Solutions to Exercises, Section 5park/Fall2013/precalculus/5.1sol.pdfInstructor’s Solutions Manual, Section 5.1 Exercise 1 Solutions to Exercises, Section 5.1 1. Find all numbers

Instructor’s Solutions Manual, Section 5.1 Exercise 1

Solutions to Exercises, Section 5.1

1. Find all numbers t such that (13 , t) is a point on the unit circle.

solution For (13 , t) to be a point on the unit circle means that the sum

of the squares of the coordinates equals 1. In other words,

(13

)2 + t2 = 1.

This simplifies to the equation t2 = 89 , which implies that t =

√8

3 or

t = −√

83 . Because

√8 = √4 · 2 = √4 · √2 = 2

√2, we can rewrite this as

t = 2√

23 or t = −2

√2

3 .


2. Find all numbers t such that (35 , t) is a point on the unit circle.

solution For (35 , t) to be a point on the unit circle means that the sum

of the squares of the coordinates equals 1. In other words,

(35

)2 + t2 = 1.

This simplifies to the equation t2 = 1625 , which implies that t = 4

5 ort = −4

5 .


3. Find all numbers t such that (t,−25) is a point on the unit circle.

solution For (t,−25) to be a point on the unit circle means that the

sum of the squares of the coordinates equals 1. In other words,

t2 + (−25

)2 = 1.


√215 or

t = −√

215 .


4. Find all numbers t such that (t,−37) is a point on the unit circle.

solution For (t,−37) to be a point on the unit circle means that the

sum of the squares of the coordinates equals 1. In other words,

t2 + (−37

)2 = 1.


√407 or

t = −√

407 . Because

√40 = √4 · 10 = √4 · √10 = 2

√10, we can rewrite

this as t = 2√

107 or t = −2

√10

7 .


5. Find the points where the line through the origin with slope 3 intersectsthe unit circle.

solution The line through the origin with slope 3 is characterized bythe equation y = 3x. Substituting this value for y into the equation forthe unit circle (x2 +y2 = 1) gives

x2 + (3x)2 = 1,

which simplifies to the equation 10x2 = 1. Thus x =√

1010 or x = −

√10

10 .Using each of these values of x along with the equation y = 3x givesthe points

(√1010 ,

3√

1010

)and

(−√1010 ,−3

√10

10

)as the points of intersection

of the line y = 3x and the unit circle.


6. Find the points where the line through the origin with slope 4 intersectsthe unit circle.

solution The line through the origin with slope 4 is characterized bythe equation y = 4x. Substituting this value for y into the equation forthe unit circle (x2 +y2 = 1) gives

x2 + (4x)2 = 1,

which simplifies to the equation 17x2 = 1. Thus x =√

1717 or x = −

√17

17 .Using each of these values of x along with the equation y = 4x givesthe points

(√1717 ,

4√

1717

)and

(−√1717 ,−4

√17

17

)as the points of intersection

of the line y = 4x and the unit circle.


7. Suppose an ant walks counterclockwise on the unit circle from thepoint (1,0) to the endpoint of the radius that forms an angle of 70◦

with the positive horizontal axis. How far has the ant walked?

solution We need to find the length of the circular arc on the unitcircle corresponding to a 70◦ angle. This length equals 70π

180 , whichequals 7π

18 .


8. Suppose an ant walks counterclockwise on the unit circle from thepoint (1,0) to the endpoint of the radius that forms an angle of 130◦

with the positive horizontal axis. How far has the ant walked?

solution We need to find the length of the circular arc on the unitcircle corresponding to a 130◦ angle. This length equals 130π

180 , whichequals 13π

18 .


9. What angle corresponds to a circular arc on the unit circle withlength π

5 ?

solution Let θ be such that the angle of θ degrees corresponds to anarc on the unit circle with length π

5 . Thus θπ180 = π

5 . Solving thisequation for θ, we get θ = 36. Thus the angle in question is 36◦.


10. What angle corresponds to a circular arc on the unit circle withlength π

6 ?

solution Let θ be such that the angle of θ degrees corresponds to anarc on the unit circle with length π

6 . Thus θπ180 = π

6 . Solving thisequation for θ, we get θ = 30. Thus the angle in question is 30◦.


11. What angle corresponds to a circular arc on the unit circle withlength 5

2?

solution Let θ be such that the angle of θ degrees corresponds to anarc on the unit circle with length 5

2 . Thus θπ180 = 5

2 . Solving this equation

for θ, we get θ = 450π . Thus the angle in question is 450

π◦, which is

approximately equal to 143.2◦.


12. What angle corresponds to a circular arc on the unit circle withlength 1?

solution Let θ be such that the angle of θ degrees corresponds to anarc on the unit circle with length 1. Thus θπ

180 = 1. Solving this equation

for θ, we get θ = 180π . Thus the angle in question is 180

π◦, which is

approximately equal to 57.3◦.


13. Find the lengths of both circular arcs on the unit circle connecting thepoints (1,0) and

(√22 ,

√2

2

).

solution The radius of the unit circle ending at the point(√2

2 ,√

22

)makes an angle of 45◦ with the positive horizontal axis. One of thecircular arcs connecting (1,0) and

(√22 ,

√2

2

)is shown below as the

thickened circular arc; the other circular arc connecting (1,0) and(√22 ,

√2

2

)is the unthickened part of the unit circle below.

The length of the thickened arc below is 45π180 , which equals π

4 . Theentire unit circle has length 2π . Thus the length of the other circulararc below is 2π − π

4 , which equals 7π4 .

45�

1

The thickened circular arc has length π4 .

The other circular arc has length 7π4 .


14. Find the lengths of both circular arcs on the unit circle connecting thepoints (1,0) and

(−√22 ,

√2

2

).

solution The radius of the unit circle ending at the point(−√2

2 ,√

22

)makes an angle of 225◦ with the positive horizontal axis. One of thecircular arcs connecting (1,0) and

(−√22 ,

√2

2

)is shown below as the

thickened circular arc; the other circular arc connecting (1,0) and(−√22 ,

√2

2

)is the unthickened part of the unit circle below.

The length of the thickened arc below is 225π180 , which equals 5π

4 . Theentire unit circle has length 2π . Thus the length of the other circulararc below is 2π − 5π


225�

1

The thickened circular arc has length 5π4 .



For each of the angles in Exercises 15–20, find the endpoint of the radiusof the unit circle that makes the given angle with the positive horizontalaxis.

15. 120◦

solution The radius making a 120◦ angle with the positive horizontalaxis is shown below. The angle from this radius to the negativehorizontal axis equals 180◦ − 120◦, which equals 60◦ as shown in thefigure below. Drop a perpendicular line segment from the endpoint ofthe radius to the horizontal axis, forming a right triangle as shownbelow. We already know that one angle of this right triangle is 60◦; thusthe other angle must be 30◦, as labeled below:

120�

60�

30�

1

The side of the right triangle opposite the 30◦ angle has length 12 ; the

side of the right triangle opposite the 60◦ angle has length√

32 . Looking

at the figure above, we see that the first coordinate of the endpoint of


the radius is the negative of the length of the side opposite the 30◦

angle, and the second coordinate of the endpoint of the radius is thelength of the side opposite the 60◦ angle. Thus the endpoint of theradius is

(−12 ,√

32

).


16. 240◦

solution The radius making a 240◦ angle with the positive horizontalaxis is shown below. The angle from this radius to the negative verticalaxis equals 270◦ − 240◦, which equals 30◦ as shown in the figure below.Draw a horizontal line segment from the endpoint of the radius to thevertical axis, forming a right triangle as shown below. We already knowthat one angle of this right triangle is 30◦; thus the other angle must be60◦, as labeled below:

240�

30�

60�

1



32 . Looking

at the figure above, we see that the first coordinate of the endpoint ofthe radius is the negative of the length of the side opposite the 30◦

angle, and the second coordinate of the endpoint of the radius is the


negative of the length of the side opposite the 60◦ angle. Thus theendpoint of the radius is

(−12 ,−

√3

2

).


17. −30◦

solution The radius making a −30◦ angle with the positivehorizontal axis is shown below. Draw a perpendicular line segmentfrom the endpoint of the radius to the horizontal axis, forming a righttriangle as shown below. We already know that one angle of this righttriangle is 30◦; thus the other angle must be 60◦, as labeled below.



32 . Looking

at the figure below, we see that the first coordinate of the endpoint ofthe radius is the length of the side opposite the 60◦ angle, and thesecond coordinate of the endpoint of the radius is the negative of thelength of the side opposite the 30◦ angle. Thus the endpoint of theradius is

(√32 ,−1

2

).

60�

30� 1


18. −150◦

solution The radius making a −150◦ angle with the positivehorizontal axis is shown below. Draw a perpendicular line segmentfrom the endpoint of the radius to the horizontal axis, forming a righttriangle as shown below. Because 180− 150 = 30, one angle of thistriangle equals 30◦, as shown below; thus the other angle must be 60◦,as also shown below.



32 . Looking

at the figure below, we see that the first coordinate of the endpoint ofthe radius is the negative of the length of the side opposite the 60◦

angle, and the second coordinate of the endpoint of the radius is thenegative of the length of the side opposite the 30◦ angle. Thus theendpoint of the radius is

(−√32 ,−1

2

).

60�

30�1


19. 390◦

solution The radius making a 390◦ angle with the positive horizontalaxis is obtained by starting at the horizontal axis, making one completecounterclockwise rotation, and then continuing for another 30◦. Theresulting radius is shown below. Drop a perpendicular line segmentfrom the endpoint of the radius to the horizontal axis, forming a righttriangle as shown below. We already know that one angle of this righttriangle is 30◦; thus the other angle must be 60◦, as labeled below.


side opposite the 60◦ angle has length√

32 . Looking at the figure below,

we see that the first coordinate of the endpoint of the radius is thelength of the side opposite the 60◦ angle, and the second coordinate ofthe endpoint of the radius is the length of the side opposite the 30◦

angle. Thus the endpoint of the radius is(√3

2 ,12

).

60�

30�

1


20. 510◦

solution The radius making a 510◦ angle with the positive horizontalaxis is obtained by starting at the horizontal axis, making one completecounterclockwise rotation, and then continuing for another 150◦

(because 510− 360 = 150). The resulting radius is shown below. Drop aperpendicular line segment from the endpoint of the radius to thehorizontal axis, forming a right triangle as shown below. One angle ofthis right triangle is 30◦ (because 180− 150 = 30), as shown below;thus the other angle must be 60◦, as also shown below.


side opposite the 60◦ angle has length√

32 . Looking at the figure below,

we see that the first coordinate of the endpoint of the radius is thenegative of the length of the side opposite the 60◦ angle, and thesecond coordinate of the endpoint of the radius is the length of the sideopposite the 30◦ angle. Thus the endpoint of the radius is

(−√32 ,

12

).


510�

30�

60�

1


For Exercises 21–26, find the angle the radius of the unit circle ending atthe given point makes with the positive horizontal axis. Among theinfinitely many possible correct solutions, choose the one with thesmallest absolute value.

21.(−1

2 ,√

32

)solution Draw the radius whose endpoint is

(−12 ,√

32

). Drop a

perpendicular line segment from the endpoint of the radius to thehorizontal axis, forming a right triangle. The hypotenuse of this righttriangle is a radius of the unit circle and thus has length 1. Thehorizontal side has length 1

2 and the vertical side of this triangle has

length√

32 because the endpoint of the radius is

(−12 ,√

32

).

120�

60�

30�

1

Thus we have a 30◦- 60◦- 90◦ triangle, with the 30◦ angle opposite thehorizontal side of length 1

2 , as labeled above. Because 180− 60 = 120,


the radius makes a 120◦ angle with the positive horizontal axis, asshown above.

In addition to making a 120◦ angle with the positive horizontal axis,this radius also makes with the positive horizontal axis angles of 480◦,840◦, and so on. This radius also makes with the positive horizontalaxis angles of −240◦, −600◦, and so on. But of all the possible choicesfor this angle, the one with the smallest absolute value is 120◦.


22.(−√3

2 ,12


(−√32 ,

12

). Drop a

perpendicular line segment from the endpoint of the radius to thehorizontal axis, forming a right triangle. The hypotenuse of this righttriangle is a radius of the unit circle and thus has length 1. Thehorizontal side has length

√3


length 12 because the endpoint of the radius is

(−√32 ,

12

).

150�

30�

60�

1

Thus we have here a 30◦- 60◦- 90◦ triangle, with the 30◦ angle oppositethe side of length 1

2 (the vertical side), as labeled above. Because180− 30 = 150, the radius makes a 150◦ angle with the positivehorizontal axis, as shown above.

In addition to making a 150◦ angle with the positive horizontal axis,this radius also makes with the positive horizontal axis angles of 510◦,870◦, and so on. This radius also makes with the positive horizontal


axis angles of −210◦, −570◦, and so on. But of all the possible choicesfor this angle, the one with the smallest absolute value is 150◦.


23.(√2

2 ,−√

22


(√22 ,−

√2

2

). Draw a


√2

2 and the vertical side of this triangle also

has length√


(√22 ,−

√2

2

).

45�

45� 1

Thus we have here an isosceles right triangle, with two angles of 45◦ aslabeled above.

In addition to making a −45◦ angle with the positive horizontal axis,this radius also makes with the positive horizontal axis angles of 315◦,675◦, and so on. This radius also makes with the positive horizontalaxis angles of −405◦, −765◦, and so on. But of all the possible choicesfor this angle, the one with the smallest absolute value is −45◦.


24.(1

2 ,−√

32


(12 ,−

√3

2

). Draw a

perpendicular line segment from the endpoint of the radius to thevertical axis, forming a right triangle. The hypotenuse of this righttriangle is a radius of the unit circle and thus has length 1. Thehorizontal side has length 1


length√


(12 ,−√3

2

).

�60�

60�

30�

1


2 (the horizontal side), as labeled above. Because90− 30 = 60, the radius makes a −60◦ angle with the positivehorizontal axis, as shown above.

In addition to making a −60◦ angle with the positive horizontal axis,this radius also makes with the positive horizontal axis angles of 300◦,660◦, and so on. This radius also makes with the positive horizontal


axis angles of −420◦, −780◦, and so on. But of all the possible choicesfor this angle, the one with the smallest absolute value is −60◦.


25.(−√2

2 ,−√

22


(−√22 ,−

√2

2

). Draw a


√2

2 and the vertical side of this triangle also

has length√


(−√22 ,−

√2

2

).

45�

45� �135�

1

Thus we have here an isosceles right triangle, with two angles of 45◦ aslabeled above. Because the radius makes a 45◦ angle with the negativehorizontal axis, it makes a −135◦ angle with the positive horizontalaxis, as shown below (because 135◦ = 180◦ − 45◦).





26.(−1

2 ,−√

32


(−12 ,−

√3

2

). Draw a

perpendicular line segment from the endpoint of the radius to thehorizontal axis, forming a right triangle. The hypotenuse of this righttriangle is a radius of the unit circle and thus has length 1. Thehorizontal side has length 1


length√


(−12 ,−√3

2

).

�120�

60�

30�

1


2 (the horizontal side), as labeled above. Because180− 60 = 120, the radius makes a −120◦ angle with the positivehorizontal axis, as shown above.





27. Find the lengths of both circular arcs on the unit circle connecting thepoint

(12 ,√

32

)and the point that makes an angle of 130◦ with the

positive horizontal axis.

solution The radius of the unit circle ending at the point(1

2 ,√

32

)makes an angle of 60◦ with the positive horizontal axis. One of thecircular arcs connecting

(12 ,√

32

)and the point that makes an angle of

130◦ with the positive horizontal axis is shown below as the thickenedcircular arc; the other circular arc connecting these two points is theunthickened part of the unit circle below.

The thickened arc below corresponds to an angle of 70◦ (because70◦ = 130◦ − 60◦). Thus the length of the thickened arc below is 70π

180 ,which equals 7π

18 . The entire unit circle has length 2π . Thus the lengthof the other circular arc below is 2π − 7π


60�130�

1





(√32 ,−1

2



solution The radius of the unit circle ending at the point(√3

2 ,−12

)makes an angle of −30◦ with the positive horizontal axis. One of thecircular arcs connecting

(√32 ,−1

2

)and the point that makes an angle of

50◦ with the positive horizontal axis is shown below as the thickenedcircular arc; the other circular arc connecting these two points is theunthickened part of the unit circle below.

The thickened arc below corresponds to an angle of 80◦ (because80◦ = 30◦ + 50◦). Thus the length of the thickened arc below is 80π




�30�

50�

1





(−√22 ,−

√2

2



solution The radius of the unit circle ending at(−√2

2 ,−√

22

)makes a

225◦ angle with the positive horizontal axis (because225◦ = 180◦ + 45◦). One of the circular arcs connecting

(−√22 ,−

√2

2

)and

the point that makes an angle of 125◦ with the positive horizontal axisis shown below as the thickened circular arc; the other circular arcconnecting these two points is the unthickened part of the unit circlebelow.

The thickened arc below corresponds to an angle of 100◦ (because100◦ = 225◦ − 125◦). Thus the length of the thickened arc below is100π180 , which equals 5π

9 . The entire unit circle has length 2π . Thus thelength of the other circular arc below is 2π − 5π


125�225�

1





(−√32 ,−1

2



solution The radius of the unit circle ending at(−√3

2 ,−12

)makes a

−150◦ angle with the positive horizontal axis. One of the circular arcsconnecting

(−√32 ,−1

2

)and the point that makes an angle of 20◦ with

the positive horizontal axis is shown below as the thickened circulararc; the other circular arc connecting these two points is theunthickened part of the unit circle below.

The thickened arc below corresponds to an angle of 170◦ (because170◦ = 150◦ + 20◦). Thus the length of the thickened arc below is 170π




�150�

20�

1




31. What is the slope of the radius of the unit circle that has a 30◦ anglewith the positive horizontal axis?

solution The radius of the unit circle that has a 30◦ angle with thepositive horizontal axis has its initial point at (0,0) and its endpoint at(√3

2 ,12

). Thus the slope of this radius is

12 − 0√

32 − 0

,

which equals 1√3, which equals

√3

3 .


32. What is the slope of the radius of the unit circle that has a 60◦ anglewith the positive horizontal axis?

solution The radius of the unit circle that has a 60◦ angle with thepositive horizontal axis has its initial point at (0,0) and its endpoint at(1

2 ,√

32

). Thus the slope of this radius is

√3

2 − 012 − 0

,

which equals√

3.

Instructor’s Solutions Manual, Section 5.1 Problem 33

Solutions to Problems, Section 5.1

For each of the angles listed in Problems 33–40, sketch the unit circleand the radius that makes the indicated angle with the positivehorizontal axis. Be sure to include an arrow to show the direction inwhich the angle is measured from the positive horizontal axis.

33. 20◦

solution

20�

1


34. 80◦

solution

80�

1


35. 160◦

solution

160�

1


36. 330◦

solution

330�

1


37. 460◦

solution

460�

1


38. −10◦

solution

�10� 1


39. −75◦

solution

�75� 1


40. −170◦

solution

�170� 1


41. Find the formula for the length of a circular arc corresponding to anangle of θ degrees on a circle of radius r .

solution Suppose 0 < θ ≤ 360 and consider a circular arc on a unitcircle of radius r corresponding to an angle of θ degrees, as shown inthe thickened part of the unit circle below:

Θ

r

The length of this circular arc equals the fraction of the entire circletaken up by this circular arc times the circumference of the entirecircle. In other words, the length of this circular arc equals θ

360 times

2πr , which equals θπr180 .

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Solutions to Exercises, Section 5park/Fall2013/precalculus/5.1sol.pdfInstructor’s Solutions Manual, Section 5.1 Exercise 1 Solutions to Exercises, Section 5.1 1. Find all numbers