Solutions to Exercises, Section 5 - Department of …park/Fall2014/precalculus/5.3sol.pdf · Instructor’s Solutions Manual, Section 5.3 Exercise 1 Solutions to Exercises, Section

Instructor’s Solutions Manual, Section 5.3 Exercise 1

Solutions to Exercises, Section 5.3

Give exact values for the quantities in Exercises 1–10. Do not use acalculator for any of these exercises—otherwise you will likely getdecimal approximations for some solutions rather than exact answers.More importantly, good understanding will come from working theseexercises by hand.

1. (a) cos 3π (b) sin 3π

solution Because 3π = 2π +π , an angle of 3π radians (as measuredcounterclockwise from the positive horizontal axis) consists of acomplete revolution around the circle (2π radians) followed by anotherπ radians (180◦), as shown below. The endpoint of the correspondingradius is (−1,0). Thus cos 3π = −1 and sin 3π = 0.

1


2. (a) cos(−3π2 ) (b) sin(−3π

2 )

solution Because −3π2 = −π − π

2 , an angle of −3π2 radians is obtained

by moving clockwise from the positive horizontal axis by half the circle(−π radians or −180◦) and then continuing clockwise for anotherone-fourth of the circle (−π2 radians or −90◦), as shown below. Theendpoint of the corresponding radius is (0,1). Thus cos(−3π

2 ) = 0 andsin(−3π

2 ) = 1.


1


3. (a) cos 11π4 (b) sin 11π

4

solution Because 11π4 = 2π + π

2 + π4 , an angle of 11π

4 radians (asmeasured counterclockwise from the positive horizontal axis) consistsof a complete revolution around the circle (2π radians) followed byanother π

2 radians (90◦), followed by another π4 radians (45◦), as shown

below. Hence the endpoint of the corresponding radius is(−√2

2 ,√

22

).

Thus cos 11π4 = −

√2

2 and sin 11π4 =

√2

2 .

1


4. (a) cos 15π4 (b) sin 15π

4

solution Because 15π4 = 2π +π + π


4 radians (asmeasured counterclockwise from the positive horizontal axis) consistsof a complete revolution around the circle (2π radians) followed by ahalf revolution around the circle (π radians), followed by another π

2radians (90◦), followed by another π

4 radians (45◦), as shown below. The

endpoint of the corresponding radius is(√2

2 ,−√

22

). Thus cos 15π

4 =√

22

and sin 15π4 = −

√2

2 .

1


5. (a) cos 2π3 (b) sin 2π

3

solution Because 2π3 = π


3 radians (as measuredcounterclockwise from the positive horizontal axis) consists of π

2radians (90◦ radians) followed by another π

6 radians (30◦), as shown

below. The endpoint of the corresponding radius is(−1

2 ,√

32

). Thus

cos 2π3 = −1

2 and sin 2π3 =

√3

2 .

1


6. (a) cos 4π3 (b) sin 4π

3

solution Because 4π3 = π + π

3 , an angle of 4π3 radians (as measured

counterclockwise from the positive horizontal axis) consists of πradians (180◦ radians) followed by another π

3 radians (60◦), as shown

below. The endpoint of the corresponding radius is(−1

2 ,−√

32

). Thus

cos 4π3 = −1

2 and sin 4π3 = −

√3

2 .

1


7. (a) cos 210◦ (b) sin 210◦

solution Because 210 = 180+ 30, an angle of 210◦ (as measuredcounterclockwise from the positive horizontal axis) consists of 180◦

followed by another 30◦, as shown below. The endpoint of thecorresponding radius is

(−√32 ,−1

2

). Thus cos 210◦ = −

√3

2 andsin 210◦ = −1

2 .

1


8. (a) cos 300◦ (b) sin 300◦

solution Because 300 = 270+ 30, an angle of 300◦ (as measuredcounterclockwise from the positive horizontal axis) consists of 270◦

followed by another 30◦, as shown below. The endpoint of thecorresponding radius is

(12 ,−

√3

2

). Thus cos 300◦ = 1

2 and

sin 300◦ = −√

32 .

1


9. (a) cos 360045◦ (b) sin 360045◦

solution Because 360045 = 360× 1000+ 45, an angle of 360045◦ (asmeasured counterclockwise from the positive horizontal axis) consistsof 1000 complete revolutions around the circle followed by another45◦. The endpoint of the corresponding radius is

(√22 ,

√2

2

). Thus

cos 360045◦ =√

22 and sin 360045◦ =

√2

2 .


10. (a) cos(−360030◦) (b) sin(−360030◦)

solution Because

−360030 = 360× (−1000)− 30,

an angle of −360030◦ consists of 1000 complete clockwise revolutionsaround the circle followed by another 30◦ in the clockwise direction.The endpoint of the corresponding radius is

(√32 ,−1

2

). Thus

cos(−360030◦) =√

32 and sin(−360030◦) = −1

2 .


11. Find the smallest number θ larger than 4π such that cosθ = 0.

solution Note that

0 = cos π2 = cos 3π2 = cos 5π

2 = . . .

and that the only numbers whose cosine equals 0 are of the form(2n+1)π

2 , where n is an integer. The smallest number of this form largerthan 4π is 9π

2 . Thus 9π2 is the smallest number larger than 4π whose

cosine equals 0.


12. Find the smallest number θ larger than 6π such that sinθ =√

22 .

solution An angle of 6π corresponds to three completecounterclockwise revolutions around the unit circle. To reach an anglewhose sine equals

√2

2 requires an additional angle of π4 . Thus 6π + π

4 ,which equals 25π

4 , is the smallest number larger than 6π whose sine

equals√

22 .


13. Find the four smallest positive numbers θ such that cosθ = 0.

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, the first coordinate of its endpoint firstbecomes 0 when θ equals π

2 (which equals 90◦), then again when θequals 3π

2 (which equals 270◦), then again when θ equals 5π2 (which

equals 360◦ + 90◦, or 450◦), then again when θ equals 7π2 (which equals

360◦ + 270◦, or 630◦), and so on. Thus the four smallest positivenumbers θ such that cosθ = 0 are π

2 , 3π2 , 5π

2 , and 7π2 .


14. Find the four smallest positive numbers θ such that sinθ = 0.

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, the second coordinate of its endpoint firstbecomes 0 after starting its travel when θ equals π (which equals 180◦),then again when θ equals 2π (which equals 360◦), then again when θequals 3π (which equals 360◦ + 180◦, or 540◦), then again when θequals 4π (which equals 720◦), and so on. Thus the four smallestpositive numbers θ such that sinθ = 0 are π , 2π , 3π , and 4π .


15. Find the four smallest positive numbers θ such that sinθ = 1.

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, then the second coordinate of its endpoint firstbecomes 1 when θ equals π


2 (which equals 360◦ + 90◦, or 450◦), then again when θ equals9π2 (which equals 2× 360◦ + 90◦, or 810◦), then again when θ equals

13π2 (which equals 3× 360◦ + 90◦, or 1170◦), and so on. Thus the four

smallest positive numbers θ such that sinθ = 1 are π2 , 5π

2 , 9π2 , and 13π

2 .


16. Find the four smallest positive numbers θ such that cosθ = 1.

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, the first coordinate of its endpoint firstbecomes 1 after starting its travel when θ equals 2π (which equals360◦), then again when θ equals 4π (which equals 720◦), then againwhen θ equals 6π , then again when θ equals 8π , and so on. Thus thefour smallest positive numbers θ such that cosθ = 1 are 2π , 4π , 6π ,and 8π .


17. Find the four smallest positive numbers θ such that cosθ = −1.

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, the first coordinate of its endpoint firstbecomes −1 when θ equals π (which equals 180◦), then again when θequals 3π (which equals 360◦ + 180◦, or 540◦), then again when θequals 5π (which equals 2× 360◦ + 180◦, or 900◦), then again when θequals 7π (which equals 3× 360◦ + 180◦, or 1260◦), and so on. Thusthe four smallest positive numbers θ such that cosθ = −1 are π , 3π ,5π , and 7π .


18. Find the four smallest positive numbers θ such that sinθ = −1.

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, the second coordinate of its endpoint firstbecomes −1 when θ equals 3π


2 (which equals 360◦ +270◦, or 630◦), then again when θ equals11π

2 , then again when θ equals 15π2 , and so on. Thus the four smallest

positive numbers θ such that sinθ = −1 are 3π2 , 7π

2 , 11π2 , and 15π

2 .


19. Find the four smallest positive numbers θ such that sinθ = 12 .

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, the second coordinate of its endpoint firstbecomes 1

2 when θ equals π6 (which equals 30◦), then again when θ

equals 5π6 (which equals 150◦), then again when θ equals 13π

6 (whichequals 360◦ + 30◦, or 390◦), then again when θ equals 17π

6 (whichequals 360◦ + 150◦, or 510◦), and so on. Thus the four smallest positivenumbers θ such that sinθ = 1

2 are π6 , 5π

6 , 13π6 , and 17π

6 .


20. Find the four smallest positive numbers θ such that cosθ = 12 .

solution Think of a radius of the unit circle whose endpoint is (1,0).If this radius moves counterclockwise, forming an angle of θ with thepositive horizontal axis, the first coordinate of its endpoint firstbecomes 1

2 when θ equals π3 (which equals 60◦), then again when θ

equals 5π3 (which equals 300◦), then again when θ equals 7π

3 (whichequals 360◦ + 60◦, or 420◦), then again when θ equals 11π

3 (whichequals 360◦ + 300◦, or 660◦), and so on. Thus the four smallest positivenumbers θ such that cosθ = 1

2 are π3 , 5π

3 , 7π3 , and 11π

3 .


21. Suppose 0 < θ < π2 and cosθ = 2

5 . Evaluate sinθ.

solution We know that

(cosθ)2 + (sinθ)2 = 1.

Thus

(sinθ)2 = 1− (cosθ)2

= 1−(2

5

)2

= 2125.

Because 0 < θ < π2 , we know that sinθ > 0. Thus taking square roots of

both sides of the equation above gives

sinθ =√

215.


22. Suppose 0 < θ < π2 and sinθ = 3

7 . Evaluate cosθ.


(cosθ)2 + (sinθ)2 = 1.

Thus

(cosθ)2 = 1− (sinθ)2

= 1−(3

7

)2

= 4049.

Because 0 < θ < π2 , we know that cosθ > 0. Thus taking square roots

of both sides of the equation above gives

cosθ =√

407

= 2√

107

.


23. Suppose π2 < θ < π and sinθ = 2

9 . Evaluate cosθ.


(cosθ)2 + (sinθ)2 = 1.

Thus


= 1−(2

9

)2

= 7781.

Because π2 < θ < π , we know that cosθ < 0. Thus taking square roots


cosθ = −√

779.


24. Suppose π2 < θ < π and sinθ = 3

8 . Evaluate cosθ.


(cosθ)2 + (sinθ)2 = 1.

Thus


= 1−(3

8

)2

= 5564.

Because π2 < θ < π , we know that cosθ < 0. Thus taking square roots


cosθ = −√

558.


25. Suppose −π2 < θ < 0 and cosθ = 0.1. Evaluate sinθ.


(cosθ)2 + (sinθ)2 = 1.

Thus


= 1− (0.1)2

= 0.99.

Because −π2 < θ < 0, we know that sinθ < 0. Thus taking square rootsof both sides of the equation above gives

sinθ = −√0.99 ≈ −0.995.


26. Suppose −π2 < θ < 0 and cosθ = 0.3. Evaluate sinθ.


(cosθ)2 + (sinθ)2 = 1.

Thus


= 1− (0.3)2

= 0.91.

Because −π2 < θ < 0, we know that sinθ < 0. Thus taking square rootsof both sides of the equation above gives

sinθ = −√0.91 ≈ −0.954.


27. Find the smallest number x such that sin(ex) = 0.

solution Note that ex is an increasing function. Because ex ispositive for every real number x, and because π is the smallest positivenumber whose sine equals 0, we want to choose x so that ex = π . Thusx = lnπ .


28. Find the smallest number x such that cos(ex + 1) = 0.

solution Note that ex is an increasing function. Because ex ispositive for every real number x, and because π

2 is the smallest positivenumber whose cosine equals 0, we want to choose x so that ex + 1 = π

2 .Thus ex = π

2 − 1, which implies that x = ln(π2 − 1).


29. Find the smallest positive number x such that

sin(x2 + x + 4) = 0.

solution Note that x2 + x + 4 is an increasing function on theinterval [0,∞). If x is positive, then x2 + x + 4 > 4. Because 4 is largerthan π but less than 2π , the smallest number bigger than 4 whose sineequals 0 is 2π . Thus we want to choose x so that x2 + x + 4 = 2π . Inother words, we need to solve the equation

x2 + x + (4− 2π) = 0.

Using the quadratic formula, we see that the solutions to this equationare

x = −1±√8π − 152

.

A calculator shows that choosing the plus sign in the equation abovegives x ≈ 1.0916 and choosing the minus sign gives x ≈ −2.0916. Weseek only positive values of x, and thus we choose the plus sign in theequation above, getting x ≈ 1.0916.


30. Find the smallest positive number x such that

cos(x2 + 2x + 6) = 0.

solution Note that x2 + 2x + 6 is an increasing function on theinterval [0,∞). If x is positive, then x2 + 2x + 6 > 6. Because 6 is a bitless than 2π , the smallest number bigger than 6 whose cosine equals 0is 2π + π

2 , which equals 5π2 . Thus we want to choose x so that

x2 + 2x + 6 = 5π2 . In other words, we need to solve the equation

x2 + 2x + (6− 5π2 ) = 0.

Using the quadratic formula, we see that the solutions to this equationare

x = −2±√10π − 202

.

A calculator shows that choosing the plus sign in the equation abovegives x ≈ 0.6894 and choosing the minus sign gives x ≈ −2.6894. Weseek only positive values of x, and thus we choose the plus sign in theequation above, getting x ≈ 0.6894.

Instructor’s Solutions Manual, Section 5.3 Problem 31

Solutions to Problems, Section 5.3

31. (a) Sketch a radius of the unit circle making an angle θ with thepositive horizontal axis such that cosθ = 6

7 .

(b) Sketch another radius, different from the one in part (a), alsoillustrating cosθ = 6

7 .

solution

(a) Because cosθ is the first coordinate of the radius of the unit circlecorresponding to the angle θ, we need to find a radius of the unit circlewhose first coordinate equals 6

7 . To do this, start with the pointcorresponding to 6

7 on the horizontal axis and then move vertically upto find a point on the unit circle whose first coordinate equals 6

7 . Thendraw the radius from the origin to that point on the unit circle:

Θ6

71

An angle θ such that cosθ = 67 .


(b) To find another radius, different from the one in part (a), alsoillustrating cosθ = 6

7 , start with the point corresponding to 67 on the

horizontal axis and then move vertically down to find a point on theunit circle whose first coordinate equals 6

7 . Then draw the radius fromthe origin to that point on the unit circle:

Θ 6

71

Another angle θ such that cosθ = 67 .


32. (a) Sketch a radius of the unit circle making an angle θ with thepositive horizontal axis such that sinθ = −0.8.

(b) Sketch another radius, different from the one in part (a), alsoillustrating sinθ = −0.8.

solution

(a) Because sinθ is the second coordinate of the radius of the unit circlecorresponding to the angle θ, we need to find a radius of the unit circlewhose second coordinate equals −0.8. To do this, start with the pointcorresponding to −0.8 on the vertical axis and then move horizontallyright to find a point on the unit circle whose second coordinate equals−0.8. Then draw the radius from the origin to that point on the unitcircle:

Θ 1

�0.8

An angle θ such that sinθ = −0.8.


(b) To find another radius, different from the one in part (a), alsoillustrating sinθ = −0.8, start with the point corresponding to −0.8 onthe vertical axis and then move horizontally left to find a point on theunit circle whose second coordinate equals −0.8. Then draw the radiusfrom the origin to that point on the unit circle:

Θ1

�0.8

Another angle θ such that sinθ = −0.8.


33. Find angles u and ν such that cosu = cosν but sinu �= sinν .

solution One easy choice is to take u = π2 and ν = −π2 . We have

cos π2 = cos(−π2 ) = 0

butsin π

2 = 1 �= −1 = sin(−π2 ).


34. Find angles u and ν such that sinu = sinν but cosu �= cosν .

solution One easy choice is to take u = 0 and ν = π . We have

sin 0 = sinπ = 0

butcos 0 = 1 �= −1 = cosπ.


35. Suppose you have borrowed two calculators from friends, but you donot know whether they are set to work in radians or degrees. Thus youask each calculator to evaluate cos 3.14. One calculator replies with ananswer of −0.999999; the other calculator replies with an answer of0.998499. Without further use of a calculator, how would you decidewhich calculator is using radians and which calculator is using degrees?Explain your answer.

solution Note that 3.14 ≈ π . Thus when using radians, a calculatorwill shown that

cos 3.14 ≈ cosπ = −1.

Thus the calculator that evaluates the cos 3.14 to be −0.999999 mustbe using radians, and the other calculator must be using degrees.


36. Suppose you have borrowed two calculators from friends, but you donot know whether they are set to work in radians or degrees. Thus youask each calculator to evaluate sin 1. One calculator replies with ananswer of 0.017452; the other calculator replies with an answer of0.841471. Without further use of a calculator, how would you decidewhich calculator is using radians and which calculator is using degrees?Explain your answer.

solution The radius of the unit circle that makes an angle of 1◦ withthe positive horizontal axis lies very close to the horizontal axis, andthus the sine of 1◦ is close to 0. However, the radius of the unit circlethat makes an angle of 1 radian (approximately 57.3◦) with the positivehorizontal axis does not lie close to the horizontal axis and thus thesine of 1 radian is not close to 0.

Thus the calculator that evaluates the sin 1 to be 0.017452 must beusing degrees, and the calculator that evaluates the sin 1 to be0.841471 must be using radians.


37. Explain why ecosx < 3 for every real number x.

solution Suppose x is a real number. Thus cosx ≤ 1. Hence

ecosx ≤ e1 = e < 3.


38. Explain why the equation

(sinx)2 − 4 sinx + 4 = 0

has no solutions.

solution We have

(sinx)2 − 4 sinx + 4 = (2− sinx)2.

Thus if (sinx)2 − 4 sinx + 4 = 0, then (2− sinx)2 = 0, which impliesthat 2− sinx = 0, which implies that sinx = 2. However, sinx ≤ 1 forevery real number x, so we cannot have sinx = 2. Thus there is no realnumber x such that (sinx)2 − 4 sinx + 4 = 0.


39. Explain why there does not exist a real number x such that esinx = 14 .

solution Suppose x is a real number. Thus sinx ≥ −1. Hence

esinx ≥ e−1 = 1e >

13 >

14 .


40. Explain why the equation

(cosx)99 + 4 cosx − 6 = 0

has no solutions.

solution Suppose x is a real number. Then cosx ≤ 1. Thus

(cosx)99 ≤ 1 and 4 cosx ≤ 4.

Adding together these two inequalities, we get

(cosx)99 + 4 cosx ≤ 5.

Thus (cosx)99 + 4 cosx cannot equal 6, which means that the equation(cosx)99 + 4 cosx − 6 = 0 has no solutions.


41. Explain why there does not exist a number θ such that log cosθ = 0.1.

solution If θ is a number such that cosθ ≤ 0, then log cosθ isundefined. If θ is a number such that cosθ > 0, then log cosθ ≤ 0because cosθ ≤ 1. Thus there does not exist a number θ such thatlog cosθ = 0.1.

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Solutions to Exercises, Section 5 - Department of …park/Fall2014/precalculus/5.3sol.pdf · Instructor’s Solutions Manual, Section 5.3 Exercise 1 Solutions to Exercises, Section