Download Solutions to Exercises, Section 5.3

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 a
calculator for any of these exercises—otherwise you will likely get
decimal approximations for some solutions rather than exact answers.
More importantly, good understanding will come from working these
exercises by hand.
1.
(a) cos 3π
(b) sin 3π
solution Because 3π = 2π + π , an angle of 3π radians (as measured
counterclockwise from the positive horizontal axis) consists of a
complete revolution around the circle (2π radians) followed by another
π radians (180◦ ), as shown below. The endpoint of the corresponding
radius is (−1, 0). Thus cos 3π = −1 and sin 3π = 0.
1
Instructor’s Solutions Manual, Section 5.3
2.
(a) cos(− 3π
2 )
Exercise 2
(b) sin(− 3π
2 )
3π
π
3π
solution Because − 2 = −π − 2 , an angle of − 2 radians is obtained
by moving clockwise from the positive horizontal axis by half the circle
(−π radians or −180◦ ) and then continuing clockwise for another
π
one-fourth of the circle (− 2 radians or −90◦ ), as shown below. The
3π
endpoint of the corresponding radius is (0, 1). Thus cos(− 2 ) = 0 and
3π
sin(− 2 ) = 1.
Instructor’s Solutions Manual, Section 5.3
Exercise 2
1
Instructor’s Solutions Manual, Section 5.3
3.
(a) cos 11π
4
Exercise 3
(b) sin 11π
4
π
π
11π
solution Because 11π
4 = 2π + 2 + 4 , an angle of 4 radians (as
measured counterclockwise from the positive horizontal axis) consists
of a complete revolution around the circle (2π radians) followed by
π
π
another 2 radians (90◦ ), followed by another 4 radians (45◦ ), as shown
√2 √2 below. Hence the endpoint of the corresponding radius is − 2 , 2 .
Thus cos
11π
4
=−
√
2
2
and sin 11π
4 =
√
2
2 .
1
Instructor’s Solutions Manual, Section 5.3
4.
(a) cos 15π
4
Exercise 4
(b) sin 15π
4
π
π
15π
solution Because 15π
4 = 2π + π + 2 + 4 , an angle of 4 radians (as
measured counterclockwise from the positive horizontal axis) consists
of a complete revolution around the circle (2π radians) followed by a
π
half revolution around the circle (π radians), followed by another 2
radians (90◦ ), followed by another π4 radians (45◦ ), as shown below. The
√ √
√2
2
15π
2
endpoint of the corresponding radius is 2 , − 2 . Thus cos 4 = 2
and sin 15π
4 =−
√
2
2 .
1
Instructor’s Solutions Manual, Section 5.3
5.
(a) cos 2π
3
Exercise 5
(b) sin 2π
3
π
π
2π
solution Because 2π
3 = 2 + 6 , an angle of 3 radians (as measured
π
counterclockwise from the positive horizontal axis) consists of 2
π
◦
◦
radians (90 radians) followed by another 6 radians (30 ), as shown
1 √3 below. The endpoint of the corresponding radius is − 2 , 2 . Thus
cos
2π
3
1
= − 2 and sin
2π
3
=
√
3
2 .
1
Instructor’s Solutions Manual, Section 5.3
6.
(a) cos 4π
3
Exercise 6
(b) sin 4π
3
π
4π
solution Because 4π
3 = π + 3 , an angle of 3 radians (as measured
counterclockwise from the positive horizontal axis) consists of π
π
radians (180◦ radians) followed by another 3 radians (60◦ ), as shown
√ 1
3
below. The endpoint of the corresponding radius is − 2 , − 2 . Thus
cos
4π
3
1
= − 2 and sin
4π
3
=−
√
3
2 .
1
Instructor’s Solutions Manual, Section 5.3
7.
(a) cos 210◦
Exercise 7
(b) sin 210◦
solution Because 210 = 180 + 30, an angle of 210◦ (as measured
counterclockwise from the positive horizontal axis) consists of 180◦
followed by another 30◦ , as√shown below. The endpoint
√ of the
3
1
3
◦
corresponding radius is − 2 , − 2 . Thus cos 210 = − 2 and
1
sin 210◦ = − 2 .
1
Instructor’s Solutions Manual, Section 5.3
8.
(a) cos 300◦
Exercise 8
(b) sin 300◦
solution Because 300 = 270 + 30, an angle of 300◦ (as measured
counterclockwise from the positive horizontal axis) consists of 270◦
followed by another 30◦ , as shown
√ below. The endpoint of the
1
3
1
corresponding radius is 2 , − 2 . Thus cos 300◦ = 2 and
sin 300◦ = −
√
3
2 .
1
Instructor’s Solutions Manual, Section 5.3
9.
(a) cos 360045◦
Exercise 9
(b) sin 360045◦
solution Because 360045 = 360 × 1000 + 45, an angle of 360045◦ (as
measured counterclockwise from the positive horizontal axis) consists
of 1000 complete revolutions around the circle followed
by another
√2 √2 ◦
45 . The endpoint of the corresponding radius is 2 , 2 . Thus
cos 360045◦ =
√
2
2
and
sin 360045◦ =
√
2
2 .
Instructor’s Solutions Manual, Section 5.3
10.
(a) cos(−360030◦ )
Exercise 10
(b) sin(−360030◦ )
solution Because
−360030 = 360 × (−1000) − 30,
an angle of −360030◦ consists of 1000 complete clockwise revolutions
clockwise direction.
around the circle followed by another 30◦ in the
√3
1
The endpoint of the corresponding radius is 2 , − 2 . Thus
cos(−360030◦ ) =
√
3
2
and
sin(−360030◦ ) = − 12 .
Instructor’s Solutions Manual, Section 5.3
Exercise 11
11. Find the smallest number θ larger than 4π such that cos θ = 0.
solution Note that
0 = cos
π
2
= cos
3π
2
= cos
5π
2
= ...
and that the only numbers whose cosine equals 0 are of the form
(2n+1)π
, where n is an integer. The smallest number of this form larger
2
9π
9π
than 4π is 2 . Thus 2 is the smallest number larger than 4π whose
cosine equals 0.
Instructor’s Solutions Manual, Section 5.3
12. Find the smallest number θ larger than 6π such that sin θ =
Exercise 12
√
2
2 .
solution An angle of 6π corresponds to three complete
counterclockwise revolutions
around the unit circle. To reach an angle
√
2
π
π
whose sine equals 2 requires an additional angle of 4 . Thus 6π + 4 ,
25π
which equals 4 , is the smallest number larger than 6π whose sine
equals
√
2
2 .
Instructor’s Solutions Manual, Section 5.3
Exercise 13
13. Find the four smallest positive numbers θ such that cos θ = 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 the
positive horizontal axis, the first coordinate of its endpoint first
π
becomes 0 when θ equals 2 (which equals 90◦ ), then again when θ
3π
5π
equals 2 (which equals 270◦ ), then again when θ equals 2 (which
7π
equals 360◦ + 90◦ , or 450◦ ), then again when θ equals 2 (which equals
◦
◦
◦
360 + 270 , or 630 ), and so on. Thus the four smallest positive
π 3π 5π
7π
numbers θ such that cos θ = 0 are 2 , 2 , 2 , and 2 .
Instructor’s Solutions Manual, Section 5.3
Exercise 14
14. Find the four smallest positive numbers θ such that sin θ = 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 the
positive horizontal axis, the second coordinate of its endpoint first
becomes 0 after starting its travel when θ equals π (which equals 180◦ ),
then again when θ equals 2π (which equals 360◦ ), then again when θ
equals 3π (which equals 360◦ + 180◦ , or 540◦ ), then again when θ
equals 4π (which equals 720◦ ), and so on. Thus the four smallest
positive numbers θ such that sin θ = 0 are π , 2π , 3π , and 4π .
Instructor’s Solutions Manual, Section 5.3
Exercise 15
15. Find the four smallest positive numbers θ such that sin θ = 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 the
positive horizontal axis, then the second coordinate of its endpoint first
π
becomes 1 when θ equals 2 (which equals 90◦ ), then again when θ
5π
equals 2 (which equals 360◦ + 90◦ , or 450◦ ), then again when θ equals
9π
◦
◦
◦
2 (which equals 2 × 360 + 90 , or 810 ), then again when θ equals
13π
◦
◦
◦
2 (which equals 3 × 360 + 90 , or 1170 ), and so on. Thus the four
π 5π 9π
13π
smallest positive numbers θ such that sin θ = 1 are 2 , 2 , 2 , and 2 .
Instructor’s Solutions Manual, Section 5.3
Exercise 16
16. Find the four smallest positive numbers θ such that cos θ = 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 the
positive horizontal axis, the first coordinate of its endpoint first
becomes 1 after starting its travel when θ equals 2π (which equals
360◦ ), then again when θ equals 4π (which equals 720◦ ), then again
when θ equals 6π , then again when θ equals 8π , and so on. Thus the
four smallest positive numbers θ such that cos θ = 1 are 2π , 4π , 6π ,
and 8π .
Instructor’s Solutions Manual, Section 5.3
Exercise 17
17. Find the four smallest positive numbers θ such that cos θ = −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 the
positive horizontal axis, the first coordinate of its endpoint first
becomes −1 when θ equals π (which equals 180◦ ), then again when θ
equals 3π (which equals 360◦ + 180◦ , or 540◦ ), then again when θ
equals 5π (which equals 2 × 360◦ + 180◦ , or 900◦ ), then again when θ
equals 7π (which equals 3 × 360◦ + 180◦ , or 1260◦ ), and so on. Thus
the four smallest positive numbers θ such that cos θ = −1 are π , 3π ,
5π , and 7π .
Instructor’s Solutions Manual, Section 5.3
Exercise 18
18. Find the four smallest positive numbers θ such that sin θ = −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 the
positive horizontal axis, the second coordinate of its endpoint first
3π
becomes −1 when θ equals 2 (which equals 270◦ ), then again when θ
7π
equals 2 (which equals 360◦ + 270◦ , or 630◦ ), then again when θ equals
11π
15π
2 , then again when θ equals 2 , and so on. Thus the four smallest
3π 7π 11π
15π
positive numbers θ such that sin θ = −1 are 2 , 2 , 2 , and 2 .
Instructor’s Solutions Manual, Section 5.3
Exercise 19
19. Find the four smallest positive numbers θ such that sin θ = 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 the
positive horizontal axis, the second coordinate of its endpoint first
1
π
becomes 2 when θ equals 6 (which equals 30◦ ), then again when θ
5π
13π
equals 6 (which equals 150◦ ), then again when θ equals 6 (which
17π
equals 360◦ + 30◦ , or 390◦ ), then again when θ equals 6 (which
◦
◦
◦
equals 360 + 150 , or 510 ), and so on. Thus the four smallest positive
1
π 5π 13π
17π
numbers θ such that sin θ = 2 are 6 , 6 , 6 , and 6 .
Instructor’s Solutions Manual, Section 5.3
Exercise 20
20. Find the four smallest positive numbers θ such that cos θ = 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 the
positive horizontal axis, the first coordinate of its endpoint first
1
π
becomes 2 when θ equals 3 (which equals 60◦ ), then again when θ
5π
7π
equals 3 (which equals 300◦ ), then again when θ equals 3 (which
11π
equals 360◦ + 60◦ , or 420◦ ), then again when θ equals 3 (which
◦
◦
◦
equals 360 + 300 , or 660 ), and so on. Thus the four smallest positive
1
π 5π 7π
11π
numbers θ such that cos θ = 2 are 3 , 3 , 3 , and 3 .
Instructor’s Solutions Manual, Section 5.3
21. Suppose 0 < θ <
π
2
Exercise 21
and cos θ = 25 . Evaluate sin θ.
solution We know that
(cos θ)2 + (sin θ)2 = 1.
Thus
(sin θ)2 = 1 − (cos θ)2
=1−
=
2 2
5
21
.
25
π
Because 0 < θ < 2 , we know that sin θ > 0. Thus taking square roots of
both sides of the equation above gives
√
21
.
sin θ =
5
Instructor’s Solutions Manual, Section 5.3
22. Suppose 0 < θ <
π
2
Exercise 22
and sin θ = 37 . Evaluate cos θ.
solution We know that
(cos θ)2 + (sin θ)2 = 1.
Thus
(cos θ)2 = 1 − (sin θ)2
=1−
=
3 2
7
40
.
49
π
Because 0 < θ < 2 , we know that cos θ > 0. Thus taking square roots
of both sides of the equation above gives
√
√
2 10
40
=
.
cos θ =
7
7
Instructor’s Solutions Manual, Section 5.3
23. Suppose
π
2
Exercise 23
< θ < π and sin θ = 29 . Evaluate cos θ.
solution We know that
(cos θ)2 + (sin θ)2 = 1.
Thus
(cos θ)2 = 1 − (sin θ)2
=1−
=
2 2
9
77
.
81
π
Because 2 < θ < π , we know that cos θ < 0. Thus taking square roots
of both sides of the equation above gives
√
77
.
cos θ = −
9
Instructor’s Solutions Manual, Section 5.3
24. Suppose
π
2
Exercise 24
< θ < π and sin θ = 38 . Evaluate cos θ.
solution We know that
(cos θ)2 + (sin θ)2 = 1.
Thus
(cos θ)2 = 1 − (sin θ)2
=1−
=
3 2
8
55
.
64
π
Because 2 < θ < π , we know that cos θ < 0. Thus taking square roots
of both sides of the equation above gives
√
55
.
cos θ = −
8
Instructor’s Solutions Manual, Section 5.3
25.
Exercise 25
Suppose − π2 < θ < 0 and cos θ = 0.1. Evaluate sin θ.
solution We know that
(cos θ)2 + (sin θ)2 = 1.
Thus
(sin θ)2 = 1 − (cos θ)2
= 1 − (0.1)2
= 0.99.
π
Because − 2 < θ < 0, we know that sin θ < 0. Thus taking square roots
of both sides of the equation above gives
√
sin θ = − 0.99 ≈ −0.995.
Instructor’s Solutions Manual, Section 5.3
26.
Exercise 26
Suppose − π2 < θ < 0 and cos θ = 0.3. Evaluate sin θ.
solution We know that
(cos θ)2 + (sin θ)2 = 1.
Thus
(sin θ)2 = 1 − (cos θ)2
= 1 − (0.3)2
= 0.91.
π
Because − 2 < θ < 0, we know that sin θ < 0. Thus taking square roots
of both sides of the equation above gives
√
sin θ = − 0.91 ≈ −0.954.
Instructor’s Solutions Manual, Section 5.3
Exercise 27
27. Find the smallest number x such that sin(ex ) = 0.
solution Note that ex is an increasing function. Because ex is
positive for every real number x, and because π is the smallest positive
number whose sine equals 0, we want to choose x so that ex = π . Thus
x = ln π .
Instructor’s Solutions Manual, Section 5.3
Exercise 28
28. Find the smallest number x such that cos(ex + 1) = 0.
solution Note that ex is an increasing function. Because ex is
π
positive for every real number x, and because 2 is the smallest positive
π
number whose cosine equals 0, we want to choose x so that ex + 1 = 2 .
π
π
Thus ex = 2 − 1, which implies that x = ln( 2 − 1).
Instructor’s Solutions Manual, Section 5.3
29.
Exercise 29
Find the smallest positive number x such that
sin(x 2 + x + 4) = 0.
solution Note that x 2 + x + 4 is an increasing function on the
interval [0, ∞). If x is positive, then x 2 + x + 4 > 4. Because 4 is larger
than π but less than 2π , the smallest number bigger than 4 whose sine
equals 0 is 2π . Thus we want to choose x so that x 2 + x + 4 = 2π . In
other words, we need to solve the equation
x 2 + x + (4 − 2π ) = 0.
Using the quadratic formula, we see that the solutions to this equation
are
√
−1 ± 8π − 15
.
x=
2
A calculator shows that choosing the plus sign in the equation above
gives x ≈ 1.0916 and choosing the minus sign gives x ≈ −2.0916. We
seek only positive values of x, and thus we choose the plus sign in the
equation above, getting x ≈ 1.0916.
Instructor’s Solutions Manual, Section 5.3
30.
Exercise 30
Find the smallest positive number x such that
cos(x 2 + 2x + 6) = 0.
solution Note that x 2 + 2x + 6 is an increasing function on the
interval [0, ∞). If x is positive, then x 2 + 2x + 6 > 6. Because 6 is a bit
less than 2π , the smallest number bigger than 6 whose cosine equals 0
π
5π
is 2π + 2 , which equals 2 . Thus we want to choose x so that
5π
x 2 + 2x + 6 = 2 . In other words, we need to solve the equation
x 2 + 2x + (6 −
5π
2 )
= 0.
Using the quadratic formula, we see that the solutions to this equation
are
√
−2 ± 10π − 20
.
x=
2
A calculator shows that choosing the plus sign in the equation above
gives x ≈ 0.6894 and choosing the minus sign gives x ≈ −2.6894. We
seek only positive values of x, and thus we choose the plus sign in the
equation 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 the
6
positive horizontal axis such that cos θ = 7 .
(b) Sketch another radius, different from the one in part (a), also
6
illustrating cos θ = 7 .
solution
(a) Because cos θ is the first coordinate of the radius of the unit circle
corresponding to the angle θ, we need to find a radius of the unit circle
6
whose first coordinate equals 7 . To do this, start with the point
6
corresponding to 7 on the horizontal axis and then move vertically up
6
to find a point on the unit circle whose first coordinate equals 7 . Then
draw the radius from the origin to that point on the unit circle:
Θ
6
7
1
6
An angle θ such that cos θ = 7 .
Instructor’s Solutions Manual, Section 5.3
Problem 31
(b) To find another radius, different from the one in part (a), also
6
6
illustrating cos θ = 7 , start with the point corresponding to 7 on the
horizontal axis and then move vertically down to find a point on the
6
unit circle whose first coordinate equals 7 . Then draw the radius from
the origin to that point on the unit circle:
Θ
6
7
1
6
Another angle θ such that cos θ = 7 .
Instructor’s Solutions Manual, Section 5.3
32.
Problem 32
(a) Sketch a radius of the unit circle making an angle θ with the
positive horizontal axis such that sin θ = −0.8.
(b) Sketch another radius, different from the one in part (a), also
illustrating sin θ = −0.8.
solution
(a) Because sin θ is the second coordinate of the radius of the unit circle
corresponding to the angle θ, we need to find a radius of the unit circle
whose second coordinate equals −0.8. To do this, start with the point
corresponding to −0.8 on the vertical axis and then move horizontally
right 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 unit
circle:
Θ
1
0.8
An angle θ such that sin θ = −0.8.
Instructor’s Solutions Manual, Section 5.3
Problem 32
(b) To find another radius, different from the one in part (a), also
illustrating sin θ = −0.8, start with the point corresponding to −0.8 on
the vertical axis and then move horizontally left 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 unit circle:
Θ
1
0.8
Another angle θ such that sin θ = −0.8.
Instructor’s Solutions Manual, Section 5.3
Problem 33
33. Find angles u and ν such that cos u = cos ν but sin u = sin ν.
solution One easy choice is to take u =
cos
π
2
π
2
and ν = − π2 . We have
π
= cos(− 2 ) = 0
but
sin
π
2
π
= 1 = −1 = sin(− 2 ).
Instructor’s Solutions Manual, Section 5.3
Problem 34
34. Find angles u and ν such that sin u = sin ν but cos u = cos ν.
solution One easy choice is to take u = 0 and ν = π . We have
sin 0 = sin π = 0
but
cos 0 = 1 = −1 = cos π .
Instructor’s Solutions Manual, Section 5.3
Problem 35
35. Suppose you have borrowed two calculators from friends, but you do
not know whether they are set to work in radians or degrees. Thus you
ask each calculator to evaluate cos 3.14. One calculator replies with an
answer of −0.999999; the other calculator replies with an answer of
0.998499. Without further use of a calculator, how would you decide
which calculator is using radians and which calculator is using degrees?
Explain your answer.
solution Note that 3.14 ≈ π . Thus when using radians, a calculator
will shown that
cos 3.14 ≈ cos π = −1.
Thus the calculator that evaluates the cos 3.14 to be −0.999999 must
be using radians, and the other calculator must be using degrees.
Instructor’s Solutions Manual, Section 5.3
Problem 36
36. Suppose you have borrowed two calculators from friends, but you do
not know whether they are set to work in radians or degrees. Thus you
ask each calculator to evaluate sin 1. One calculator replies with an
answer of 0.017452; the other calculator replies with an answer of
0.841471. Without further use of a calculator, how would you decide
which 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◦ with
the positive horizontal axis lies very close to the horizontal axis, and
thus the sine 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 close to the horizontal axis and thus the
sine of 1 radian is not close to 0.
Thus the calculator that evaluates the sin 1 to be 0.017452 must be
using degrees, and the calculator that evaluates the sin 1 to be
0.841471 must be using radians.
Instructor’s Solutions Manual, Section 5.3
Problem 37
37. Explain why ecos x < 3 for every real number x.
solution Suppose x is a real number. Thus cos x ≤ 1. Hence
ecos x ≤ e1 = e < 3.
Instructor’s Solutions Manual, Section 5.3
Problem 38
38. Explain why the equation
(sin x)2 − 4 sin x + 4 = 0
has no solutions.
solution We have
(sin x)2 − 4 sin x + 4 = (2 − sin x)2 .
Thus if (sin x)2 − 4 sin x + 4 = 0, then (2 − sin x)2 = 0, which implies
that 2 − sin x = 0, which implies that sin x = 2. However, sin x ≤ 1 for
every real number x, so we cannot have sin x = 2. Thus there is no real
number x such that (sin x)2 − 4 sin x + 4 = 0.
Instructor’s Solutions Manual, Section 5.3
Problem 39
39. Explain why there does not exist a real number x such that esin x = 14 .
solution Suppose x is a real number. Thus sin x ≥ −1. Hence
esin x ≥ e−1 =
1
e
>
1
3
1
> 4.
Instructor’s Solutions Manual, Section 5.3
Problem 40
40. Explain why the equation
(cos x)99 + 4 cos x − 6 = 0
has no solutions.
solution Suppose x is a real number. Then cos x ≤ 1. Thus
(cos x)99 ≤ 1
and
4 cos x ≤ 4.
Adding together these two inequalities, we get
(cos x)99 + 4 cos x ≤ 5.
Thus (cos x)99 + 4 cos x cannot equal 6, which means that the equation
(cos x)99 + 4 cos x − 6 = 0 has no solutions.
Instructor’s Solutions Manual, Section 5.3
Problem 41
41. Explain why there does not exist a number θ such that log cos θ = 0.1.
solution If θ is a number such that cos θ ≤ 0, then log cos θ is
undefined. If θ is a number such that cos θ > 0, then log cos θ ≤ 0
because cos θ ≤ 1. Thus there does not exist a number θ such that
log cos θ = 0.1.