Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 69 pdf

20.5 Applying Trigonometry to a General Triangle: The Law of Cosines and the Law of Sines 661
C an acute angle
C
C
AA
a
a
c
B
B
y
y
y
x
(x, y)
(x, y)
(b, 0)
(b, 0)
b
y
x
b
(i)
C an obtuse angle
(ii)
Figure 20.39
The coordinates of C are (0, 0), the coordinates of A are (b,0), and we’ll denote the
coordinates of B by (x, y). Check that regardless of whether we use part (i) or part (ii)
of Figure 20.39, we have
cos C =
x

a
, or, equivalently, x = a cos C
sin C =
y
a
, or, equivalently, y = a sin C.
The distance between points A and B is c.
c = the distance between A and B
=

(x − b)
2
+ (y − 0)
2
(Using the formula for the distance between points)
c =

(a cos C − b)
2
+ (a sin C − 0)
2
Squaring both sides gives
c
2
= (a cos C − b)
2
+ (a sin C)
2
c
2

= a
2
cos
2
C − 2ab cos C + b
2
+ a
2
sin
2
C
c
2
= a
2
(cos
2
C + sin
2
C) − 2ab cos C + b
2
c
2
= a
2
+ b
2
− 2ab cos C.
We have proven the Law of Cosines.
When Might We Use the Law of Cosines?

When we know the lengths of all the sides of a triangle, we can use the Law of Cosines
to find any angle (SSS).
When we know the lengths of two sides and the angle between them, we can use the
Law of Cosines to find the length of the third side of the triangle (SAS).
Scenario II
An astronomer sights a planet in her telescope. She measures the angle of elevation of
her line of sight. She travels 5 miles and repeats the procedure. Can she determine how far
away the planet is using the information she has gathered? Figure 20.40, on the following
page, is not drawn to scale (and we are ignoring the curvature of the earth).
662 CHAPTER 20 Trigonometry—Circles and Triangles
B
C
A
c
b
a = 5
planet
π – θ
2
θ
2
θ
1
θ
1
and θ
2
are the angles of
elevation of her line of sight
from the two locations.

Figure 20.40
From her measurements of θ
1
and θ
2
(see Figure 20.40) she can determine two angles
(and hence the third as well) of the triangle drawn. Knowing the length of the side between
them determines the triangle. The Law of Cosines is not useful in this case, but the Law of
Sines will be. The Law of Sines says
sin A
a
=
sin B
b
=
sin C
c
.Wecanfind A, B, and C and we
know a,soweuse
sin A
a
=
sin C
c
. This allows us to solve for c.
Before proving the Law of Sines we will look at an area formula for a triangle. It is
interesting in its own right, and the Law of Sines follows from it with alacrity.
Obtuse Angles and the Area of an Oblique Triangle
Obtuse Angles
In Section 19.4 we looked at trigonometric functions of any angle by placing the angle in

standard position. If the point (x, y) lies on the terminal side of the angle, we scaled to a
unit circle, obtaining the point

x
√
x
2
+y
2
,
y
√
x
2
+y
2

, and from this we could read off the
cosine and sine values. While this is completely adequate, it can be convenient, when we
have an obtuse angle in the context of a triangle, not to have to refer back to a coordinate
system. Suppose C is an obtuse angle in triangle ABC as shown in Figure 20.41.
B
B = (x, y)
A
C
a
y
c
C
b

|x||x|
y
x
C
a
y
π – C
If C is obtuse, (π – C) is acute
Figure 20.41
We see that a =

x
2
+ y
2
,so
sin C =
y
a
= sin(π − C) and cos C =
x
a
=−cos(π − C).
These are in fact identities, true for any value of C.
sin x = sin(π − x)
cos x =−cos(π − x)
20.5 Applying Trigonometry to a General Triangle: The Law of Cosines and the Law of Sines 663
EXERCISE 20.4 Verify the identities sin x = sin(π − x) and cos x =−cos(π − x) using either the unit
circle or what you know about shifting and flipping the graphs of sine and cosine.
Area of a Triangle

The area of any triangle is given by
area =
1
2
base · height.
B
B
C
C
A
D
D
b
h
h
A
b
(i)
(ii)
Figure 20.42
EXERCISE 20.5 Refer to Figure 20.42 above and verify that the formula area =
1
2
b · h for an oblique triangle
can be derived using areas of right triangles. In Figure 20.42 part (i) verify that
area of ABC = area of ABD + area of BCD =
1
2
b · h.
For Figure 20.42 part (ii) verify that

area of ABC = area of BCD − area of ABD =
1
2
b · h.
Suppose that we know the measures of two sides of a triangle and the included angle.
Let’s say we know the measure of sides a and b and angle C. 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 below.
B(x, y)
B(x, y)
CC
A
xx
b
cca
a
A
b
(i)(ii)
y
y
yy
Figure 20.43
Denote the coordinates of B by (x, y).
sin C =
y
a
, or, equivalently, y = a sin C.
664 CHAPTER 20 Trigonometry—Circles and Triangles
But y = the height of the triangle. Therefore,

area =
1
2
b · h
=
1
2
b · a sin C
area =
1
2
ab sin C.
The area of a triangle with sides of length a and b and included angle C is given by
area of triangle ABC =
1
2
ab sin C.
Notice that we could have given a coordinate-free proof using the result that sin x =
sin(π − x). We simply chose to keep the proof in line with that of the Law of Cosines.
14
Proof of the Law of Sines
To prove the Law of Sines,
sin A
a
=
sin B
b
=
sin C
c

,
we refer to Figure 20.43. Using the formula for the area of a triangle, we see that the area of
triangle ABC is given by
1
2
ab sin C. It is also given by
1
2
ac sin B and
1
2
bc sin A. Therefore,
1
2
ab sin C =
1
2
ac sin B =
1
2
bc sin A.
Multiplying by 2 gives
ab sin C = ac sin B = bc sin A,
and dividing by abc gives
sin C
c
=
sin B
b
=

sin A
a
.
When Might We Use the Law of Sines?
When we know two angles of a triangle and the length of one side, we can use the Law
of Sines to find the length of the other sides (AAS or ASA).
When we know the lengths of two sides of the triangle and know the angle opposite one
of these sides, we can try to find the length of the other side and other angles (SSA).
Recall that this latter case is the ambiguous case.
Figure 20.44 illustrates why the ambiguous case has this name. Sides a and c are of
given lengths and angle C is fixed. In cases (b) and (d) the triangle is completely determined.
In case (c) there are two possible options for the triangle, and in case (a) the triangle is
impossible to construct.
14
The Law of Cosines can be proven in a coordinate-free manner.
20.5 Applying Trigonometry to a General Triangle: The Law of Cosines and the Law of Sines 665
B
C
a
c
(a) (b)
c is too short
to make a triangle
B
A
C
a
c
c is just long
enough for a right triangle

(c)
B
AA
C
a
c
c
two possibilities
(d)
B
A
C
a
c
c ≥ a
only one possibility
The Ambiguous Case: two sides and the angle opposite one of them are known.
Measures of side a and c and angle C are known.
Figure 20.44
How is the ambiguity reflected in the mathematics when we apply the Law of Sines to
the side-side-angle-opposite situation?
In the case in which the triangle is impossible to construct (like case a) we end up with
an equation with no solution, such as sin A = 1.4.
In the case in which the triangle is completely determined or there are two possible
triangles, we will end up with an equation such as
sin A =
1
√
3
.

If we simply say A = arcsin(
1
√
3
), we end up with A =
π
3
= 60
◦
. But we know that A
could be 60
◦
or 120
◦
. We must try both options. Either A = 120
◦
is a possibility or it
is not.
In other words, we know that sin A =
a
c
sin C,butAis not necessarily arcsin

a
c
sin C

.
The problem is that


a
c
sin C

is always positive, so arcsin

a
c
sin C

will give an angle
between 0 and π/2, yet the angle in the triangle in question could be obtuse. We must
consider this possibility independently. Try θ and (π − θ).
This complication did not arise when we used the Law of Cosines to solve for an angle
because the cosine of an obtuse angle is negative.
PROBLEMS FOR SECTION 20.5
1. Find the area of the triangles below.
3
4
10
12
2
2
120°
70°
60°
(a) (b) (c)
666 CHAPTER 20 Trigonometry—Circles and Triangles
2. Find sin θ, cos θ, and tan θ.
3

4
θ
3. What methods would you employ to solve each of the triangles given.
4
4
4
2
3
3
6
10
25°
30°
120°
25°
(d)(c)(b)(a)
4. Solve the triangles in the previous problem. Label the lengths of the unidentified sides
and the measures of the unidentified angles. Round off your answers to the nearest
tenth.
5. After a hurricane a tree is left standing but makes an angle of 10
◦
with its former upright
position. Suppose the tree is tilting away from the sun and casting a shadow 25 feet
long. If the angle of elevation of the sun is 30
◦
, how long is the tree?
10°
6. Three football players, Dante, Anthony, and Cliff, are on a playing field. There are 20
yards between Dante and Anthony, 30 yards between Dante and Cliff, and 40 yards
between Anthony and Cliff. Dante fakes a pass to Anthony but throws the ball instead

to Cliff. Through what angle must he have turned if he was first facing Anthony and is
now facing Cliff?
7. A surveyor is interested in finding the distance across a lake from point P to point
Q. He takes the measurements indicated in the figure on the following page. Find the
distance from P to Q.
20.6 Trigonometric Identities 667
3 mi
2.5 mi
Lake
Q
P
A
30°
8. In a certain region land is valued at $25/square foot. What is the value of a triangular
piece of property whose sides measure 90 feet, 140 feet, and 120 feet?
9. A boat is sailing up the Nile at a steady pace. The river is straight and the boat maintains
a distance of 60 feet from the shoreline. At a given instant the boat’s bearing to a temple
on the shore is N 25
◦
E (see picture). Fifteen minutes later the bearing is N 35
◦
E. How
fast is the boat going? Note: N 25
◦
E means 25
◦
East of due N.
boat
boat
25°

25°
35°
temple
E
N
E
N
North 25° East
N 25° E
10. When designing the one-third-of-a-mile-long Georgia World Congress Center, the
building that housed nearly one-fifth of the events of the 1996 Olympics, engineers
had to take into account the curvature of the earth (Sports Illustrated, August 5, 1996).
Assuming a constant curvature of the earth, how many feet would it curve in one-third
of a mile? In other words, assume a cross-section of the earth is a perfect circle and
draw a tangent line to the curve of this circle at one end of the building. How far away
would the tangent line be from the circle itself at the other end of the building? (Use
3960 miles as the radius of the earth.)
20.6
TRIGONOMETRIC IDENTITIES
The six trigonometric functions are intimately related to one another. These relationships
allow us to transform certain trigonometric expressions into equivalent expressions that may
be easier for us to handle. Knowing some basic trigonometric identities can be useful when
working with trigonometric expressions and solving trigonometric equations. Recall that
an identity is an equation that is true for all values of x for which it makes sense. Some
668 CHAPTER 20 Trigonometry—Circles and Triangles
of these identities, like the Pythagorean identity discussed in Section 19.1, you should
know off the top of your head. Others, which can be interpreted as horizontal shifts of
trigonometric functions, can be quickly gleaned from the graphs of trigonometric functions.
Other relationships are more complex. As a minimum you need to know of their existence
and where you can look them up.

Among the more complex relationships are the addition formulas. It is crucial to realize
that sin(A + B) =sin A + sin B. The following example illustrates this.
◆
EXAMPLE 20.14 Show that sin(A + B) =sin A + sin B.
SOLUTION Any one counterexample will suffice to show that sin(A + B) is not equal to sin A + sin B.
We could let A and B be almost anything. Let A =
π
3
and B =
π
3
.
sin

π
3
+
π
3

?
=
sin

π
3

+ sin

π

3

sin

2π
3

?
=
√
3
2
+
√
3
2
√
3
2
=
√
3
so sin(A + B) =sin A + sin B.
vertical
postion
doesn't
double
v
u
P()

2π
3
P()
π
3
Figure 20.45 ◆
We can verify that sin 2A = 2 sin A intuitively. Think about a Ferris wheel. If you ride
twice the distance, does your vertical position double? No! Does your horizontal position
double if you ride twice the distance? Again, the answer is no.
In fact, it can be shown that
sin(A + B) =sin A cos B + sin B cos A,
cos(A + B) =cos A cos B − sin A sin B.
These are called the addition formulas for sine and cosine. From the addition formulas
we can derive subtraction formulas as well as important double-angle formulas and power-
reducing formulas. Work carefully through Exercises 20.6–20.9 below.
EXERCISE 20.6 Use the addition formulas and the symmetry properties of sine and cosine to verify the
following subtraction formulas.
sin(A − B) =sin A cos B − sin B cos A
cos(A − B) =cos A cos B + sin A sin B
20.6 Trigonometric Identities 669
EXERCISE 20.7 From the addition formulas, derive the following double-angle formulas.
(a) sin 2x = 2 sin x cos x
(b) cos 2x = cos
2
x − sin
2
x = 2 cos
2
x − 1 = 1 − 2 sin
2

x
EXERCISE 20.8 From the formula for cos 2x, derive the following power-reducing formulas.
(a) cos
2
x =
1
2
(1 + cos 2x)
(b) sin
2
x =
1
2
(1 − cos 2x)
EXERCISE 20.9 From the Pythagorean identity sin
2
x + cos
2
x = 1 derive the following identities.
(a) tan
2
x + 1 = sec
2
x
(b) 1 + cot
2
x = csc
2
x
Summary of Important Identities

Definitions tan x =
sin x
cos x
cot x =
cos x
sin x
sec x =
1
cos x
csc x =
1
sin x
Pythagorean sin
2
x + cos
2
x = 1 (20.1)
identities tan
2
x + 1 = sec
2
x Divide both sides of (20.1) by cos
2
x.
1 + cot
2
x = csc
2
x Divide both sides of (20.1) by sin
2

x.
Symmetry-based sin(−x) =−sin x Sine is odd.
cos(−x) = cos x Cosine is even.
Addition/Subtraction sin(A ± B) = sin A cos B ± sin B cos A
formulas cos(A ± B) = cos A cos B ∓ sin A sin B
Double-angle sin 2x = 2 sin x cos x Deduced from addition formulas above
formulas cos 2x = cos
2
x − sin
2
x (20.2) Now substitute (1 − cos
2
x) for sin
2
x.
cos 2x = 2 cos
2
x − 1 (20.3)
Power-reducing cos
2
x =
1
2
(1 + cos 2x) Solve (20.3) for cos
2
x.
identities sin
2
x =
1

2
(1 − cos 2x) Substitute (1 − sin
2
x) for cos
2
x in (20.2)
and solve for sin
2
x.
Periodicity-based sin x = sin(x + k · 2π), k any integer Sine and cosine have period 2π.
tan x = tan(x + kπ), k any integer Tangent has period π .
Horizontal shift Looking at the graphs of sine and cosine and knowing that
they are horizontal translates can provide you with any of these.
Cofunction identities For complementary angles A and B
sin A = cos B
tan A = cot B
sec A = csc B
There are also half-angle formulas for sin(x/2) and cos(x/2) and product identities for
sin A cos B, sin A sin B, and the like. You can look these up in a reference book if you need
them. Those listed above are enough for our purposes.
670 CHAPTER 20 Trigonometry—Circles and Triangles
PROBLEMS FOR SECTION 20.6
1. True or false: If the equation is not always true, give a counterexample.
(a) sin(A − B) =sin(A) − sin(B) (b) cos(A + B) =cos A + cos B
2. Compute the following exactly. Do not use calculator approximations.
(a) cos

π
12


= cos

π
4
−
π
6

(b) sin

−π
12

= sin

π
6
−
π
4

(c) tan

π
12

3. Using the addition formulas and what you know about even and odd trigonometric func-
tions, find expressions for sin(x − 2y) and cos(x − 2y) in terms of cos x, cos y, sin x
and sin y.
4. Match with each of the expressions on the right with the equivalent expression on the

left.
(a) cos
2
x (i) 1 − cos
2
x
(b) sin
2
x (ii) 1 − sin
2
x
(c) tan
2
x (iii)
1
2
−
1
2
cos 2x
(d) sin x (iv) −1 + sec
2
x
(e) cos x (v) 1 − sin
2
(x + π/2)
(f) − cos x (vi) sin(x − π/2)
(vii)
1
2

+
1
2
cos 2x
(viii) cos(x − π/2)
5. Use the addition formulas for sin x and cos x to derive the addition and subtraction
formulas for tan x.
tan(A + B) =
tan A + tan B
1 − tan A tan B
tan(A − B) =
tan A − tan B
1 + tan A tan B
6. Use the addition formula for tan(A + B) to show that
tan 2x =
2 tan x
1 − tan
2
x
.
7. Solve.
(a) cos
2
x − cos 2x = 0 x ∈ (−∞, ∞)
(b) sin x cos x =
√
3 x ∈ [0, 2π]
(c) sin
2
x − cos 2x = 0 x ∈ [0, 2π]

(d) − cos
2
x +
1
2
sin x + 1 = 0 x ∈ [0, 2π]
8. Write a formula for cos 3x entirely in terms of sums and powers of cos x.
9. Use the power-reducing formulas for sin
2
x and cos
2
x to show that
tan
2
x =
1 − cos 2x
1 + cos 2x
.