Solutions to Exercises, Section 5park/Fall2014/precalculus/5.7sol.pdf · Instructor’s Solutions...

Instructor’s Solutions Manual, Section 5.7 Exercise 1

Solutions to Exercises, Section 5.7

1. Evaluate cos−1 12 .

solution cos π3 = 12 ; thus cos−1 1

2 = π3 .


2. Evaluate sin−1 12 .

solution sin π6 = 1

2 ; thus sin−1 12 = π

6 .


3. Evaluate tan−1(−1).

solution tan(−π4 ) = −1; thus

tan−1(−1) = −π4 .


4. Evaluate tan−1(−√3).

solution tan(−π3 ) = −√

3; thus tan−1(−√3) = −π3 .


u

Ν

a

bc

Use the right triangle above for Exercises 5–12. This triangle is notdrawn to scale corresponding to the data in the exercises.

5. Suppose a = 2 and c = 3. Evaluate u in radians.

solution Because the cosine of an angle in a right triangle equals thelength of the adjacent side divided by the length of the hypotenuse, wehave cosu = 2

3 . Using a calculator working in radians, we then have

u = cos−1 23 ≈ 0.841 radians.


6. Suppose a = 3 and c = 4. Evaluate u in radians.

solution Because the cosine of an angle in a right triangle equals thelength of the adjacent side divided by the length of the hypotenuse, wehave cosu = 3


u = cos−1 34 ≈ 0.722734 radians.


7. Suppose a = 2 and c = 5. Evaluate ν in radians.

solution Because the sine of an angle in a right triangle equals thelength of the opposite side divided by the length of the hypotenuse, wehave sinν = 2


ν = sin−1 25 ≈ 0.412 radians.


8. Suppose a = 3 and c = 5. Evaluate ν in radians.

solution Because the sine of an angle in a right triangle equals thelength of the opposite side divided by the length of the hypotenuse, wehave sinν = 3


ν = sin−1 35 ≈ 0.643501 radians.


9. Suppose a = 5 and b = 4. Evaluate u in degrees.

solution Because the tangent of an angle in a right triangle equals thelength of the opposite side divided by the length of the adjacent side,we have tanu = 4

5 . Using a calculator working in degrees, we then have

u = tan−1 45 ≈ 38.7◦.


10. Suppose a = 5 and b = 6. Evaluate u in degrees.

solution Because the tangent of an angle in a right triangle equals thelength of the opposite side divided by the length of the adjacent side,we have tanu = 6


u = tan−1 65 ≈ 50.1944◦.


11. Suppose a = 5 and b = 7. Evaluate ν in degrees.

solution Because the tangent of an angle in a right triangle equals thelength of the opposite side divided by the length of the adjacent side,we have tanν = 5


ν = tan−1 57 ≈ 35.5◦.


12. Suppose a = 7 and b = 6. Evaluate ν in degrees.

solution Because the tangent of an angle in a right triangle equals thelength of the opposite side divided by the length of the adjacent side,we have tanν = 7


ν = tan−1 76 ≈ 49.3987◦.


13. Find the smallest positive number t such that 10cos t = 6.

solution The equation above implies that cos t = log 6. Thus we taket = cos−1(log 6) ≈ 0.67908.


14. Find the smallest positive number t such that 10sin t = 7.

solution The equation above implies that sin t = log 7. Thus we taket = sin−1(log 7) ≈ 1.00675.


15. Find the smallest positive number t such that etan t = 15.

solution The equation above implies that tan t = ln 15. Thus we taket = tan−1(ln 15) ≈ 1.21706.


16. Find the smallest positive number t such that etan t = 500.

solution The equation above implies that tan t = ln 500. Thus wetake t = tan−1(ln 500) ≈ 1.41125.


17. Find the smallest positive number y such that cos(tany) = 0.2.

solution The equation above implies that we should choosetany = cos−1 0.2 ≈ 1.36944. Thus we should choosey ≈ tan−1 1.36944 ≈ 0.94007.


18. Find the smallest positive number y such that sin(tany) = 0.6.

solution The equation above implies that we should choosetany = sin−1 0.6 ≈ 0.643501. Thus we should choosey ≈ tan−1 0.643501 ≈ 0.571793.


19. Find the smallest positive number x such that

sin2x − 3 sinx + 1 = 0.

solution Write y = sinx. Then the equation above can be rewrittenas

y2 − 3y + 1 = 0.

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

y = 3+√52

≈ 2.61803

and

y = 3−√52

≈ 0.38197.

Thus sinx ≈ 2.61803 or sinx ≈ 0.381966. However, there is no realnumber x such that sinx ≈ 2.61803 (because sinx is at most 1 forevery real number x), and thus we must have sinx ≈ 0.381966. Thusx ≈ sin−1 0.381966 ≈ 0.39192.



sin2x − 4 sinx + 2 = 0.

solution Write y = sinx. Then the equation above can be rewrittenas

y2 − 4y + 2 = 0.

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

y = 2+√2 ≈ 3.41421

andy = 2−√2 ≈ 0.585786.

Thus sinx ≈ 3.41421 or sinx ≈ 0.585786. However, there is no realnumber x such that sinx ≈ 3.41421 (because sinx is at most 1 forevery real number x), and thus we must have sinx ≈ 0.585786. Thusx ≈ sin−1 0.585786 ≈ 0.62585.



cos2x − 0.5 cosx + 0.06 = 0.

solution Write y = cosx. Then the equation above can be rewrittenas

y2 − 0.5y + 0.06 = 0.

Using the quadratic formula or factorization, we find that the solutionsto this equation are

y = 0.2 and y = 0.3.

Thus cosx = 0.2 or cosx = 0.3, which suggests that we choosex = cos−1 0.2 or x = cos−1 0.3. Because arccosine is a decreasingfunction, cos−1 0.3 is smaller than cos−1 0.2. Because we want to findthe smallest positive value of x satisfying the original equation, wechoose x = cos−1 0.3 ≈ 1.2661.



cos2x − 0.7 cosx + 0.12 = 0.

solution Write y = cosx. Then the equation above can be rewrittenas

y2 − 0.7y + 0.12 = 0.

Using the quadratic formula or factorization, we find that the solutionsto this equation are

y = 0.3 and y = 0.4.

Thus cosx = 0.3 or cosx = 0.4, which suggests that we choosex = cos−1 0.3 or x = cos−1 0.4. Because arccosine is a decreasingfunction, cos−1 0.4 is smaller than cos−1 0.3. Because we want to findthe smallest positive value of x satisfying the original equation, wechoose x = cos−1 0.4 ≈ 1.15928.

Instructor’s Solutions Manual, Section 5.7 Problem 23

Solutions to Problems, Section 5.7

23. Explain whycos−1 3

5 = sin−1 45 = tan−1 4

3 .

[Hint: Take a = 3 and b = 4 in the triangle above. Then find c andconsider various ways to express u.]

solution Consider a right triangle whose nonhypotenuse sides havelengths 3 and 4. By the Pythagorean Theorem, the hypotenuse haslength

√32 + 42, which equals 5. Let u denote the angle formed by the

side of length 3 and the hypotenuse, as shown below:

u3

45

Looking at the right triangle above, we see that

cosu = 35 and sinu = 4

5 and tanu = 43 .

Thusu = cos−1 3

5 and u = sin−1 45 and u = tan−1 4

3 .

Thus cos−1 35 = sin−1 4

5 = tan−1 43 because all three quantities equal u.


24. Explain whycos−1 5

13 = sin−1 1213 = tan−1 12

5 .

solution Consider a right triangle whose nonhypotenuse sides havelengths 5 and 12. By the Pythagorean Theorem, the hypotenuse haslength

√52 + 122, which equals 13. Let u denote the angle formed by

the side of length 5 and the hypotenuse, as shown below:

u5

1213

Looking at the right triangle above, we see that

cosu = 513 and sinu = 12

13 and tanu = 125 .

Thus


u = cos−1 513 and u = sin−1 12

13 and u = tan−1 125 .

Thus cos−1 513 = sin−1 12

13 = tan−1 125 because all three quantities equal u.


25. Suppose a and b are numbers such that

cos−1 a = π7 and sin−1 b = π

7 .

Explain why a2 + b2 = 1.

solution The equations

cos−1 a = π7 and sin−1 b = π

7 .

mean thatcos π7 = a and sin π

7 = b.

Because (cos π7 )2 + (sin π

7 )2 = 1, we have a2 + b2 = 1.


26. Without using a calculator, sketch the unit circle and the radius thatmakes an angle of cos−1 0.1 with the positive horizontal axis.

solution

0.1 1


27. Without using a calculator, sketch the unit circle and the radius thatmakes an angle of sin−1(−0.1) with the positive horizontal axis.

solution

�0.1

1


28. Without using a calculator, sketch the unit circle and the radius thatmakes an angle of tan−1 4 with the positive horizontal axis.

solution The radius shown in the figure below has slope 4:

1


29. Find all numbers t such that

cos−1 t = sin−1 t.

solution Suppose t is a number such that cos−1 t = sin−1 t. Leta = cos−1 t = sin−1 t. Thus

cosa = t and sina = t.

Because cos2a+ sin2a = 1, the equations above imply that t2 + t2 = 1,which implies that

t = ± 1√2= ±

√2

2.

Choosing the plus sign gives t =√

22 , and indeed we have

cos−1

√2

2= sin−1

√2

2

because both sides of the equation above equal π4 .

Choosing the minus sign in the expression above for t gives t = −√

22 ,

but

cos−1(−√

22

)= 3π

4and sin−1

(−√

22

)= −π

4.

Thus cos−1(−√

22

)�= sin−1

(−√

22

). Hence the only value of t satisfying

cos−1 t = sin−1 t is t =√

22 .


30. There exist angles θ such that cosθ = − sinθ (for example, −π4 and 3π4

are two such angles). However, explain why there do not exist anynumbers t such that

cos−1 t = − sin−1 t.

solution Suppose t is a number such that cos−1 t = − sin−1 t. Leta = cos−1 t = − sin−1 t. Thus −a = sin−1 t, which implies thatt = sin(−a) = − sina. Hence

cosa = t and sina = −t.

Because cos2a+ sin2a = 1, the equations above imply thatt2 + (−t)2 = 1, which implies that

t = ± 1√2= ±

√2

2.

Choosing the plus sign gives t =√

22 , but

cos−1

√2

2= π

4and − sin−1

√2

2= −π

4.

Thus cos−1√

22 �= − sin−1

√2

2 .

Choosing the minus sign in the expression above for t gives t = −√

22 ,

but

cos−1(−√

22

)= 3π

4and − sin−1

(−√

22

)= π

4.


Thus cos−1(−√

22

)�= − sin−1

(−√

22

).

Hence there are no values of t satisfying cos−1 t = − sin−1 t.

Solutions to Exercises, Section 5park/Fall2014/precalculus/5.7sol.pdf · Instructor’s Solutions...

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