Solutions to Exercises, Section 5park/Fall2013/precalculus/5.4sol.pdf · Instructor’s Solutions...

Instructor’s Solutions Manual, Section 5.4 Exercise 1

Solutions to Exercises, Section 5.4

1. Find the four smallest positive numbers θ such that tanθ = 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 first and second coordinates of itsendpoint first become equal (which is equivalent to having tanθ = 1)when θ equals π

4 (which equals 45◦), then again when θ equals 5π4

(which equals 225◦), then again when θ equals 9π4 (which equals

360◦ + 45◦, or 405◦), then again when θ equals 13π4 (which equals

360◦ + 225◦, or 585◦), and so on.

Thus the four smallest positive numbers θ such that tanθ = 1 are π4 ,

5π4 , 9π

4 , and 13π4 .


2. Find the four smallest positive numbers θ such that tanθ = −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 first coordinate equals the negative ofthe second coordinate (which is equivalent to having tanθ = −1) whenθ equals 3π

4 (which equals 135◦), then again when θ equals 7π4 (which

equals 315◦), then again when θ equals 11π4 (which equals 360◦ + 135◦,

or 495◦), then again when θ equals 15π4 (which equals 360◦ + 315◦, or

675◦), and so on.

Thus the four smallest positive numbers θ such that tanθ = −1 are 3π4 ,

7π4 , 11π

4 , and 15π4 .


3. Suppose 0 < θ < π2 and cosθ = 1

5 . Evaluate:

(a) sinθ (b) tanθ

solution The figure below gives a sketch of the angle involved in thisexercise:

Θ

1

The angle between 0 and π2 whose cosine

equals 15 .

(a) We know that(cosθ)2 + (sinθ)2 = 1.

Thus(1

5

)2 + (sinθ)2 = 1. Solving this equation for (sinθ)2 gives

(sinθ)2 = 2425.

The sketch above shows that sinθ > 0. Thus taking square roots ofboth sides of the equation above gives

sinθ =√

245

=√

4 · 65

= 2√

65.


(b)

tanθ = sinθcosθ

=2√

6515

= 2√

6.


4. Suppose 0 < θ < π2 and sinθ = 1

4 . Evaluate:

(a) cosθ (b) tanθ


Θ

1

The angle between 0 and π2 whose sine

equals 14 .


Thus(cosθ)2 + (1

4

)2 = 1. Solving this equation for (cosθ)2 gives

(cosθ)2 = 1516.

The sketch above shows that cosθ > 0. Thus taking square roots ofboth sides of the equation above gives

cosθ =√

154.


(b)

tanθ = sinθcosθ

=14√154

= 1√15

=√

1515

.


5. Suppose π2 < θ < π and sinθ = 2

3 . Evaluate:

(a) cosθ (b) tanθ


Θ

1

The angle between π2 and π whose sine

equals 23 .


Thus (cosθ)2 + (23


(cosθ)2 = 59.

The sketch above shows that cosθ < 0. Thus taking square roots ofboth sides of the equation above gives

cosθ = −√

53.


(b)

tanθ = sinθcosθ

= −23√5

3

= − 2√5= −2

√5

5.


6. Suppose π2 < θ < π and sinθ = 3

4 . Evaluate:

(a) cosθ (b) tanθ


Θ

1

The angle between π2 and π whose sine

equals 34 .


Thus (cosθ)2 + (34


(cosθ)2 = 716.

The sketch above shows that cosθ < 0. Thus taking square roots ofboth sides of the equation above gives

cosθ = −√

74.


(b)

tanθ = sinθcosθ

= −34√7

4

= − 3√7= −3

√7

7.


7. Suppose −π2 < θ < 0 and cosθ = 45 . Evaluate:

(a) sinθ (b) tanθ


Θ 1

The angle between −π2 and 0 whosecosine equals 4

5 .


Thus(4

5


(sinθ)2 = 925.

The sketch above shows that sinθ < 0. Thus taking square roots ofboth sides of the equation above gives

sinθ = −35.


(b)

tanθ = sinθcosθ

= −3545

= −34.


8. Suppose −π2 < θ < 0 and cosθ = 15 . Evaluate:

(a) sinθ (b) tanθ


Θ 1

The angle between −π2 and 0 whosecosine equals 1

5 .


Thus(1

5


(sinθ)2 = 2425.

The sketch above shows that sinθ < 0. Thus taking square roots ofboth sides of the equation above gives

sinθ = −√

245

= −2√

65.


(b)

tanθ = sinθcosθ

= −2√

6515

= −2√

6.


9. Suppose 0 < θ < π2 and tanθ = 1

4 . Evaluate:

(a) cosθ (b) sinθ


Θ

1

The angle between 0 and π2 whose

tangent equals 14 .

(a) Rewrite the equation tanθ = 14 in the form sinθ

cosθ = 14 . Multiplying both

sides of this equation by cosθ, we get

sinθ = 14 cosθ.

Substitute this expression for sinθ into the equation(cosθ)2 + (sinθ)2 = 1, getting

(cosθ)2 + 116(cosθ)2 = 1,

which is equivalent to


(cosθ)2 = 1617.


cosθ = 4√17

= 4√

1717

.

(b) We have already noted that sinθ = 14 cosθ. Thus

sinθ =√

1717

.


10. Suppose 0 < θ < π2 and tanθ = 2

3 . Evaluate:

(a) cosθ (b) sinθ


Θ

1

The angle between 0 and π2 whose

tangent equals 23 .

(a) Rewrite the equation tanθ = 23 in the form sinθ

cosθ = 23 . Multiplying both

sides of this equation by cosθ, we get

sinθ = 23 cosθ.


(cosθ)2 + 49(cosθ)2 = 1,



(cosθ)2 = 913.


cosθ = 3√13

= 3√

1313

.

(b) We have already noted that sinθ = 23 cosθ. Thus

sinθ = 2√

1313

.


11. Suppose −π2 < θ < 0 and tanθ = −3. Evaluate:

(a) cosθ (b) sinθ


Θ 1

The angle between −π2 and 0 whosetangent equals −3.

(a) Rewrite the equation tanθ = −3 in the form sinθcosθ = −3. Multiplying

both sides of this equation by cosθ, we get

sinθ = −3 cosθ.


(cosθ)2 + 9(cosθ)2 = 1,



(cosθ)2 = 110.


cosθ = 1√10

=√

1010

.

(b) We have already noted that sinθ = −3 cosθ. Thus

sinθ = −3√

1010

.


12. Suppose −π2 < θ < 0 and tanθ = −2. Evaluate:

(a) cosθ (b) sinθ


Θ 1

The angle between −π2 and 0 whosetangent equals −2.

(a) Rewrite the equation tanθ = −2 in the form sinθcosθ = −2. Multiplying

both sides of this equation by cosθ, we get

sinθ = −2 cosθ.


(cosθ)2 + 4(cosθ)2 = 1,



(cosθ)2 = 15.


cosθ = 1√5=√

55.

(b) We have already noted that sinθ = −2 cosθ. Thus

sinθ = −2√

55.


Given that

cos 15◦ =√

2 + √3

2and sin 22.5◦ =

√2 − √2

2,

in Exercises 13–22 find exact expressions for the indicated quantities.[These values for cos 15◦ and sin 22.5◦ will be derived in Examples 4and 5 in Section 6.3.]

13. sin 15◦

solution We know that

(cos 15◦)2 + (sin 15◦)2 = 1.

Thus

(sin 15◦)2 = 1− (cos 15◦)2

= 1−(√2+√3

2

)2

= 1− 2+√34

= 2−√34

.

Because sin 15◦ > 0, taking square roots of both sides of the equationabove gives

sin 15◦ =√

2−√32

.


14. cos 22.5◦


(cos 22.5◦)2 + (sin 22.5◦)2 = 1.

Thus

(cos 22.5◦)2 = 1− (sin 22.5◦)2

= 1−(√2−√2

2

)2

= 1− 2−√24

= 2+√24

.

Because cos 22.5◦ > 0, taking square roots of both sides of the equationabove gives

cos 22.5◦ =√

2+√22

.


15. tan 15◦

solution

tan 15◦ = sin 15◦

cos 15◦

=√

2−√3√2+√3

=√

2−√3√2+√3

·√

2−√3√2−√3

= 2−√3√4− 3

= 2−√3


16. tan 22.5◦

solution

tan 22.5◦ = sin 22.5◦

cos 22.5◦

=√

2−√2√2+√2

=√

2−√2√2+√2

·√

2−√2√2−√2

= 2−√2√4− 2

= 2−√2√2

= √2− 1


17. cot 15◦

solution

cot 15◦ = 1tan 15◦

= 1

2−√3

= 1

2−√3· 2+√3

2+√3

= 2+√34− 3

= 2+√3


18. cot 22.5◦

solution

cot 22.5◦ = 1tan 22.5◦

= 1√2− 1

= 1√2− 1

·√

2+ 1√2+ 1

= √2+ 1


19. csc 15◦

solution

csc 15◦ = 1sin 15◦

= 2√2−√3

= 2√2−√3

·√

2+√3√2+√3

= 2√

2+√3√4− 3

= 2√

2+√3


20. csc 22.5◦

solution

csc 22.5◦ = 1sin 22.5◦

= 2√2−√2

= 2√2−√2

·√

2+√2√2+√2

= 2√

2+√2√4− 2

= 2√

2+√2√2

·√

2√2

=√

4+ 2√

2


21. sec 15◦

solution

sec 15◦ = 1cos 15◦

= 2√2+√3

= 2√2+√3

·√

2−√3√2−√3

= 2√

2−√3√4− 3

= 2√

2−√3


22. sec 22.5◦

solution

sec 22.5◦ = 1cos 22.5◦

= 2√2+√2

= 2√2+√2

·√

2−√2√2−√2

= 2√

2−√2√4− 2

= 2√

2−√2√2

·√

2√2

=√

4− 2√

2


Suppose u and ν are in the interval (0, π2 ), with

tan u = 2 and tan ν = 3.

In Exercises 23–32, find exact expressions for the indicated quantities.

23. cotu

solution cotu = 1tanu

= 12


24. cotν

solution

cotν = 1tanν

= 13


25. cosu


2 = tanu

= sinucosu

.

To find cosu, make the substitution sinu = √1− (cosu)2 in theequation above (this substitution is valid because we know that0 < u < π

2 and thus sinu > 0), getting

2 =√

1− (cosu)2

cosu.

Now square both sides of the equation above, then multiply both sidesby (cosu)2 and rearrange to get the equation

5(cosu)2 = 1.

Because 0 < u < π2 , we see that cosu > 0. Thus taking square roots of

both sides of the equation above gives cosu = 1√5, which can be

rewritten as cosu =√

55 .


26. cosν


3 = tanν

= sinνcosν

.

To find cosν , make the substitution sinν = √1− (cosν)2 in theequation above (this substitution is valid because we know that0 < ν < π

2 and thus sinν > 0), getting

3 =√

1− (cosν)2

cosν.

Now square both sides of the equation above, then multiply both sidesby (cosν)2 and rearrange to get the equation

10(cosν)2 = 1.

Because 0 < ν < π2 , we see that cosν > 0. Thus taking square roots of

both sides of the equation above gives cosν = 1√10

, which can be

rewritten as cosν =√

1010 .


27. sinu

solution

sinu =√

1− (cosu)2

=√

1− 15

=√

45

= 2√5

= 2√

55


28. sinν

solution Because 0 < ν < π2 , we know that sinν > 0. Thus

sinν =√

1− (cosν)2

=√

1− 110

=√

910

= 3√10

= 3√

1010

.


29. cscu

solution cscu = 1sinu

=√

52


30. cscν

solution cscν = 1sinν

=√

103


31. secu

solution secu = 1cosu

= √5


32. secν

solution

secν = 1cosν

= √10


33. Find the smallest number x such that tan 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 tangent equals 0, we want to choose x so that ex = π .Thus x = lnπ ≈ 1.14473.


34. Find the smallest number x such that tan ex is undefined.

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 tangent is undefinded, we want to choose x so thatex = π

2 . Thus x = ln π2 ≈ 0.45158.

Instructor’s Solutions Manual, Section 5.4 Problem 35

Solutions to Problems, Section 5.4

35. (a) Sketch a radius of the unit circle making an angle θ with thepositive horizontal axis such that tanθ = 1

7 .

(b) Sketch another radius, different from the one in part (a), alsoillustrating tanθ = 1

7 .

solution

(a) Because tanθ is the slope of the radius of the unit circle correspondingto the angle θ, we sketch a radius of the unit circle whose slope equals17 :

Θ

1

An angle θ such that tanθ = 17 .


(b) To find another radius, different from the one in part (a), alsoillustrating tanθ = 1

7 , draw the other radius of the unit circle that hasslope 1

7 . This second radius lies on the same line as the first radius:

Θ

1

Another angle θ such that tanθ = 17 .


36. (a) Sketch a radius of the unit circle making an angle θ with thepositive horizontal axis such that tanθ = 7.

(b) Sketch another radius, different from the one in part (a), alsoillustrating tanθ = 7.

solution

(a) Because tanθ is the slope of the radius of the unit circle correspondingto the angle θ, we sketch a radius of the unit circle whose slope equals7:

Θ

1

An angle θ such that tanθ = 7.

(b) To find another radius, different from the one in part (a), alsoillustrating tanθ = 7, draw the other radius of the unit circle that hasslope 7. This second radius lies on the same line as the first radius:


Θ

1

Another angle θ such that tanθ = 7.


37. Suppose a radius of the unit circle makes an angle with the positivehorizontal axis whose tangent equals 5, and another radius of the unitcircle makes an angle with the positive horizontal axis whose tangentequals −1

5 . Explain why these two radii are perpendicular to each other.

solution The first radius has slope 5 and the second radius has slope−1

5 . The product of these two slopes equals −1. Thus these two radiiare perpendicular (see Section 2.1).


38. Explain why

tan(θ + π2 ) = −

1tanθ

for every number θ that is not an integer multiple of π2 .

solution Suppose θ is not an integer multiple of π2 . Note that tanθ

equals the slope of the radius of the unit circle that makes an angle ofθ with the positive horizontal axis, and tan(θ + π

2 ) equals the slope ofthe radius of the unit circle that makes an angle of θ + π

2 with thepositive horizontal axis. These two radii are perpendicular to eachother (because θ + π

2 is obtained by adding π2 to θ), and thus the

product of their slopes equals −1 (see Section 2.1). In other words,

(tanθ)(tan(θ + π

2 )) = −1,

which is equivalent to the equation

tan(θ + π2 ) = −

1tanθ

.


39. Explain why the previous problem excluded integer multiples of π2

from the allowable values for θ.

solution If θ =mπ2 for some integer m, then either tanθ is

undefined (this happens if m is odd) or tan(θ + π2 ) is undefined (this

happens if m is even). Thus these values of θ must be excluded whenconsidering expressions involving both tanθ and tan(θ + π

2 ).


40. 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 tan 89.9. One calculator replies with ananswer of −2.62; the other calculator replies with an answer of 572.96.Without further use of a calculator, how would you decide whichcalculator is using radians and which calculator is using degrees?Explain your answer.

solution Note that tan 90◦ is undefined, and the tangent of an angleslightly less than 90◦ is a large positive number. Thus the calculatorthat evaluates the tan 89.9 to be 572.96 must be using degrees, and thecalculator that evaluates tan 89.9 to be −2.62 must be using radians.


41. 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 tan 1. One calculator replies with ananswer of 0.017455; the other calculator replies with an answer of1.557408. 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 has slope close to 0. Hence the tangent of 1◦ is close to 0.However, the radius of the unit circle that makes an angle of 1 radian(approximately 57.3◦) with the positive horizontal axis does not lie closeto the horizontal axis and thus the tangent of 1 radian is not close to 0.

Thus the calculator that evaluates the tan 1 to be 0.017455 must beusing degrees, and the calculator that evaluates the tan 1 to be1.557408 must be using radians.


42. Find a number θ such that the tangent of θ degrees is larger than50000.

solution The idea here is that an angle slightly less than 90◦ has avery large tangent. Thus we experiment using a calculator, finding that

tan 89.99◦ ≈ 5730.

Because 89.99◦ is not close enough to 90◦ to give our desired result, wetry getting even closer to 90◦, finding that

tan 89.999◦ ≈ 57296.

Thus 89.999◦ is an angle whose tangent is larger than 50000. Of coursethere are also many other correct answers.


43. Find a positive number θ such that the tangent of θ degrees is lessthan −90000.

solution The idea here is that an angle slightly more than 90◦ has anegative tangent with a large absolute value. Thus we experiment usinga calculator, finding that

tan 90.01◦ ≈ −5730.

Because 90.01◦ is not close enough to 90◦ to give our desired result, wetry getting even closer to 90◦, finding that

tan 90.001◦ ≈ −57296.

Because 90.001◦ is also not close enough to 90◦ to give our desiredresult, we try getting even closer to 90◦, finding that

tan 90.0001◦ ≈ −572958.

Thus 90.0001◦ is an angle whose tangent is less than −90000. Ofcourse there are also many other correct answers.


44. Explain why| sinθ| ≤ | tanθ|

for all θ such that tanθ is defined.

solution Suppose θ is not an odd multiple of π2 , which implies that

tanθ is defined. Because | cosθ| ≤ 1, we have

| sinθ| ≤ | sinθ|| cosθ| =

∣∣∣∣ sinθcosθ

∣∣∣∣ = | tanθ|.


45. Suppose θ is not an odd multiple of π2 . Explain why the point (tanθ,1)

is on the line containing the point (sinθ, cosθ) and the origin.

solution The slope of the line containing the points (sinθ, cosθ) and(0,0) is (cosθ)−0

(sinθ)−0 , which equals 1tanθ . Thus the equation of the line in the

xy-plane that contains the points (sinθ, cosθ) and (0,0) is

y = 1tanθ

x.

Taking x = tanθ in the equation above gives y = 1. In other words, thepoint (tanθ,1) is on this line.


46. In 1768 the Swiss mathematician Johann Lambert proved that if θ is arational number in the interval (0, π2 ), then tanθ is irrational. Explainwhy this result implies that π is irrational.

[Lambert’s result provided the first proof that π is irrational.]

solution Recall thattan π

4 = 1.

Because 1 is a rational number, Lambert’s result implies that π4 is

irrational. Thus 4 · π4 is irrational (see Problem 7 in Section 0.1). Inother words, π is irrational.

Solutions to Exercises, Section 5park/Fall2013/precalculus/5.4sol.pdf · Instructor’s Solutions...

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