Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 68 pdf

20.4 Solving Trigonometric Equations 651
8. Let f(x)=cos x and g(x) = arctan x. Find the following, where a and b are positive
constants. Your answers should be exact and as simple as possible.
(a) g(f (π))
(b) f

g

−a
b

In Problems 9 through 11, simplify the expressions given that x ∈

0,
π
2

.
9. (a) sin
−1
(sin x) (b) cos
−1
(cos x)
10. (a) sin
−1
(sin(−x)) (b) cos
−1
(cos(−x))
11. (a) tan
−1
(tan(x)) (b) tan

−1
(tan(−x))
In Problems 12 through 14, simplify the expressions given that x ∈

π
2
,2π

.
12. (a) arcsin(sin x) (b) arccos(cos x)
13. (a) arcsin(sin(−x)) (b) arccos(cos(−x))
14. (a) arctan(tan(x)) (b) arctan(tan(−x))
20.4 SOLVING TRIGONOMETRIC EQUATIONS
A trigonometric equation is an equation in which the variable to be solved for is the argument
of a trigonometric function. If we can get the trigonometric equation into the form sin u = k,
cos u = k, or tan u = k, where k is a constant, then we can use the inverse trigonometric
functions to help solve for u.
◆
EXAMPLE 20.9 Solve for x.
4 cos x + 1 =−1
SOLUTION
4 cos x + 1 =−1
4cos x =−2
cos x =−
1
2
One solution to this equation is cos
−1

−

1
2

. But there are actually two solutions for
x ∈ [0, 2π] and infinitely many solutions due to the periodicity of the cosine function. For
this problem we can turn to a triangle we know and love. We know cos(π/3) =
1
2
.Tohave
anegative cosine the angle must be in the second or third quadrant. So, the solutions are
x =
2π
3
+ 2πn or x =
4π
3
+ 2πn,
where n is any integer. See Figure 20.30 on the following page for illustration.
652 CHAPTER 20 Trigonometry—Circles and Triangles
π
3
2
2
1
2
1
2π
3
P(
v

)
4π
3
P(
)
ux
–
2
1
–
y = cos x
y =
–2π 2π–ππ
π
2
3π
solutions
Figure 20.30 ◆
◆
EXAMPLE 20.10 Solve for θ.
2 sin
2
θ − sin θ = 1
SOLUTION This is a quadratic equation in sin θ. We can either work with the sin θ or we can begin the
problem with the substitution u = sin θ, solve for u, and then return to sin θ.We’ll take this
latter approach.
2u
2
− u = 1
2u

2
− u − 1 = 0
(u − 1)(2u + 1) = 0
u − 1 = 0or2u+1=0
u=1oru=−
1
2
sin θ = 1 or sin θ =−
1
2
So
θ =
π
2
+ 2πn or θ =−
π
6
+ 2πn or θ =−
5π
6
+ 2πn,
where n is an integer.
π
6
π
2
P(
v
)
5π

6
P(
)
u
x
sin x
–2π 2π
1
–
–
7π
6
P(
) or
P(2π
)
π
6
–
π
6
P(
) or
1
2
Figure 20.31 ◆
20.4 Solving Trigonometric Equations 653
◆
EXAMPLE 20.11 Solve for x.
2 cos

2
x + sin x = 1
SOLUTION We begin by converting the cosines into sines so we have a quadratic in sin x.
2 cos
2
x + sin x = 1
2(1 − sin
2
x) + sin x = 1
2 − 2 sin
2
x + sin x − 1 = 0
−2 sin
2
x + sin x + 1 = 0
2 sin
2
x − sin x − 1 = 0
This is the equation we began with in the previous example, so the solutions are the
same.
◆
◆
EXAMPLE 20.12 Find all x in the interval [−2π ,2π]such that
tan x · sec x = tan x.
SOLUTION CAUTION Don’t divide both sides of the equation by tan x and leave yourself with only
sec x = 1. The equation is true if tan x = 0. Lobbing off the tan x means you’ll miss a
couple of solutions.
x
y = cos x
x

y = tan x
–2π –π 2ππ
tan x sec x = tan x
tan x sec x – tan x = 0
tan x (sec x – 1) = 0
tan x = 0

x = –2π, –π, 0, π, 2π
sec x –1 = 0
sec x = 1
cos x = 1
x = –2π, 0, 2π
or
or
Figure 20.32
Putting this all together, we get
x =−2π, −π,0,π,2π.
◆
◆
EXAMPLE 20.13 Find all x ∈ [0, 2π] such that cos(2x) = 0.3. Give exact answers and then give numerical
approximations.
SOLUTION Let’s begin by letting w = 2x. Then we can solve cos w = 0.3.
We want 0 ≤ x ≤ 2π,
so 2 · 0 ≤ 2x ≤ 2 · 2π
and 0 ≤ w ≤ 4π .
654 CHAPTER 20 Trigonometry—Circles and Triangles
We can get one solution to cos w = 0.3 by using inverse cosine.
w = arccos(0.3) ≈ 1.266
From Figure 20.33 we see that there is a second point, Q, on the unit circle with a
u-coordinate of 0.3. Let’s locate Q with an angle between 0 and 2π.

Q = 2π − arccos(0.3)
P(1.266)
Q
u
v
.3 1
1
Figure 20.33
We want all w ∈ [0, 4π] such that cos w = 0.3.
w = arccos(0.3), and w = arccos(0.3) + 2π,

corresponding
to point P

w = 2π − arccos(0.3), and w = 2π − arccos(0.3) + 2π.

corresponding
to point Q

x =
w
2
,so
x=
1
2
arccos(0.3), x =
1
2
arccos(0.3) + π,

x = π −
1
2
arccos(0.3), x = π −
1
2
arccos(0.3) + π = 2π −
1
2
arccos(0.3).
Numerical approximations give us
x ≈ 0.63 or 3.77 or 2.51 to 5.65.
A sketch of y = cos(2x) and y = 0.3 on [0, 2π] supports our solutions (and assures us
that we have found all the solutions requested).
20.4 Solving Trigonometric Equations 655
π
2
π
2π
3π
2
y = .3
y
x
Figure 20.34
◆
PROBLEMS FOR SECTION 20.4
1. Find all x between 0 and 2π such that
(a) 4 cos
2

x = 3.
(b) 2 sin
2
x − sin x − 1 = 0. (Hint: this is a quadratic in sin x.)
(c) sin x = 2/3. (Give exact answers, as well as numerical approximations.)
2. Find all solutions.
(a) sin
2
x = 0.25.
(b) cos
2
x + 2 cos x + 1 = 0. (Hint: Let u = cos x and first find u.)
(c) cos
2
x + 4 cos x + 3 = 0.
3. Find all solutions to the following equations:
(a) sec
2
x = 2.
(b) cos
2
x = 0.2 cos x. Why can’t you cancel the cos x from both sides of the equation?
(c) sin
2
x = 3 cos x + 1.
4. Below is the graph of f(x)=cos(x
2
) on [−π, π]. Find the exact coordinates of the
x-intercepts.
y

x
For each of the equations in Problems 5 through 13, find all solutions in the interval
[0, 2π]. Give exact answers (as opposed to numerical approximations).
656 CHAPTER 20 Trigonometry—Circles and Triangles
Use a graph to check that you have found all solutions in this interval. (Check
f(x)=0.5 on [0, 2π] by graphing y = f(x)and y = 0.5 on [0, 2π] and looking for
points of intersection or by graphing y = f(x)−0.5 on [0, 2π] and looking for zeros.)
5. sec x =−1
6. cos(2x) =1
7. 2 sin
2
(2x) =1
8. 3 tan
2
x − 1 = 0
9. cos
2
x + 4 sin x = 4
10. cos(3x) =0.5
11. 2 sin(3x) =−1
12. tan
2
t − 4 tan t = 5
13. cos x = 1 + sin x
(Hint: Square both sides so you’ll be able to convert cosines to sines. But be sure
to check all your answers, because squaring both sides of an equation can introduce
extraneous roots.)
14. Determine how many solutions each of the following equations has, and approximate
the solutions. These equations involve both trigonometric and algebraic functions and
therefore cannot be solved exactly using analytic methods. Instead, take a graphical

approach.
(a) sin x − x = 0
(b) 2 cos x = x
(c) sin x −
x
3
= 0
15. Find all t ∈ [0, π] such that 4 sec
2
(2t) − 3 =0.
16. Find all t ∈ [−π, π] such that 2 sin
2
t − 3 sin t + 1 = 0.
17. Find all x ∈ [0, 2π] such that
cos 3x =−
1
√
2
.
Give exact answers. Verify graphically that you have given the correct number of
solutions.
18. Find all solutions to
5 cos(x) = 6 cos
3
(x) − sin(x) cos(x)
in the interval [0, 2π].
20.5 Applying Trigonometry to a General Triangle: The Law of Cosines and the Law of Sines 657
19. If x is a solution to cos x sin x cos(2x) sin(2x) =0, find all possible values of cos(x).
20. Find all values of x in the interval [0, 2π] such that
sin(4x) =

1
√
2
.
Explain how you know that you have all the values requested.
21. In the Alfred Hitchcock film “Rear Window,” Jimmy Stewart plays a man who watches
his neighbor’s apartment through a camera with a telephoto lens because he believes
that the man has committed a murder. Suppose that the neighbor’s window is across
a 50-foot courtyard and is at a height 10 feet below the location of Jimmy Stewart’s
camera as placed on the tripod.
At what angle should Jimmy Stewart aim his camera in order to see into his
neighbor’s window? Give an exact answer and a degree measurement correct to one
decimal place.
22. A population of deer in a forest displays regular fluctuations in size. Scientists have
chosen to model the population size with a sinusoidal function. At its height the deer
population is 7000, while at its low it is 2000. The time between highs and lows is 6
months. The population at time t = 0 is 4500 and decreasing.
(a) Model the deer population as a function of time t in months. A picture should
accompany your answer.
(b) If t = 0 is now, what is the deer population 3 months from now?
(c) When is the first time in the future that the deer population will reach 3000? Give
an exact answer and then a numerical approximation.
(d) Call your answer to part (c) t
∗
.Giveany one time other than your answer to part
(c) at which the deer population is also 3000. (There are infinitely many correct
answers.) Give an exact answer in terms of t
∗
.
20.5 APPLYING TRIGONOMETRY TO A GENERAL

TRIANGLE: THE LAW OF COSINES AND THE
LAW OF SINES
We began our study of triangle trigonometry by observing that if the measure of one acute
angle of a right triangle is determined, then all the angles are determined and the ratio of
the lengths of the sides are determined. This is the basis of trigonometry. We are able to
apply trigonometry so that if we are given enough information to geometrically determine a
triangle we can solve the triangle; that is, we can find the measure of all sides and all angles.
Suppose we are dealing with a triangle that is not a right triangle. Such a triangle is
referred to as an oblique triangle. There are various configurations of information that will
determine the triangle. The question we will look at in this section is how trigonometry
can aid us in solving a generic triangle. Oblique triangles arise in many contexts, such as
surveying, astronomy, and navigation.
658 CHAPTER 20 Trigonometry—Circles and Triangles
In order to determine a generic triangle we need to know the length of at least one side
of the triangle and we need at least two other measurements. This means we need one of
the following four configurations of information.
1. Three sides (SSS)
2. Two sides and the included angle (SAS)
3. One side and any two angles (SAA)
4. Two sides and the angle opposite one of them (SSA)
In the first three cases, a unique triangle is determined by the information given. In the
last case, the case of knowing the lengths of two sides and the angle opposite one of them, the
triangle may not be uniquely determined. There may be a unique triangle, or two different
triangles, or even no triangles corresponding to the specifications. Due to the ambiguities,
we’ll refer to this last case as the “ambiguous case.”
There are two laws, the Law of Cosines and the Law of Sines, that allow us to export
our knowledge from right triangles to oblique triangles. We will first state them, then give
particular scenarios in which each can be applied, then prove them and discuss general
applications of the laws.
B

a
C
b
A
c
a triangle
without an
obtuse angle
B
a
C
b
A
c
a triangle
with an
obtuse angle
Figure 20.35
We’ll denote the angles of the generic triangle by uppercase A, B , and C and denote by
lowercase a, b, and c the lengths of the sides opposite A, B, and C, respectively.
The Law of Cosines
The Law of Cosines generalizes the Pythagorean Theorem; it says
c
2
= a
2
+ b
2
− 2ab cos C.
We’ll do a reality check to make sure that this makes sense. We can think of the Law of

Cosines as the Pythagorean Theorem with an adjustment term, −2ab cos C, tagged on.
First suppose that angle C is a right angle. Then the Law of Cosines just boils down to the
Pythagorean Theorem, because cos C = 0ifCis a right angle. Now, keeping the lengths of
a and b fixed, let angle C change. As angle C decreases the length of c decreases. On the
other hand, as angle C increases c increases as well.
20.5 Applying Trigonometry to a General Triangle: The Law of Cosines and the Law of Sines 659
If angle C = π/2, then cos C = 0 so the adjustment term, −2ab cos C,is0;
we’ve got the Pythagorean Theorem.
If angle C ∈ (0, π/2), then cos C is positive, making the adjustment term,
−2ab cos C, negative. The smaller C, the more negative the
adjustment term. This matches our intuition.
If angle C ∈ (π/2, π), then cos C is negative, making the adjustment term
−2ab cos C, positive. The larger C, the larger the adjustment
term. Again, this matches our intuition.
B
a
C
A
b
c
B
a
C
b
A
c
(ii ) C (0, )(i) C =
π
2
π

2
(iii) C ( , π)
π
2
B
a
C
b
A
c
Figure 20.36
The Law of Sines
sin A
a
=
sin B
b
=
sin C
c
We can rewrite
sin A
a
=
sin B
b
as
sin A
sin B
=

a
b
. Certainly in the case of a right triangle this makes
sense, because if C is a right angle then
sin A
sin B
=
a
c
b
c
=
a
b
.
Why this is true in a generic triangle will be clarified in the proof.
a
b
c
A
C
B
Figure 20.37
660 CHAPTER 20 Trigonometry—Circles and Triangles
Scenario I
A ship is at sea and is heading toward port A. The captain knows the distance from his ship
to port A, the distance from his ship to port B, and the distance between the two ports. Due
to weather conditions he wants to change course and head to port B. Through what angle
must the ship turn?
B

a
C
b
A
c
Port
Port
Figure 20.38
Geometrically, we are looking at a triangle, the two ports and the ship representing the
vertices A, B, and C respectively. Let’s call the three angles A, B, and C and denote by a,
b, and c, respectively, the lengths of the sides opposite these angles. The captain knows a,
b, and c. This is enough information to determine the triangle, that is, to determine all three
of the angles. Since these angles are determined, we ought to be able to use our knowledge
of trigonometry to find out what they are. The Law of Cosines allows us to do this. It tells us
that c
2
= a
2
+ b
2
− 2ab cos C. We can solve for cos C. Using arccosine we can find angle
C. The range of arccosine is [0, π] and angle C must be between 0 and π, so we will have
solved the captain’s problem.
EXERCISE 20.3 Suppose that in Scenario I given above the captain has the following data. His ship is 60
miles from port A and 40 miles from port B. Ports A and B are 30 miles apart. Through
what angle must the ship turn?
ANSWER arccos

43
48


≈ 0.46 radians or ≈ 26.38
◦
Proof of the Law of Cosines
We know that the Law of Cosines, c
2
= a
2
+ b
2
− 2ab cos C, is true for C = π/2, so we
need only look at the two cases that follow, the case where C is an acute angle and the case
where C is an obtuse angle.
Place triangle ABC so that angle C is in standard position with its vertex at the origin
and side b is lying along the positive x-axis as shown on the following page.