points for the graph of by adding 1 to the coordinate of each key point of f x . y-g x f x ,g x g x 1 cos x f x 1,

2 , 13

2 , 0 , 1

2, 00, 1

y-

2

b

2

1 2

−1

g

f

x π π 2

y f x cos x

25. The graph of g is a horizontal shift units to the right of

the graph of f.

26. Shift the graph of f two units upward to obtain the graph

of g.

27.

Period:

Amplitude: 2

Symmetry: origin

Key points: Intercept Minimum Intercept Maximum Intercept

Since generate key points for the graph of by multiplying

the coordinate of each key point of by

2. f x y-

g x g x 4 sin x 2 f x ,

2 , 03

2, 0 , 0 2

, 20, 0

2

b

2

1 2

x

−

π π 32 2

5

4

3

−5

f

g

y f x 2 sin x

22.

g is a vertical shift of f two units

downward.

f x cos 4 x , g x 2 cos 4 x 23. The graph of g has twice the

amplitude as the graph of f. The

period is the same.

24. The period of g is one-third the

period of f .




31.

Period:

Amplitude:

Symmetry: origin

Key points: Intercept Minimum Intercept Maximum Intercept

Since the graph of is the graph of but translated upward by three units.

Generate key points for the graph of by adding 3 to the coordinate of each key point of f x . y-g x

f x ,g x g x 3 1

2 sin

x

2 3 f x ,

4 , 03 ,1

22 , 0 , 1

20, 0

1

2

2

b

2

12 4

−1

1

2

3

4

5

f

g

x −π π 3

y f x 1

2 sin

x

2

32.

Period:

Amplitude: 4

Symmetry: origin

Key points: Intercept Maximum Intercept Minimum Intercept

Since the graph of is the graph of but translated downward by three units.

Generate key points for the graph of by subtracting 3 from the coordinate of each key point of f x . y-g x f x ,g x g x 4 sin x 3 f x 3,

2, 03

2, 21, 01

2, 20, 0

2

b

2

2

−8

2

4 f

g

1 x

y f x 4 sin x

30.

Period:

Amplitude: 2

Symmetry: axis


Since the graph of is the graph of but

i) shrunk horizontally by a factor of 2,

ii) shrunk vertically by a factor of and

iii) reflected about the axis.

Generate key points for the graph of by

i) dividing the coordinate of each key point of by 2, and

ii) dividing the coordinate of each key point of by 2. f x y-

f x x -

g x

x -

12,

f x ,g x g x cos 4 x 12 f 2 x ,

, 23

4, 0 2

, 2 4, 00, 2

y-

2

b

2

2

−2

2

π

f

g

x

y f x 2 cos 2 x




35.

Period:

Amplitude:

Key points:

3

2, 3, 2 , 0

0, 0, 2, 3, , 0,

3

2

y

x π π 32 2

−−

π

2π 32

−4

1

2

3

4

y 3 sin x

37.

Period:

Amplitude:

Key points:

3

2, 0, 2 ,

1

3

0,1

3, 2, 0, , 1

3,

13

2

y

x π π

2π 2

1

−1

2

3

4

3

1

32

3

4

3

−

−

−

y 1

3 cos x 38.

Period:

Amplitude:

Key points:

3

2, 0, 2 , 4

0, 4, 2, 0, , 4,

4

2

y

x π π 2π −2 −π

−2

−4

4

y 4 cos x

36.

Period:

Amplitude:

Key points:

3

2,

1

4, 2 , 0

0, 0, 2,

1

4, , 0,

1

4

2

y

x π π 2π −2 −π

−1

−2

1

2

y 1

4 sin x

33.

Period:

Amplitude: 2

Symmetry: axis


Since the graph of is the graph of but with a phase shift (horizontal translation)

of Generate key points for the graph of by shifting each key point of units to the left. f x g x .

f x ,g x g x 2 cos x f x ,

2 , 23

2, 0 , 2 2

, 00, 2

y-

2

b

2

1 2

−3

3

f

g

x

π π 2

y f x 2 cos x

34.

Period:

Amplitude: 1

Symmetry: axis

Key points: Minimum Intercept Maximum Intercept Minimum

Since the graph of is the graph of but with a phase shift (horizontal translation)

of Generate key points for the graph of by shifting each key point of units to the right. f x g x .

f x ,g x g x cos x f x ,

2 , 13

2, 0 , 1 2

, 00, 1

y-

2

b

2

1 2

−2

2

π π 2

f

g

x

y f x cos x




45.

Period:

Amplitude: 1

Shift: Set

Key points: 4, 0, 3

4, 1, 5

4, 0, 7

4, 1, 9

4, 0

x

4 x

9

4

x

4 0 and x

4 2

2

−3

−2

1

2

3

π π −

x

y y sin x

4; a 1, b 1, c

4

44.

Period:

Amplitude: 10

Key points:

1284−4−12

12

8

4

−12

x

y

0, 10, 3, 0, 6, 10, 9, 0, 12, 10

2

6 12

y 10 cos x

643.

Period:

Amplitude: 1

Key points:

−1 2 3

−3

−2

2

3

x

y

0, 0, 3

4, 1, 3

2, 0, 9

4, 1, 3, 0

2

2 3 3

y sin2 x

3; a 1, b

2

3, c 0

39.

Period:

Amplitude: 1

Key points:

3 , 0, 4 , 10, 1,

, 0, 2

,

1,

4

y

x π 4π 2π −2

−1

−2

2

y cos x

2 40.

Period:

Amplitude: 1

Key points:

3

8, 1, 2

, 0

0, 0,

8, 1,

4, 0,

2

y

x

−2

1

2

π

4

y sin 4 x

41.

Period:

Amplitude: 1

Key points:

0, 1, 1

4, 0, 1

2, 1, 3

4, 0

2

2 1

1 2

−2

1

2

x

y y cos 2 x 42.

Period:

Amplitude: 1

Key points:

6, 1, 8, 0

0, 0, 2, 1, 4, 0,

2

4 8

2

1

−2

62−2−6 x

y y sin x

4






55.

Period:

Amplitude:

Shift:

Key points: 2,

2

3, 3

2, 0, 5

2, 2

3 , 7

2, 0, 9

2,

2

3

x

2 x

9

2

x

2

4 0 and

x

2

4 2

2

3

4

−4

−3

−2

−1

1

2

3

4

π π 4 x

y

y

2

3 cos x

2

4; a

2

3, b

1

2, c

4

53.

Period:

Amplitude: 3

Shift: Set

Key points: , 0,

2, 3, 0, 6, 2

, 3, , 0

x x

x 0 and x 2

2

−8

2

4

π π 2 x

y y 3 cos x 3

54.

Period:

Amplitude: 4

Shift: Set

Key points:

4, 8, 4

, 4, 3

4, 0, 5

4, 4, 7

4, 8

x

4 x

7

4

x

4 0 and x

4 2

2

−4

2

4

6

10

x

y

π 3π 2π −2 − π π

y

4 cos x

4

4

51.

Period:

Amplitude:

Vertical shift two units upwardKey points:

0.20.1−0.1 0

1.8

2.2

x

y

0, 2.1, 1

120, 2, 1

60, 1.9, 1

40, 2, 1

30, 2.1

1

10

2

60

1

30

y 2 1

10 cos 60 x 52.

Period:

Amplitude: 2

Key points:

−7

−6

−5

−4

1

−

x

y

π π π 2

0, 1,

2, 3, , 5, 3

2 , 3, 2 , 1

2

y 2 cos x 3




63.

Amplitude:

Vertical shift one unit upward of

Thus, f x 2 cos x 1.

g x 2 cos x ⇒ d 1

1

23 1 2 ⇒ a 2

f x a cos x d 64.

Amplitude:

a 2, d 1

d 1 2 1

1 2 cos 0 d

1 32

2

f x a cos x d

65.

Amplitude:

Since is the graph of reflected in the

x -axis and shifted vertically four units upward, we haveThus, f x 4 cos x 4.a 4 and d 4.

g x 4 cos x f x

1

28 0 4

f x a cos x d 66.

Amplitude:

Reflected in the axis:

a 1, d 3

d 3

4 1 cos 0 d

a 1 x -

2 4

2 1

f x a cos x d

56.

Period:

Amplitude: 3

Shift: Set

Key points:

6, 3,

12, 0, 0, 3, 12

, 0, 6, 3

x

6 x

6

6 x 0 and 6 x 2

2

6

3

2

3

x

y

π

y 3 cos6 x

57.

−6 6

−4

4

y 2 sin4 x 58.

−12

−8

12

8

y 4 sin2

3 x

3 59.

−3

−1

3

3

y cos2 x

2 1

60.

−6

−6

6

2

y 3 cos x

2

2 2 61.

−20

−0.12

20

0.12

y 0.1 sin x

10 62.

−0.03

−0.02

0.03

0.02

y 1

100 sin 120 t




74.

(a) Period

(b)1 cycle

4 seconds

60 seconds

1 minute 15 cycles per minute

2

2 4 seconds

v

1.75 sin

t

2

71.

In the interval

when x 5

6,

6,

7

6,

11

6.sin x

1

2

2 , 2 ,

y2

1

2

−2

2

2

−2

y1 sin x 72.

y1 y

2 when x ,

y2 1

−2

2

2

−2

y1 cos x

73.

(a) Time for one cycle

(b) Cycles per min cycles per min

(c) Amplitude: 0.85; Period: 6

Key points: 0, 0, 3

2, 0.85, 3, 0, 9

2, 0.85, 6, 0

60

6 10

2

3 6 sec

t 2 4 8 10

0.25

0.50

0.75

1.00

−0.25

−1.00

v y 0.85 sin t

3

(c)

1 3 5 7

−2

−3

1

2

3

t

v

69.

Amplitude:

Period:

Phase shift:

Thus, y 2 sin x

4.

1

4 c 0 ⇒ c

4

bx c 0 when x

4

2 ⇒ b 1

a 2

y a sinbx c 70.

Amplitude:

Period: 2

Phase shift:

a 2, b

2, c

2

c

b

1 ⇒ c

2

2

b 4 ⇒ b

2

2 ⇒ a 2

y a sinbx c

67.

Amplitude:

Since the graph is reflected in the x -axis, we have

Period:

Phase shift:

Thus, y 3 sin 2 x .

c 0

2

b ⇒ b 2

a 3.

a 3

y a sinbx c 68.

Amplitude:

Period:

Phase shift:

a 2, b 1

2, c 0

c 0

2

b 4 ⇒ b

1

2

4

2 ⇒ a 2

y a sinbx c




78. (a) and (c)

Reasonably good fit

(d) Period is 29.6 days.

(e) March 12

The Naval observatory says that 50% of the moon’s face will be illuminated on

March 12, 2007.

y 0.44 44% ⇒ x 71.

y

x 10 20 30 40

0.2

0.4

0.6

0.8

1.0

P e r c e n t o f m o o n ’ s

f a c e i l l u m i n a t e d

Day of the year

(b)

y 12 1

2 sin0.21 x 0.92

C 0.92

Horizontal shift: 0.213 7.4 C 0

b 2

29.6 0.21

2

b 47.4 29.6

Period: 8 8 7 6 8

5 7.4 (average length of interval in data)

Amplitude: 12

⇒ a 12

Vertical shift: 1

2 ⇒ d

1

2

75.

(a) Period:

(b) f 1

p 440 cycles per second

2

880

1

440 seconds

y 0.001 sin 880 t 76.

(a) Period:

(b)1 heartbeat

65 seconds

60 seconds

1 minute 50 heartbeats per minute

2

5 3

6

5 seconds

P 100 20 cos5 t

3

77. (a)

(b)

The model is a good fit.

0

0

12

100

C t 56.55 26.95 cos t

6 3.67

d 1

2high low

1

283.5 29.6 56.55

c

b 7 ⇒ c 7 6 3.67

b 2

p

2

12

6

p 2high time low time 27 1 12

a 1

2high low

1

283.5 29.6 26.95 (c)


(d) Tallahassee average maximum:

Chicago average maximum:

The constant term, gives the average maximum

temperature.

(e) The period for both models is months.

This is as we expected since one full period is one

year.

(f) Chicago has the greater variability in temperature

throughout the year. The amplitude, a, determines this

variability since it is .12high temp low temp

2

6 12

d ,

56.55

77.90

0

0

12

100




85. Since the graphs are the

same, the conjecture is that

.sin x cos x

22

1

−2

f = g

x π π 32 2

−

π 32

y

86. f x sin x , g x cos x

2

x 0

0 1 0 0

0 1 0 01cos x

2

1sin x

2 3

2

2

83. True.

Since and so is a reflection in the x -axis of y sin x

2.cos x sin x

2, y cos x sin x

2,

84. Answers will vary.

Conjecture: sin x cos x

22

1

−2

f = g

x

y

π π 32 2

π 32

−

79.

(a) Period

Yes, this is what is expected because there are

365 days in a year.(b) The average daily fuel consumption is given by the

amount of the vertical shift (from 0) which is given

by the constant 30.3.

(c)

The consumption exceeds 40 gallons per day when

124 < x < 252.

0

0

365

60

2

2

365

365

C 30.3 21.6 sin2 t

365 10.9 80. (a) Period

The wheel takes 12 minutes to revolve once.

(b) Amplitude: 50 feet

The radius of the wheel is 50 feet.

(c)

0

0 20

110

2

6 12 minutes

81. False. The graph of is the graph oftranslated to the left by one period, and the graphs are

indeed identical.

sin( x )sin( x 2 ) 82. False. has an amplitude that is half that

of For y a cos bx , the amplitude is a. y cos x .

y 12 cos 2 x




87. (a)

The graphs are nearly the same for

(b)

The graphs are nearly the same for

2 < x <

2.

−2

−2 2

2

2 < x <

2.

−2

−2 2

2 (c)

The graphs now agree over a wider range, 3

4 < x <

3

4.

−2

−2 2

2

−2

2 −2

2

cos x 1 x 2

2!

x 4

4!

x 6

6!

sin x x x 3

3!

x 5

5!

x 7

7!

88. (a)

(by calculator)

(c)

(by calculator)

(e)

(by calculator)cos 1 0.5403

cos 1 1 1

2!

1

4! 0.5417

sin

6 0.5

sin

6 1

63

3!

65

5! 0.5000

sin1

2 0.4794

sin1

2

1

2

123

3!

125

5! 0.4794 (b)

(by calculator)

(d)

(by calculator)

(f)

(by calculator)cos

4 0.7071

cos

4 1

42

2!

42

4! 0.7074

cos0.5 0.8776

cos0.5 1 0.52

2!

0.54

4! 0.8776

sin 1 0.8415

sin 1 1 1

3!

1

5! 0.8417

The error in the approximation is not the same in each case. The error appears to increase as x moves farther away from 0.

89. log10

x 2 log10

x 212 12 log

10 x 2

91. lnt 3

t 1 ln t 3 lnt 1 3 ln t lnt 1

93.

log10

xy

12 log

10 x log

10 y

12 log

10 xy

90.

2 log2 x log

2 x 3

log2 x 2 x 3 log

2 x 2 log

2 x 3

92.

1

2 ln z

1

2 ln z2 1

ln z

z2 1

1

2 ln z

z2 1 1

2ln z ln z2 1

94.

log2 x 3 y

log2 x 2( xy)

2 log2 x log

2 xy log

2 x 2 log

2 xy

95.

ln3 x

y4ln 3 x 4 ln y ln 3 x ln y4




96.

ln x 2 2 x

ln x 3 2 x

x 2

ln 2 x

x 2 ln x 3

1

2ln2 x

x 2 ln x 3

1

2ln 2 x 2 ln x 3 ln x

1

2ln 2 x ln x 2 ln x 3

Section 4.6 Graphs of Other Trigonometric Functions

■ You should be able to graph

■ When graphing or you should first graph orbecause

(a) The x -intercepts of sine and cosine are the vertical asymptotes of cosecant and secant.

(b) The maximums of sine and cosine are the local minimums of cosecant and secant.

(c) The minimums of sine and cosine are the local maximums of cosecant and secant.

■ You should be able to graph using a damping factor.

y a sinbx c y a cosbx c y a cscbx c y a secbx c

y a cscbx c y a secbx c

y a cotbx c y a tanbx c


1.

Period:

Matches graph (e).

2

2

y sec 2 x 3.

Period:

Matches graph (a).

1

y 1

2 cot x 2.

Period:

Asymptotes:

Matches graph (c).

x , x

b

12 2

y tan x

2

4.

Period:

Matches graph (d).

2

y csc x 5.

Period:

Asymptotes:

Matches graph (f).

x 1, x 1

2

b

2

2 4

y 12

sec

x 2

6.

Period:

Asymptotes:

Reflected in x -axis

Matches graph (b).

x 1, x 1

2

b

2

2 4

y 2 sec

x 2

Vocabulary Check

1. vertical 2. reciprocal 3. damping 4.

5. 6. 7. 2

,

1

1,

x

n



Section 4.6 Graphs of Other Trigonometric Functions 387

7.

Period:

Two consecutive asymptotes:

x

2 and x

2

1

2

3

x π π

y

−

y 1

3 tan x

9.

Period:


3 x

2 ⇒ x

6

3 x

2 ⇒ x

6

3

x

4

3

2

1

− π

3π

3

y y tan 3 x

x 0

y 01

3

1

3

4

4

x 0

y 0 11

12

12

8.

Period:


x

2, x

2

−3

1

2

3

π π −

x

y y 1

4 tan x

x 0

y 01

4

1

4

4

4

10.

Period:

Two consecutive

asymptotes:

x 12

, x 12

1

2

−8

−4

x

y

y 3 tan x

x 0

y 3 0 3

1

4

1

4

13.

Period:

Two consecutive

asymptotes:

x 0, x 1

2

2

x

4

3

2

1

−3

−4

21−1−2

y y csc x

11.

Period:

Two consecutiveasymptotes:

x

2, x

2

2

1

2

3

x π

y

− π

y 1

2 sec x 12.

Period:

Two consecutive

asymptotes:

x

2, x

2

2

π

3

2 x

y

y 1

4 sec x

14.

Period:

Two consecutive

asymptotes:

x 0, x

4

2

4

2

8

6

4

2

−2− π

4π

4

x

y y 3 csc 4 x

0

11

21 y

3

3 x 0

1

2

1

4

1

2 y

3

3 x

2 1 2 y

5

6

1

2

1

6 x

6 3 6 y

5

24

8

24 x








39.

−6

−0.6

6

0.6 y 0.1 tan x

4

4

41.

x π π 2

2

y

x 7

4,

3

4,

4,

5

4

tan x 1

40.

−6

−2

6

2

y 1

3 cos x

2

2 ⇒ y

1

3 sec x

2

2

42.

x π π 2

2

1

x 5

3,

2

3,

3,

4

3

tan x 3 43.

x

y

π

2

3π

2

1

2

3

−3

x 4

3,

3,

2

3,

5

3

cot x 3

3

31.

−5

5 −5

5

y tan x

3 33.

−4

2

2−

4

y 2 sec 4 x 2

cos 4 x 32.

−3

3

4

3

4

3−

y tan 2 x

34.

−3

−2

3

2

y sec x ⇒ y 1

cos x 35.

−3

−

2

3

2

3

3

y tan x

4 36.

−3

2

3

2

3−

3

1

4 tan x

2

y 1

4 cot x

2

37.

y 1

sin4 x

−3

−

2

2

3 y csc4 x 38.

y 2

cos2 x

−4

−

4 y 2 sec(2 x ) ⇒




44.

x π

2

π

2

3π

23π

2− −

2

3

−3

x 7

4,

3

4,

4,

5

4

cot x 1 45.

x π π 2π π 2

1

−−

y

x ±2

3, ±

4

3

sec x 2 46.

x π π π 2π 2

1

−−

y

x 5

3,

3,

3,

5

3

sec x 2

47.

x 7

4,

5

4,

4,

3

4

x π 3ππ3π

2 2 2 2− −

1

2

3

−1

ycsc x 2 48.

x 2

3,

3,

4

3,

5

3

x π 3ππ3π

2 2 2 2− −

1

2

3

csc x 2 3

3

49.

Thus, is an even function and the graph has

axis symmetry. y-

f x sec x

f x

1

cos x

1

cos x

f x sec x

y

x

3

4

π π π 2−

f x sec x 1

cos x 50.

Thus, the function is odd

and the graph of

is symmetric about the origin.

y tan x

tan x tan x

x

2

3

−3

3π

2−

π

2−

π

2

3π

2

f x tan x

51.

(a)

(b) on the interval,

(c) As and

since is

the reciprocal of f x .g x g x

12 csc x → ±

f x 2 sin x → 0 x → ,

6< x <

5

6 f > g

1

−1

2

3

f

g

x π π π 32 4

π

4

y

g x 1

2 csc x

f x 2 sin x 52.

(a)

(b) The interval in which

(c) The interval in which

which is the same interval as part (b).

2 f < 2g is 1,13,

f < g is 1,13.

−3

g

f

3

1−1

f x tan x

2, g x

1

2 sec

x

2




57.

As

Matches graph (d).

x → 0, f x → 0 and f x > 0.

f x x cos x 58.

Matches graph (a) as x → 0, f x → 0.

f x x sin x

60.

Matches graph (c) as x → 0, g x → 0.

g x x cos x 59.

As

Matches graph (b).

x → 0, g x → 0 and g x is odd.

g x x sin x

61.

The graph is the line y 0.

f x g x

g x 0

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y f x sin x cos x

2 62.

It appears that

That is,

sin x cos x

2 2 sin x .

f x g x .

g x 2 sin x

−4

2

4

x π π −

y

f x sin x cos x

2

63.

f x g x

g x 1

21 cos 2 x

–1

2

3

y

x π π −

f x sin2 x 64.

It appears that

That is,

cos2 x

2

1

21 cos x .

f x g x .

g x 1

21 cos x

−1

−3 3 6−6

2

3

x

y f x cos2 x

2

53.

The expressions are equivalent except when

and y1 is undefined.

sin x 0

sin x csc x sin x 1

sin x 1, sin x 0

−2

3−3

2

y1 sin x csc x and y

2 1 54.

The expressions are equivalent.

sin x sec x sin x 1

cos x

sin x

cos x tan x

−4

−2 2

4

y1 sin x sec x , y

2 tan x

55.


−4

2 −2

4cot x cos x

sin x

y1

cos x

sin x and y

2 cot x

1

tan x 56.


tan2 x sec2 x 1

1 tan2 x sec2 x

−1

−

3

2

3

2

3

y1 sec2 x 1, y

2 tan2 x




65.

The damping factor is

As x → , g x →0.

y e x 22.

e x 22≤ g x ≤ e x 22

−1

8−8

1g x e x 22 sin x

66.

Damping factor:

As x → , f x → 0.

−3 6

−3

3

e x

f x e x cos x 67.

Damping factor:

As x →, f x → 0.

−9 9

−6

6

y 2 x 4.

2 x 4 ≤ f x ≤ 2 x 4

f x 2 x

4

cos x 68.

Damping factor:

As x → , h x → 0.

−8

−1

8

1

2 x 24

h x 2 x 2

4 sin x

69.

As x → 0, y → .

0

−2

8

6

y 6

x cos x , x > 0 71.

As x → 0, g x → 1.

−1

6 −6

2

g x sin x

x 70.

As x → 0, y → .

0

−2

6

6

y 4

x sin 2 x , x > 0

72.

As x → 0, f ( x ) → 0.

−1

−6 6

1

f ( x ) 1 cos x

x

73.

As oscillates

between and 1.1

x → 0, f x

−2

−

2

f x sin1

x

74.

As oscillates. x → 0, h( x )

−1

−

2

h( x ) x sin1

x

75.

Grounddistance

x

14

10

6

2

−2

−6

−10

−14

Angle of elevation

d

π π π 32 4

π

4

d 7tan x

7 cot x

tan x 7

d 76.

x

20

40

60

80

Angle of camera

D i s t a n c e

d

0 π

2π

4π

4π

2− −

d 27cos x

27 sec x ,

2 < x <

2

cos x 27

d






83. As from the left,

As from the right, f x tan x → . x →

2

f x tan x → . x →

282. True.

If the reciprocal of is translated units to the

left, we have

y 1

sin x

2 1

cos x sec x .

2 y sin x

y sec x 1

cos x

84. As from the left, .

As from the right, . f ( x ) csc x → x →

f ( x ) csc x → x →

85.

(a)

The zero between 0 and 1 occurs at x 0.7391.

3

−2

−3

2

f x x cos x

(b)

This sequence appears to be approaching the zero

of f : x 0.7391.

x 9 cos 0.7504 0.7314

x 8 cos 0.7221 0.7504

x 7 cos 0.7640 0.7221

x 6 cos 0.7014 0.7640

x 5 cos 0.7935 0.7014

x 4 cos 0.6543 0.7935

x 3 cos 0.8576 0.6543

x 2 cos 0.5403 0.8576

x 1

cos 1 0.5403

x 0 1

x n cos x

n1

86.

The graphs are nearly the

same for 1.1 < x < 1.1.

y x 2 x 3

3!

16 x 5

5!

−6

−

2

3

2

3

6 y tan x 87.

The graph appears to

coincide on the interval

1.1 ≤ x ≤ 1.1.

y2 1

x 2

2!

5 x 4

4!

−6

−

6

2

3

2

3

y1 sec x

88. (a)

—CONTINUED—

−3 3

−2

2

y2

−3 3

−2

2

y1

y2 4

sin x 13

sin 3 x 15

sin 5 x y1 4

sin x 13

sin 3 x





Section 4.7 Inverse Trigonometric Functions 397

■ You should know the definitions, domains, and ranges of y arcsin x , y arccos x , and y arctan x .

Function Domain Range

y arcsin x ⇒ x sin y

y arccos x ⇒ x cos y

y arctan x ⇒ x tan y

■ You should know the inverse properties of the inverse trigonometric functions.

■ You should be able to use the triangle technique to convert trigonometric functions of inverse trigonometric functions

into algebraic expressions.

tanarctan x x and arctantan y y,

2 < y <

2

cosarccos x x and arccoscos y y, 0 ≤ y ≤

sinarcsin x x and arcsinsin y y,

2 ≤ y ≤

2

2 < x <

2 < x <

0 ≤ y ≤ 1 ≤ x ≤ 1

2

≤ y ≤ 2

1 ≤ x ≤ 1

Vocabulary Check

Alternative

Function Notation Domain Range

1.

2.

3.

2 < y <

2 < x < y tan1 x y arctan x

0 ≤ y ≤ 1 ≤ x ≤ 1 y cos1 x y arccos x

2 ≤ y ≤

21 ≤ x ≤ 1 y sin1 x y arcsin x

1.

2 ≤ y ≤

2 ⇒ y

6 y arcsin

1

2 ⇒ sin y

1

2 for 2. y arcsin 0 ⇒ sin y 0 for

2 ≤ y ≤

2 ⇒ y 0

3. 0 ≤ y ≤ ⇒ y

3 y arccos

1

2 ⇒ cos y

1

2 for 4. y arccos 0 ⇒ cos y 0 for 0 ≤ y ≤ ⇒ y

2

5.

2 < y <

2 ⇒ y

6

y arctan 3

3 ⇒ tan y

3

3 for 6.

2 < y <

2 ⇒ y

4

y arctan1 ⇒ tan y 1 for

Section 4.7 Inverse Trigonometric Functions






36.

cos 6 3

2

arccos1

2 2

3

arccos1 37.

tan arctan x

4 θ

4

x

tan x

4

38.

arccos4

x

cos 4

x 39.

sin arcsin x 2

5 θ

5 x + 2

sin x 2

5

40.

arctan x 1

10

tan x 1

1041.

arccos x 3

2 x θ

x + 3

2 x

cos x 3

2 x

42.

x 1

arctan 1

x 1

tan x 1

x 2 1

1

x 143. sinarcsin 0.3 0.3 44. tanarctan 25 25

45. cosarccos0.1 0.1 46. sinarcsin0.2 0.2 47.

Note: 3 is not in the range of

the arcsine function.

arcsinsin 3 arcsin0 0

48.

Note: is not in the range of the arccosine function.7

2

arccoscos7

2 arccos 0

2 49. Let

and sin y 3

5.

tan y 3

4, 0 < y <

2,

x y

53

4

y

y arctan3

4,

50. Let

secarcsin4

5 sec u 5

3.

sin u 4

5, 0 < u <

2,

3

45

u

u arcsin4

5

, 51. Let

and cos y 1

5 5

5.

tan y 2 2

1, 0 < y <

2,

x

y

5 2

1

y y arctan 2,




52. Let

1

25

u

sinarccos 5

5 sin u 2

5

2 5

5.

cos u 5

5, 0 < u <

2,

u arccos 5

5, 53. Let

and

x y

5

12

13

y

cos y 12

13.sin y

5

13, 0 < y <

2,

y arcsin5

13,

55. Let

and

x y

34−3

5

y

sec y

345

.tan y 35

, 2 < y < 0,

y arctan3

5,54. Let

13

12

−5u

cscarctan 5

12 csc u 13

5.

tan u 512

, 2

< u < 0,

u arctan 5

12,

56. Let

4−3

7

u

tanarcsin3

4 tan u 3

7

3 7

7.

sin u 3

4,

2 < u < 0,

u arcsin3

4, 57. Let

and

x y

3

−2

5

y

sin y 5

3.cos y

2

3,

2 < y < ,

y arccos2

3,




60. Let

x + 12

1

x

u

sinarctan x sin u x

x 2 1 .

tan u x x

1,u arctan x , 61. Let

and cos y 1 4 x 2.

sin y 2 x 2 x

1,

1 − 4 x 2

2 x

y

1

y arcsin2 x ,

62. Let

secarctan 3 x sec u 9 x 2 1.

tan u 3 x 3 x

1,

1

3 x 9 + 1 x 2

u

u arctan 3 x , 63. Let

and sin y 1 x 2.

cos y x x

1, 1 − x 2

x

y

1

y arccos x ,

64. Let

x −1

2 x x − 2

1

u

secarcsin x 1 sec u 1

2 x x 2.

sin u x 1 x 1

1,

u arcsin x 1, 65. Let

and tan y 9 x 2

x .

cos y x

3, 9 − x 2

x

y

3

y arccos x

3,

66. Let

cotarctan1

x cot u x .

tan u 1

x , 1

u x

x + 12

u arctan1

x , 67. Let

and csc y x 2 2

x .

tan y x

2,

x 2 + 2

x

y

2

y arctan x

2,

59. Let

and cot y 1

x .

tan y x x

1,

x 2 + 1 x

y

1

y arctan x ,58. Let

cotarctan5

8 cot u 8

5.

tan u 5

8, 0 < u <

2, 5

8

89

u

u arctan5

8,




69.

They are equal. Let

and

The graph has horizontal asymptotes at y ±1.

g x 2 x

1 4 x 2 f x

1 + 4 x 2

2 x

y

1

sin y 2 x

1 4 x 2.

tan y 2 x 2 x

1,

y arctan 2 x , −3 3

−2

2

f x sinarctan 2 x , g x 2 x

1 4 x 2

70.

Asymptote:

These are equal because:

Let

Thus, f x g x .

4 x 2

x g x

f x tanarccos x

2 tan u

u arccos x

2.

2

u x

4 − x 2

x 0

g x 4 x 2

x

−3

−2

3

2

f x tanarccos x

2 71. Let

Thus,

x 2 + 81

x

y

9

arcsin y 9

x 2 81, x < 0.

arcsin y 9

x 2 81, x > 0;

tan y 9

x and sin y

9

x 2 81, x > 0;

9

x 2 81, x < 0.

y arctan9

x .

72. If

then

arcsin 36 x 2

6 arccos x

6.

sin u 36 x 2

6,

6

u x

36 − x 2

arcsin 36 x 2

6 u, 73. Let Then,

and

Thus,

( x − 1)2 + 9

3

y

x − 1

y arcsin x 1 x 2 2 x 10

.

sin y x 1 x 12 9.

cos y 3

x 2 2 x 10

3

x 12 9

y arccos3

x 2 2 x 10.

68. Let

cosarcsin x h

r cos u r 2 x h2

r .

sin u x h

r , r

x h−

r x h− −( )22

u

u arcsin x h

r ,




75.

Domain:

Range:

This is the graph of with a factor of 2.

−2 −1 1 2

π

π 2

x

y

f x arccos x

0 ≤ y ≤ 2

1 ≤ x ≤ 1

y 2 arccos x

76.

Domain:

Range:

This is the graph of

with a

horizontal stretch of a

factor of 2.

f x arcsin x

2 ≤ y ≤

2

2 ≤ x ≤ 2

1 2−2 x

π

π

y

−

y arcsin x

2 77.

Domain:

Range:


shifted

one unit to the right.

g x arcsin x

2 ≤ y ≤

2

0 ≤ x ≤ 2

−1 1 2 3

π

π

y

x

−

f x arcsin x 1

78.

Domain:

Range:


shifted

two units to the left.

y arccos t

0 ≤ y ≤

3 ≤ t ≤ 1

−4 −3 −2 −1t

π

ygt arccost 2 79.

Domain: all real numbers

Range:


with a

horizontal stretch of a

factor of 2.

g x arctan x

2 < y <

2−4 −2 2 4

π

π

y

x

−

f x arctan 2 x

80.

Domain: all real numbers

Range:


shifted

upward units. 2

y arctan x

0 < y ≤

−4 −2 2 4 x

π

y

f x

2 arctan x 81.

Domain:

Range: all real numbers

−2 1 2

π

y

v

1 − v2

v

y

1

1 ≤ v ≤ 1, v 0

hv tanarccos v 1 v2

v

74. If

then

2

u

x − 2

4 x x − 2

arccos x 2

2 arctan

4 x x 2

x 2.

cos u x 2

2,

arccos x 2

2 u,






92. (a)

(b) When

When

arctan1200

750 1.01 58.0.

s 1200,

arctan300

750

0.38 21.8.

s 300,

arctans

750

tan s

75093.

(a)

(b) is maximum when feet.

(c) The graph has a horizontal asymptote at

As x increases, decreases.

0.

x 2

0

−0.5

6

1.5

arctan3 x

x 2 4

94. (a)

(b)

feeth 20 tan 20 11

17 12.94

tan hr h

20

r 1

240 20

arctan11

17 0.5743 32.9

tan 11

17 95.

(a)

(b)

h 50 tan 26 24.39 feet

tan 26 h

50

arctan20

41 26.0

tan 20

41

20 ft

41 ft

θ

96. (a)

(b)

arctan6

1 1.41 80.5

x 1 mile

arctan6

7 0.71 40.6

x 7 miles

arctan6

x

tan 6

x 97. (a)

(b)

x 12: arctan12

20 31.0

x 5: arctan 520

14.0

arctan x

20

tan x

20

98. False.

is not in the range of

arcsin1

2

6

arcsin x .5

6

99. False.

is not in the range of the arctangent function.

arctan 1

4

5

4

100. False.

is defined for all real x , but and require

Also, for example,

Since , but undefined.arcsin 1

arccos 1

2

0 arctan 1

4

arctan 1 arcsin 1

arccos 1.

1 ≤ x ≤ 1.arccos x arcsin x arctan x













Section 4.8 Applications and Models 411

7. Given:

B 90 72.08 17.92

A arccos 1652 72.08º

cos A 16

52

2448 12 17 49.48

a 522 162

b = 16

c = 52a

AC

Bb 16, c 52 8. Given:

8.03

arcsin 1.329.45 a

b = 1.32

c = 9.45

AC

Bsin B

b

c ⇒ B arcsin

b

c

cos A b

c ⇒ A arccos

b

c arccos

1.32

9.45 81.97

a c2 b2 87.5601 9.36

b 1.32, c 9.45

9. Given:

b

c = 430.5a

AC

B

12°15′

b 430.5 cos 1215 420.70

cos 1215 b

430.5

a 430.5 sin 1215 91.34

sin 1215 a

430.5

B 90 12 15 77 45

A 12 15 , c 430.5 10. Given:

a = 14.2

b

c

AC

B

65°12′

tan B ba

⇒ b a tan B 14.2 tan 65 12 30.73

cos B a

c ⇒ c

a

cos B

14.2

cos 65 12 33.85

A 90 B 90 65 12 24 48

B 65 12 , a 14.2

11.

12

b12

b

b

h

θ θ

h

1

24 tan 52

2.56 inches

tan h

12b ⇒ h

1

2b tan 12.

12

b12

b

b

h

θ θ

h

1

210 tan 18 1.62 meters

tan h

12b ⇒ h

1

2b tan

13.

12

b12

b

b

h

θ θ

h 1

246 tan 41 19.99 inches

tan h

12b ⇒ h

1

2 b tan 14.

12

b12

b

b

h

θ θ

h 1

211 tan 27 2.80 feet

tan h

12b ⇒ h

1

2b tan






26. (a) Since the airplane speed is

after one minute its distance travelled is 16,500 feet.

18°

16500a

a 16,500 sin 18 5099 ft

sin 18 a

16,500

275ft

sec60sec

min 16,500ft

min,

(b)

275s 10,000feet

18°

117.7 seconds

s 10,000

275(sin 18)

sin 18 10,000

275s

27.

x 4 sin 10.5 0.73 mile

10.5°

4 x sin 10.5

x

4

28.

Angle of grade:

Change in elevation:

2516.3 feet

21,120 sinarctan 0.12

y 21,120 sin

sin y

21,120

arctan 0.12 6.8

tan 12 x

100 x

100 x

12 = x y4 miles = 21,120 f eet

θ 29. The plane has traveled

N

S

EW

90052°

38°

a

b

cos 38 b

900 ⇒ b 709 miles east

sin 38 a

900 ⇒ a 554 miles north

1.5600 900 miles.

30. (a) Reno is miles N of Miami.

Reno is miles W of Miami.

(b) The return heading isN

S

EW

280°

10°

280.

2472 cos 10 2434

N

S

EW

100°

80°

10° Miami

Reno2472 mi

2472 sin 10 429

25.

arctan 950

26,400

2.06

tan 950

26,400

5 miles 5 miles5280 feet

1 mile 26,400 feet

5 miles

950feet

θ

θ

Not drawn to scale1200 feet 150 feet 400 feet 950 feet






39.

17,054 ft

a cot 16 a cot 57 55

6 ⇒ a 3.23 miles

cot 16 a cot 57 556a

tan 16 a

a cot 57 556

tan 16 a

x 556

57°16°

x 550

60

H

P1 P2

a

tan 57 a

x ⇒ x a cot 57

41.

arctan 5 78.7

tan

1 32

1

132

52

12

5

L2: 3 x y 1 ⇒ y x 1 ⇒ m

2 1

L1: 3 x 2 y 5 ⇒ y

3

2 x

5

2 ⇒ m

1

3

240.

5410 feet

h 17

1

tan 2.5

1

tan 9 1.025 miles

h

tan 2.5

h

tan 9 17

x h

tan 9 17

tan 9

h

x 17

x h

tan 2.5

x − 1717

2.5° 9°

x

h

Not drawn to scale

tan 2.5 h

x

42.

arctan9

7 52.1

arctanm

2 m

1

1 m2m

1 arctan 15 2

1 152

tan m

2 m

1

1 m2m

1

L2 x 5 y 4 ⇒ m

2 1

5

L1 2 x y 8 ⇒ m

1 2 43. The diagonal of the base has a length of

Now, we have

35.3.

arctan1

2

a

2a

θ

tan a

2a

1

2

a2 a2 2a.

44.

arctan 2 54.7º

θ

a

a 2

tan a 2

a 2 45.

Length of side:

36°

25

d

2d 29.4 inches

sin 36 d

25 ⇒ d 14.69

38.

Distance between towns:

d D

28°

28°

55°

55°10 km

T 2T 1

D d 18.8 7 11.8 kilometers

cot 28 D

10 ⇒ D 18.8 kilometers

cot 55 d

10 ⇒ d 7 kilometers




46.

25 inches

Length of side 2a 212.5

a 25 sin 30 12.5

sin 30 a

25

2530°

a 47.

y 2b 2 3r

2 3r

b 3r

2

b r cos 30

cos 30 b

r

br

r

30°

2

x

y

48.

Distance 2a 9.06 centimeters

4.53

a c sin 15 17.5 sin 15

sin 15 a

c

c 35

2 17.5

c

15°

a

49.

a 10

cos 35 12.2

cos 35 10

a

b 10 tan 35 7

tan 35 b

10

35° 35°

ba

10 10

10

10

10

10

50.

c 10.82 7.22 13 feet

b 6sin

7.2 feet

sin 6

b

90 33.7 56.3

f 21.6

2 10.8 feet

a 18

cos 21.6 feet

6

6bc

a f

θ φ

936

cos 18

a

arctan2

3 0.588 rad 33.7

tan 12

18

51.

Use since

Thus, d 4 sin t .

2

2 ⇒

d 0 when t 0.d a sin t

d 0 when t 0, a 4, period 2 52. Displacement at is

Amplitude:

Period:

d 3 sin t 3

2

6 ⇒

3

a 3

0 ⇒ d a sin t .t 0

53.

Use since

Thus, d 3 cos4

3t 3 cos4 t

3 .

2

1.5 ⇒

4

3

d 3 when t 0.d a cos t

d 3 when t 0, a 3, period 1.5 54. Displacement at is

Amplitude:

Period:

d 2 cos t 5

2

10 ⇒

5

a 2

2 ⇒ d a cos t .t 0




55.

(a) Maximum displacement amplitude

(b)

cycles per unit of time

(c)

(d) 8 t

2 ⇒ t

1

16

d 4 cos 40 4

4

Frequency

2

8

2

4

d 4 cos 8 t 56.

(a) Maximum displacement:

(b) Frequency: cycles per unit of time

(c)

(d) Least positive value for t for which

t

2

1

20

1

40

20 t

2

20 t arccos 0

cos 20 t 0

1

2 cos 20 t 0

d 0

t 5 ⇒ d 1

2 cos 100 1

2

2

20

2 10

a 1

2 1

2

d 1

2 cos 20 t

57.

(a) Maximum displacement amplitude

(b)

cycles per unit of time

(c)

(d) 120 t ⇒ t 1

120

d 1

16 sin 600 0

60

Frequency

2

120

2

1

16

d 1

16 sin 120 t

58.

(a) Maximum displacement:

(b) Frequency: cycles per unit of time

(c)

(d) Least positive value for t for which

t

792

1

792

792 t

792 t arcsin 0

sin 792 t 0

1

64 sin 792 t 0

d 0

t 5 ⇒ d 1

64 sin3960 0

2

792

2 396

a 1

64 1

64

d 1

64 sin 792 t

59.

2 264 528

264

2

Frequency

2

d a sin t 60. At buoy is at its high point

Returns to high point every 10 seconds:

Period:

d 7

4 cos

t

5

2

10 ⇒

5

a 74

Distance from high to low 2a 3.5

⇒ d a cos t .t 0,




61.

(a)

t

1

−1

y

π π 38 8

π 4

π 2

y 1

4 cos 16t , t > 0

(b) Period:

(c)1

4cos 16t 0 when 16t

2 ⇒ t

32

2

16

8

62. (a)

(c) L L1 L

2

2

sin

3

cos

(b)

The minimum length of the elevator is 7.0 meters.

(d)

From the graph, it appears that the minimum length is

7.0 meters, which agrees with the estimate of part (b).

−12

−2 2

12

0.1 23.0

0.2 13.1

0.3 9.9

0.4 8.43

cos 0.4

2

sin 0.4

3

cos 0.3

2

sin 0.3

3

cos 0.2

2

sin 0.2

3

cos 0.1

2

sin 0.1

L1 L

2 L

2 L

1

0.5 7.6

0.6 7.2

0.7 7.0

0.8 7.13

cos 0.8

2

sin 0.8

3

cos 0.7

2

sin 0.7

3

cos 0.6

2

sin 0.6

3

cos 0.5

2

sin 0.5

L1 L

2 L

2 L

1

63. (a) and (b)

Base 1 Base 2 Altitude Area

8 22.1

8 42.5

8 59.7

8 72.7

8 80.5

8 83.1

8 80.7

The maximum occurs when and is approximately

83.1 square feet.

60

8 sin 70º8 16 cos 70º

8 sin 60º8 16 cos 60º

8 sin 50º8 16 cos 50º

8 sin 40º8 16 cos 40º

8 sin 30º8 16 cos 30º

8 sin 20º8 16 cos 20º

8 sin 10º8 16 cos 10º

(c)

(d)

The maximum of 83.1 square feet occurs when

3 60.

0

900

100

641 cos sin

16 16 cos 4 sin

A 8 8 16 cos 8 sin

2




66. False. One period is the time for one complete cycle of

the motion.

68. Aeronautical bearings are always taken clockwise from

North (rather than the acute angle from a north-south line).

64. (a)

(c) Period:

This corresponds to the 12 months in a year. Since the

sales of outerwear is seasonal, this is reasonable.

2

6 12

Month (1 January)↔

Averagesales

(inmillionsofdollars)

t

S

2 4 121086

3

6

9

12

15

(b)

Shift:

Note: Another model is


(d) The amplitude represents the maximum displacement from

average sales of 8 million dollars. Sales are greatest in

December (cold weather Christmas) and least in June.

S 8 6.3 sin t 6

2.

S 8 6.3 cos t 6

S d a cos bt

d 14.3 6.3 8

2

b 12 ⇒ b

6

t 2 4 121086

3

6

9

12

15

A v e r a g e s a l e s

( i n m i l l i o n s o f d o l l a r s )

Month (1 ↔ January)

S

a 1

214.3 1.7 6.3

65. False. Since the tower is not exactly vertical, a right

triangle with sides 191 feet and d is not formed.

67. No. N 24 E means 24 east of north.

69. passes through

y 4 x 6

y 2 4 x 4

y 2 4 x 1

y

x −1−2−3−4 1 2 3 4−1

1

2

3

5

6

7

1, 2m 4, 70. Linear equation through

y 12 x

16

b 16

0 16 b

0 12

13 b

y

x −1−2−3 2 3

−1

−2

−3

1

2

3

y 12 x b

13, 0m

12

71. Passes through and

y 4

5 x

22

5

y 6 45 x 8

5

y 6 4

5 x 2

y

x −1−2 1 2 3 4 5−1

1

2

3

4

6

7

m 2 6

3 2

4

5

3, 22, 6 72. Linear equation through and

y 4

3 x

1

3

y 2

3

4

3 x 1

4

4

3

1

34

y

x −1−2−3 2 3

−1

−2

−3

1

2

3

m 13 23

12 14

1

2,

1

31

4,

2

3



Review Exercises for Chapter 4


1. 0.5 radian 2. 4.5 radians

3.

(a)

(b) The angle lies in Quadrant II.

(c) Coterminal angles:

3

4 2

5

4

11

4 2

3

4

11

4

π

x

y

11

44.

(a)

(b) Quadrant I

(c)

2

9 2

16

9

2

9 2

20

9

x

y

2

9

π

2

95.

(a)

(b) The angle lies in Quadrant II.


4

3 2

10

3

4

3 2

2

3

−4

3

π

x

y

4

3

6.

(a)

(b) Quadrant I

(c)

23

3 2 17

3

23

3 8

3

x

−

y

23

3

π

23

3 7.

(a)

(b) The angle lies in Quadrant I.


70 360 290

70 360 430

70°

y

x

70 8.

(a)

(b) Quadrant IV

(c)

280 360 80

280 360 640

x

280°

y

280



Review Exercises for Chapter 4 421

9.

(a)

(b) The angle lies in Quadrant III.


110 360 470

110 360 250

−110°

x

y

110 10.

(a)

(b) Quadrant IV

(c)

405 360 45

405 720 315

x

−405°

y

405 11.

8.378 radians

8

3 radians

480 480 rad

180

12. 127.5

180 2.225 13.

3

16 radian 0.589 radian

33 45 33.75 33.75 rad

180

14. 196 77 196 77

60

180 3.443 15.

5 rad

7

5 rad

7

180

rad 128.571

16. 11

6

180

330.000 17. 3.5 rad 3.5 rad

180

rad 200.535

18. 5.7 180

326.586 19. radians

inchess r 2023

30 48.17

138 138

180

23

30 20.

meterss 11.52

11

3 meters

s r 11 60

180

60 60

180 radians

21. (a)

(b)

400 inches per minute

Linear speed 666

23 inches

1 minute

6623 radians per minute

Angular speed 33

132 radians

1 minute 22.

12.05 miles per hour

212.1 inches per second

67.5 inches per second

5 rads 13.5 inches

linear speed angular speed radius

23. radians

square inches A 1

2r 2

1

21822

3 339.29

120 120

180

2

324.

A 55.31 square millimeters

A 1

2 r 2

1

25

6 6.52

26. t 3

4, x , y 2

2, 2

2 25. corresponds to the point .1

2, 3

2 t 2

3




27. corresponds to the point 3

2,

1

2.t 5

6 28. t

4

3, x , y 1

2, 3

2

29. corresponds to the point

cot7

6

x

y 3tan

7

6

y

x

1

3 3

3

sec7

6

1

x

2 3

3cos

7

6 x

3

2

csc7

6

1

y

2sin7

6

y 1

2

3

2,

1

2.t 7


cot

4

x

y 1tan

4

y

x 1

sec

4

1

x 2cos

4 x

2

2

csc

4

1

y

2sin

4

y 2

2

2

2, 2

2 .t

4


cot2

3 x

y 3

3tan2

3 y

x 3

sec2

3 1

x 2cos2

3 x

1

2

csc2

3 1

y

2 3

3sin2

3 y 3

2

1

2,

3

2 .t 2


is undefined.

is undefined.cot 2 x

ytan 2

y

x 0

sec 2 1

x

1cos 2 x 1

csc 2 1

ysin 2 y 0

1, 0.t 2

33. sin11

4 sin

3

4

2

234. cos 4 cos 0 1 35. sin17

6 sin5

6 1

2

36. cos13

3 cos5

3 1

2 37. tan 33 75.3130 38. csc 10.5

1

sin 10.5 1.1368

39. sec12

5 1

cos12

5 3.2361 40. sin

9 0.3420

41.

cot adj

opp

5

4tan

opp

adj

4

5

cos adj

hyp

5

41

5 41

41 sec

hyp

adj

41

5

sin opp

hyp

4

41

4 41

41 csc

hyp

opp 41

4

opp 4, adj 5, hyp 42 52 41 42.

cot adj

opp

6

6 1tan

opp

adj

6

6 1

sec hyp

adj

6 2

6 2cos

adj

hyp

6

6 2 2

2

csc hyp

opp

6 2

6 2sin

opp

hyp

6

6 2 2

2

hyp 62 62 6 2

adj 6, opp 6

43.

cot adj

opp

4

4 3 3

3tan

opp

adj

4 3

4 3

cos adj

hyp

4

8

1

2 sec

hyp

adj

8

4 2

sin opp

hyp

4 3

8

3

2 csc

hyp

opp

8

4 3

2 3

3

adj 4, hyp 8, opp 82 42 48 4 3




44.

cot adj

opp

2 14

5

sec hyp

adj

9

2 14

9 14

28

csc hyp

opp

9

5

tan opp

adj

5

2 14

5 14

28

cos adj

hyp

2 14

9

sin opp

hyp

5

9

adj 92 52 2 14

opp 5, hyp 9 45.

(a)

(b)

(c)

(d) tan sin

cos

13

2 23

1

2 2 2

4

sec 1

cos

3

2 2

3 2

4

cos 2 2

3

cos 89

cos2 8

9

cos2 1 1

9

1

3

2

cos2 1

sin2 cos2 1

csc 1

sin 3

sin 1

3

46.

(a)

(b)

(c)

(d) csc

1

cot

2

1

1

16 17

4

cos 1

sec

1

17 17

17

sec 1 tan2 1 16 17

cot 1

tan

1

4

tan 4 47.

(a)

(b)

(c)

(d) tan sin

cos

14

154

1

15 15

15

sec 1

cos

4

15

4 15

15

cos 15

4

cos 15

16

cos2 15

16

cos2 1 1

16

1

42

cos2 1

sin2 cos2 1

sin 1

csc

1

4

csc 4

48.

(a)

(b) cot csc2 1 25 1 2 6

sin 1

csc

1

5

csc 5

(c)

(d) sec90 csc 5

tan 1

cot

1

2 6 6

12

49. tan 33 0.6494 51. sin 34.2 0.562150. csc 11 1

sin 11 5.2408




55.

kilometer or 71.3 meters

1 10'° x

Not drawn to scale

3. 5 k m

x 3.5 sin 1 10 0.07

sin 1 10 x

3.5 56.

x 25tan 52

19.5 feet

tan 52 25

x

52°

x

25

57.

cot x y 3

4tan y

x 4

3

sec r

x

5

3cos

x

r

3

5

csc r

y

5

4sin

y

r

4

5

x 12, y 16, r 144 256 400 20 58.

cot x

y

3

4tan

y

x

4

3

sec r

x

5

3

cos x

r

3

5

csc r

y

5

4sin

y

r

4

5

r 32 42 5

x , y 3, 4

59.

cot x

y

23

52

4

15tan

y

x

52

23

15

4

sec r

x 2416

23

241

4cos

x

r

23

2416

4

241

4 241

241

csc r

y 2416

52

2 241

30

241

15sin

y

r

52

2416

15

241

15 241

241

r 2

32

5

22

241

6

x 2

3, y

5

2

52. sec 79.3 1

cos 79.3 5.3860 53.

3.6722

cot 15 14 1

tan15 1460

54.

0.2045

cos 78 11 58 cos7811

60

58

3600

60.

cot x

y

103

23 5tan

y

x 23

103

1

5

sec

r

x

2 263

103

26

5cos

x

r

103

2 263

5 26

26

csc r

y

2 263

23 26sin

y

r

23

2 263

26

26

r 10

3 2

2

32

2 26

3

x , y 10

3,

2

3




61.

cot x

y

0.5

4.5

1

9tan

y

x

4.5

0.5 9

sec r

x

822

0.5

82cos x

r

0.5

822

82

82

csc r

y 822

4.5

82

9sin

y

r

4.5

822

9 82

82

r 0.52 4.52 20.5 82

2

x 0.5, y 4.5

62.

cot x y 0.3

0.4 3

4 0.75tan y

x 0.4

0.3 4

3 1.33

sec r

x

0.5

0.3

5

3 1.67cos

x

r

0.3

0.5

3

5 0.6

csc r

y

0.5

0.4

5

4 1.25sin

y

r

0.4

0.5

4

5 0.8

r 0.32 0.42 0.5

x , y 0.3, 0.4

64.

cot x

y

2 x

3 x

2

3

sec r

x 13 x

2 x

13

2

csc r

y 13 x

3 x

13

3

tan y

x

3 x

2 x

3

2

cos x

r

2 x

13 x

2 13

13

sin y

r

3 x

13 x

3 13

13

r 2 x 2 3 x 2 13 x

x , y 2 x , 3 x , x > 0 65. is in Quadrant IV.

cot 5 11

11

sec 6

5

csc r

y

6 11

11

tan y

x

11

5

cos x

r

5

6

sin y

r

11

6

r 6, x 5, y 36 25 11

sec 6

5, tan < 0 ⇒

63.

cot x

y x

4 x

1

4tan y

x

4 x

x 4

sec r

x 17 x

x 17cos

x

r

x

17 x 17

17

csc r

y 17 x

4 x

17

4sin

y

r

4 x

17 x

4 17

17

r x 2 4 x 2 17 x

x x , y 4 x

x , 4 x , x > 0




66.

is in Quadrant II.

cot 1

tan

5

2

sec 1

cos

3 5

5

tan sin

cos

2 5

5

cos

1

sin2

5

3

sin 1

csc

2

3

csc 3

2, cos < 0 67. is in Quadrant II.

cot 55

3

sec 8

55

8 55

55

csc 8

3

tan y

x

3

55

3 55

55

cos x

r

55

8

sin y

r

3

8

y 3, r 8, x 55

sin 3

8, cos < 0 ⇒

68.

is in Quadrant III.

cot

1

tan

4

5

csc 1

sin

41

5

sin 1 cos2 1 16

41

5 41

41

cos 1

sec

4 41

41

sec 1 tan2 1 25

16 41

4

tan 5

4

, cos < 0 69.

cot

x

y 2

21

2 21

21

sec r

x

5

2

5

2

csc r

y

5

21

5 21

21

tan y

x

21

2

sin y

r 21

5

sin > 0 ⇒ is in Quadrant II ⇒ y 21

cos x

r

2

5 ⇒ y2 21

70.

is in Quadrant IV.

cot 1

tan 3

tan sin

cos

3

3

sec 1

cos

2 3

3

cos 1 sin2 1 1

4 3

2

csc 1

sin 2

sin 2

4

1

2, cos > 0 71.

′θ

264°

x

y

264 180 84

264




72.

′θ

635°

x

y

85

635 720 85 73.

′θ x

y

6

5

π −

4

5

5

6

5 2

4

5

6

5 74.

′θ

x

y

17

3

π

3

6

3

17

3

18

3

3

75.

tan

3 3

cos 3 1

2

sin

3

3

2 76.

tan

4

2

22 1

cos 4 2

2

sin

4 2

277.

tan7

3 tan

3 3

cos7 3 cos

3 1

2

sin7

3 sin

3

3

2

78.

2

22 1

tan 5

4 tan

4

cos 5

4 cos

4

2

2

sin 5

4 sin

4

2

2 79.

tan 495 tan 45 1

cos 495 cos 45 2

2

sin 495 sin 45 2

280.

tan150 12

32 3

3

cos150 3

2

sin150 1

2

81.

tan240 tan 60 3

cos240 cos 60 1

2

sin240 sin 60 3

282.

tan315 22

22 1

cos315 2

2

sin315 2

2

83. sin 4

0.7568 84. tan 3

0.1425 85. sin

3.2 0.0584

86. cot4.8 1

tan4.8 0.0878 87. sec12

5 1

cos12

5 3.2361 88. tan25

7 4.3813




89.

2

1

−2

x

− π π 3

22

y

Period: 2

Amplitude: 1

y sin x 91.

Amplitude: 5

Period:

−6

−2

2

4

6

x

π 6

y

2

25 5

f x 5 sin2 x

590.

Amplitude: 1

Period:

x

2

−1

−2

2π π π −

2

y cos x

92.

Amplitude: 8

Period:

−8

−6

−4

8

x

y

π 8π 4

2 14

8

f x 8 cos x

4 93.

Shift the graph of

two units upward.

4

3

2

−1

−2

x

π π π 2

y

−

y sin x

y 2 sin x 94.

Amplitude: 1

Period:

321−2−3 x

−1

−2

−3

−5

−6

−1

2

2

y 4 cos x

96.

Amplitude: 3

Period:

−4

−3

1

2

3

4

t π

y

2

gt 3 cost 97.

(a)

(b)

264 cycles per second.

f 1

1264

y 2 sin528 x

2

b

1

264 ⇒ b 528

a 2,

y a sin bx

98. (a)

0

14

12

22

S t 18.09 1.41 sin t

6 4.60

95.

Amplitude:

Period:

−4

−3

−2

−1

1

3

4

t π

y

2

5

2

gt 5

2 sint

(b)

so this is expected.

(c) Amplitude: 1.41

The amplitude represents the maximum change in the time

of sunset from the average time .d 18.09

12 months 1 year,

Period 2

6 26 12






119. arccos 0.324 1.24 radians118. cos1 3

2

6 120. radiansarccos0.888 2.66

122. radianstan18.2 1.45121. tan1 1.5 0.98 radian

127.

Use a right triangle. Let

then

and cos 45.

tan 34 arctan

34

θ

5

4

3

cosarctan34

45

126.

−1.5 1.5

2

−

2

f x arcsin 2 x 125.

−4 4

2

−

2

f x arctan x

2 tan1 x

2

123.

−1.5 1.5

−

f x 2 arcsin x 2 sin1 x 124.

−1.5 1.5

3

0

y 3 arccos x

128. Let

tanarccos35 tan u

43

4

u

3

5

u arccos35.

129.

Use a right triangle. Let

then and sec 135 .tan

125

arctan125

θ

5

1213

secarctan125

135 130. Let

cot arcsin1213 cot u

512

u

−1213

5u arcsin1213.

131. Let Then

and

2

y

x

4 − x 2

tan y tanarccos x

2 4 x 2

x .cos y

x

2

y arccos x

2. 132.

x −11

θ

12 − ( x − 1)2

sec 1

x 2 x

cos 12 x 12 x 2 x sin

x

1

arcsin x 1 ⇒

2 ≤ ≤

2

secarcsin x 1




133.

arctan70

30 66.8

tan 70

30 134.

h 25 tan 21 9.6 feeth

25

21°

tan 21 h

25

135.

The distance is 1221 miles and the bearing is 85.6.

sec 4.4 D

1217 ⇒ D 1217 sec 4.4 1221

tan 93

1217 ⇒ 4.4

sin 25 d

4

810 ⇒ d

4 342

cos 48 d

3

650 ⇒ d

3 435

cos 25 d

2

810 ⇒ d

2 734

sin 48 d

1

650

⇒ d 1 483

48°

48°

65°25°

d 3d 4

d d 1 2

810650

D

B

C A θ

N

S

WE

d 1 d

2 1217

d 3 d

4 93

136. Amplitude:

Period: 3 seconds

d 0.75 cos2 t

3

b 2

3

a 0.75

d a cos bt

1.5

2 0.75 inches 137. False. The sine or cosine functions

are often useful for modeling

simple harmonic motion.

138. True. The inverse sine,

is defined where

and

2 ≤ y ≤

2.

1 ≤ x ≤ 1

y arcsin x ,

139. False. For each there

corresponds exactly one

value of . y

140. False. The range of arctan is

so arctan1

4.

2,

2,

141.

Matches graph d

Period: 2

Amplitude: 3

y 3 sin x

142. matches graph (a).

Period:

Amplitude: 3

2

y 3 sin x 143.

Matches graph bPeriod: 2

Amplitude: 2

y 2 sin x 144. matches graph (c).

Period:

Amplitude: 2

4

y 2 sin x

2

145. is undefined at the zeros of since sec 1

cos .g cos f sec




Problem Solving for Chapter 4

1. (a)

revolutions

radians or

(b) s r 47.255.5 816.42 feet

990 11

4 2 11

2

132

48

11

4

8:57 6:45 2 hours 12 minutes 132 minutes 2. Gear 1:

Gear 2:

Gear 3:

Gear 4:

Gear 5:24

19

360 454.737 7.94 radians

40

32360 450

5

2 radians

24

22360 392.727 6.85 radians

24

26360 332.308 5.80 radians

24

32360 270

3

2 radians

3. (a)

(b)

x 3000

tan 39 3705 feet

tan 39 3000

x

d 3000

sin 39 4767 feet

sin 39 3000

d (c)

w 3000 tan 63 3705 2183 feet

3000 tan 63 w 3705

tan 63 w 3705

3000

146. (a) (b) tan

2 cot 0.1 0.4 0.7 1.0 1.3

0.27760.64211.18722.36529.9666cot

0.27760.64211.18722.36529.9666tan

2

147. The ranges for the other four trigonometric functions

are not bounded. For the range

is For the range is

, 1 1, . y sec x and y csc x ,, .

y tan x and y cot x ,

148.

(a) A is changed from The displacement is

increased.

(b) k is changed from The friction damps the

oscillations more rapidly.

(c) b is changed from 6 to 9: The frequency of oscilla-

tion is increased.

110 to

13 :

15 to

13 :

y Aekt cos bt 15et 10 cos 6t

149.

(a)

As r increases, the area function increases more rapidly.

0

0 6

4

A s

s r 0.8 0.8r , r > 0

A 12 r 20.8 0.4r 2, r > 0

A 12 r 2 , s r


(b)

0

3

30

As

0

s 10 , > 0

A 12 102 50 , > 0



Problem Solving for Chapter 4 433

5. (a)

h is even.

(b)

h is even.

−1

−2

3

2

h x sin2 x

−1

−2

3

2

h x cos2 x

7. If we alter the model so that we can use

either a sine or a cosine model.

For the cosine model we have:

For the sine model we have:

Notice that we needed the horizontal shift so that the sine

value was one when .

Another model would be:

Here we wanted the sine value to be 1 when t 0.

h 51 50 sin8 t 3

2 t 0

h 51 50 sin8 t

2h 51 50 cos8 t

b 8

d 1

2max min

1

2101 1 51

a 1

2max min

1

2101 1 50

h 1 when t 0, 8.

(a)

(b)

This is the time between heartbeats.

(c) Amplitude: 20

The blood pressure ranges between

and

(d) Pulse rate

(e) Period

64 60

2 b ⇒ b

64

60 2

32

15

60

64

15

16 sec

60 secmin34 secbeat

80 beatsmin

100 20 120.

100 20 80

Period 2

8 3

6

8

3

4 sec

0

70

5

130

P 100 20 cos8 3

t

6. Given: f is an even function and g is an odd function.

(a)

since f is even

Thus, h is an even function.(b)

since g is odd

Thus, h is an even function.

Conjecture: The square of either an even function or an

odd function is an even function.

h x

g x 2

g x 2

h x g x 2

h x g x 2

h x

f x 2

h x f x 2

h x f x 2

4. (a) are all similar triangles since they all have the same angles. is part of

all three triangles and Thus,

(b) Since the triangles are similar, the ratios of corresponding sides are equal.

(c) Since the ratios: it does not matter which triangle is used to calculate sin A.

Any triangle similar to these three triangles could be used to find sin A. The value of sin A would not change.

(d) Since the values of all six trigonometric functions can be found by taking the ratios of the sides of a right triangle,

similar triangles would yield the same values.

opp

hyp

BC

AB

DE

AD

FG

AF sin A

BC

AB

DE

AD

FG

AF

B D F .C E G 90.

A ABC , ADE , and AFG




9. Physical (23 days):

Emotional (28 days):

Intellectual (33 days):

(a)

(b) Number of days since birth until September 1, 2006:

5 11 31 1

20 years leap years remaining August days day in

July days September

All three drop early in the month, then peak toward the middle of the month, and drop again

toward the latter part of the month.

(c) For September 22, 2006, use

I 0.945

E 0.901

P 0.631

t 7369.

−2

7349 7379

2

P

I

E

t 7348

t 365 20

−2

7300 7380

2

P I E

I sin2 t

33, t ≥ 0

E sin2 t

28, t ≥ 0

P sin2 t

23, t ≥ 0

10.

(a)

(b) The period of

The period of

(c) is periodic since the sine

and cosine functions are periodic.

h x A cos x B sin x

g x is .

f x is 2 .

−6

−

6

g

f

g x 2 cos 2 x 3 sin 4 x

f x 2 cos 2 x 3 sin 3 x 11. (a) Both graphs have a period of 2 and intersect when

They should also intersect when

and

(b) The graphs intersect when

(c) Since and

the graphs will intersect again at these values. Therefore

f 13.35 g4.65.

4.65 5.35 5213.35 5.35 42

x 5.35 32 0.65.

x 5.35 2 7.35. x 5.35 2 3.35

x 5.35.



Problem Solving for Chapter 4 435

12. (a) is true since this is a two period

horizontal shift.

(b) is not true.

is a horizontal translation of

is a doubling of the period of

(c) is not true.

is a horizontal

translation of by half a period.

For example, sin1

2 2 sin1

2 .

f 1

2t

f 1

2 t c f 1

2t

1

2 c

f 1

2 t c f 1

2t

f t . f 12

t

f t . f t 1

2c

f t 1

2c f 1

2t

f t 2c f t 13.

(a)

(b)

(c) feet

(d) As you more closer to the rock, decreases, which

causes y to decrease, which in turn causes d todecrease.

1

d y x 3.46 1.71 1.75

tan 1

y

2 ⇒ y 2 tan 60 3.46 feet

tan 2

x

2 ⇒ x 2 tan 40.52 1.71 feet

2 40.52

sin 2

sin 1

1.333

sin 601.333

0.6497

sin 1

sin 2 1.333

2 ft

x y

d

θ

θ 1

2

14.

(a)

The graphs are nearly the same for 1 < x < 1.

−2

2

2−

2

arctan x x x 3

3

x 5

5

x 7

7

(b)

The accuracy of the approximation improved slightly by

adding the next term x 99.

−2

2

2−

2



Chapter 4 Practice Test


1. Express 350° in radian measure. 2. Express in degree measure.5 9

3. Convert to decimal form.135 14 12 4. Convert form.22.569 to D M S

5. If use the trigonometric identities to find tan . cos 23, 6. Find given .sin 0.9063

7. Solve for in the figure below.

20°

35

x

x 8. Find the reference angle for . 6 5

9. Evaluate .csc 3.92 10. Find .sec given that lies in Quadrant III and tan 6

11. Graph y 3 sin x

2. 12. Graph y 2 cos x .

13. Graph y tan 2 x . 14. Graph . y csc x

4

15. Graph using a graphing calculator. y 2 x sin x , 16. Graph using a graphing calculator. y 3 x cos x ,

17. Evaluate .arcsin 1 18. Evaluate arctan3.

19. Evaluate sinarccos

4

35. 20. Write an algebraic expression for .cosarcsin

x

4For Exercises 21–23, solve the right triangle.

B

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