Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 66 ppsx

20.1 Right-Triangle Trigonometry: The Definitions 631
x
p
A
C
u
v
B = (1, 0)
(0, –1)
(0, 1)
O
(–1, 0)
1
sin x = length of CP
cos x = length of OC
tan x = length of AB
sec x = length of OA
Figure 20.6
Trigonometry of Right Triangles
From times of antiquity, scientists and engineers have used triangle trigonometry to con-
struct edifices and to estimate distances and heights—from estimating the height of the great
Egyptian pyramids in Giza to estimating the sizes and distances of the sun and the moon.
A triangle has three sides and three angles. To solve a triangle means to find measures for
all three sides and all three angles from information given.
In the case of a right triangle, if we know the measure of one of the acute angles, then
we know the measures of all the angles. This, together with the length of any one side,
enables us to solve the triangle. If the lengths of any two particular sides of a right triangle
are known, we can use the Pythagorean Theorem to determine the length of the third and
trigonometry to determine the measure of the angles. Knowing the measures of all three
angles does not determine the triangle, but knowing the length of all three sides does.
5

(Think about why this is so.)
Examples involving solving right triangles often include the terms “angle of elevation”
and “angle of depression.” Angle of elevation refers to the angle from the horizontal up to
an object; angle of depression refers to the angle from the horizontal down to an object.
◆
EXAMPLE 20.1 A little girl flying a kite on a taut 350-foot string asks her father for the height of her kite.
Her father estimates the angle of elevation of the kite to be 55
◦
. Give an estimate for the
height of the kite. Assume that the girl is holding the string 3 feet above the ground and her
father is measuring the angle of elevation from this height.
350 ft.
350 ft.
55°
55°
3 ft. 3 ft.
x
Figure 20.7
5
To “ determine” a triangle means that these specifications determine all three sides and all three angles.
632 CHAPTER 20 Trigonometry—Circles and Triangles
SOLUTION Let x + 3 = the height of the kite.
We’re looking for a trigonometric function that relates x and the known parts of the
triangle. x is the side opposite the 55
◦
angle and we know the length of the hypotenuse.
sin 55
◦
=
x

350
sin(55
◦
) =
x
350
x = 350 · sin(55
◦
)
x ≈ 286.7
The kite is flying at a height of approximately 290 feet.
◆
◆
EXAMPLE 20.2 When the sun is 38
◦
above the horizon, a totem pole casts a shadow of 25 feet. How tall is
the totem pole?
SOLUTION Let T be the height of the totem pole.
Totem
Shadow
38°
25
T
Figure 20.8
Using the diagram, we can write a trigonometric equation that relates the information
we know with the information we want to find.
tan 38
◦
=
T

25
Thus, T = 25 tan 38
◦
≈ 19.5 feet.
◆
◆
EXAMPLE 20.3 A clock tower sitson top of a tall building. From a point 300 feet from the base of the building
the angle of elevation to the base of the clock tower is 40
◦
and the angle of elevation to the
top of the tower is 49
◦
. How tall is the clock tower?
SOLUTION Let x = the height (in feet) of the clock tower.
Let y = the height (in feet) of the building up to the base of the tower.
20.1 Right-Triangle Trigonometry: The Definitions 633
x
y
x + y
40°
40°
9°
300 ft
300
300
49°
Figure 20.9
We’ ll use a trigonometric function that relates the lengths of the sides opposite and adjacent
to the known angle.
Using the smaller triangle we can solve for y.

tan 40
◦
=
y
300
y = 300 tan 40
◦
Using the larger triangle we have
tan 49
◦
=
x + y
300
x + y = 300 tan 49
◦
x = 300 tan 49
◦
− y = 300 tan 49
◦
− 300 tan 40
◦
x = 300(tan 49
◦
− tan 40
◦
)
x ≈ 93.4.
The clock tower is approximately 93.4 feet tall.
◆
PROBLEMS FOR SECTION 20.1

1. Given each triangle below, write expressions for the six trigonometric functions.
(a)
θ
4
3
(b)
θ
Q
1
(c)
θ
R
1
2. As illustrated on the following page, a tree casts a shadow of 20 feet when the sun’s
rays make an angle of 0.7 radians with the ground. How tall is the tree?
634 CHAPTER 20 Trigonometry—Circles and Triangles
ground
.7
3. A 25-foot ladder is leaning against a straight wall. If the base of the ladder is 7 feet
from the wall, what angle is the ladder making with the ground and how high up the
wall does it go?
4. You’re in an apartment looking across a 25-foot boulevard at a building across the way.
The angle of depression to the foot of the building is 15
◦
and the angle of elevation to
the top of the building is 50
◦
. How tall is the building?
5. You are standing on an overpass, 35 feet above street level, waving to a friend who is
at the window of a high-rise dormitory. The angle of depression to the bottom of the

dormitory (on street level) is 15
◦
. The angle of elevation to your friend’s window is
45
◦
. What is your friend’s elevation?
6. We are standing on flat ground in Monument Valley trying to estimate the height of the
edifices. We have surveying equipment and take all of our measurements from a height
of 5 feet. We find the angle of elevation to the top of one structure is 23
◦
. We move 500
feet closer to the structure and find that the angle of elevation is now 29
◦
. How tall is
the structure?
23°
29°
5 ft. 5 ft.
7. Graph each of the following. Be sure to display at least one full period of the function.
For parts (a), (b), and (e), use what you know about graphing
1
f(x)
when given the graph
of f(x).
(a) y = 3 sec x (b) y =−cot x (c) y = tan

x
2

(d) y =−3tan x (e) y = 2 csc x

8. In several ancient civilizations trigonometry was a highly developed field. This can in
part be attributed to ancient astronomers. How could an ancient astronomer who could
measure angles of elevation use trigonometry to estimate the height of the moon?
20.2 Triangles We Know and Love, and the Information They Give Us 635
9. Peter is measuring the height of a church steeple. He stands on level ground 500 feet
from the base of the church and determines that the angle of elevation from the ground
to the base of the steeple is 23
◦
. From the same spot he measures the angle of elevation
to the highest point of the steeple and finds it is 29
◦
.
(a) How high is the church, from the base of the church at ground level to the tip of
the steeple? Give an exact answer and then give a numerical approximation.
(b) How high is the steeple? Give an exact answer and then give a numerical approx-
imation.
20.2 TRIANGLES WE KNOW AND LOVE, AND THE
INFORMATION THEY GIVE US
In Section 20.1 we used the ratios of the lengths of sides of a right triangle to define
sin θ, cos θ, tan θ , and their reciprocals for any acute angle θ . Generally these definitions
aren’t computationally practical if our aim is very accurate computation of the values of
trigonometric functions. We don’t relish the thought of sitting down with a protractor
and ruler, constructing some angle θ, drawing a right triangle, measuring the sides, and
computing the ratio, do we? No! We’d much prefer to whip out a calculator and push some
buttons.
6
In the absence of a calculator but in the presence of a calibrated unit circle we
can use the circle to approximate the value of the trigonometric function, but the degree of
accuracy we can expect is limited. There are, however, two special triangles that are very
useful to us in obtaining the exact values of the trigonometric functions of the angles they

contain, as well as the exact values of trigonometric functions of their close relatives. These
two triangles are triangles all students of trigonometry come to know and love.
The 45
◦
,45
◦
Right Triangle
7
Draw an isosceles right triangle.
8
Its acute angles are equal; both are π/4 radians (or
45
◦
). Let’s label the legs. We can call them anything. Let’s use 1 for simplicity. Using
the Pythagorean Theorem we see that the hypotenuse has length
√
2. From the triangle we
see that
9
sin(π/4) =
1
√
2
, cos(π/4) =
1
√
2
, tan(π/4) = 1.
6
You may wonder “How does a calculator do it? How does it evaluate sin 0.1, for instance?” This is a very good and interesting

question. The short (and somewhat accurate) answer is that the calculator uses a polynomial to approximate sin x near x = 0 and
evaluates that polynomial at x = 0.1. The concept is along the lines of what we do when we take tangent-line approximations, but
it is done with much more sophistication. We will study this in Chapter 30.
7
These triangles tend to be identified by their degree-angle measurements; they are easier to say aloud, and the study of these
triangles is traditionally first undertaken in a trigonometry class as opposed to a calculus class. Radians show their advantage with
calculus, as we will see in the next chapter.
8
An isosceles triangle has two sides of equal length.
9
In some places the dogma is that writing a fraction with a radical in the denominator, like 1/
√
2, is a mortal sin. Relax.
Writing 1/
√
2 doesn’t disturb us. Use 1/
√
2or
√
2/2, whichever makes you happiest.
636 CHAPTER 20 Trigonometry—Circles and Triangles
1
1
π
4
π
4
√2
Figure 20.10
This information could also be obtained from the unit circle. P(π/4) is the point of

intersection of the line u = v and the unit circle u
2
+ v
2
= 1. Substituting v = u into the
equation of the circle we obtain
u
2
+ u
2
= 1
2u
2
= 1
u
2
=
1
2
u =±
1
√
2
.
If u =
1
√
2
then v =
1

√
2
;ifu=−
1
√
2
then v =−
1
√
2
.
v
u
π
4
π
4
(a)
1
P(
(
u
2
+ v
2
= 1
v
u
(b)
π

4
(
5π
4
P(
(
√2
1
(
(
–
√2
1
–
,
√2
1
(
(
√2
1
,
= P(
=
Figure 20.11
At no extra cost we obtain the coordinates of a second point on the unit circle. Using the
symmetry properties of the circle we can identify the coordinates of two more. (See Figure
20.12.) This allows us to compute all the trigonometric functions for any x that takes us
to one of these four points. Combined with the u- and v-intercepts, this gives us a total of
eight points on the unit circle whose exact coordinates we can associate with P(x).From

Figure 20.12, we can read off the following information.
20.2 Triangles We Know and Love, and the Information They Give Us 637
P(x)=

1
√
2
,
1
√
2

for x = π/4 +2πn, P(x)=

−1
√
2
,
−1
√
2

for x = 5π/4 +2πn,
P(x)=

−1
√
2
,
1

√
2

for x = 3π/4 +2πn, P(x)=

1
√
2
,
−1
√
2

for x = 7π/4 +2πn,
where n is any integer.
π
4
7π
4
P(
√2
1
(
(
√2
1
,
P(
) =
√2

1
(
(
√2
1
,
) =
3π
2
P(
) = (0, –1)
π
2
P(
) = (0, 1)
v
u
P(0) = (1, 0)
P(π) = (–1, 0)
–
3π
4
P(
√2
1
(
(
√2
1
,

) =
–
5π
4
P(
√2
1
(
(
√2
1
,
) =
––
Figure 20.12
In general, if we know the coordinates of a point P on the unit circle, we can identify
the coordinates of points Q, R, and S in the other three quadrants. Quadrants are labeled
I, II, III, and IV, working counterclockwise from the first quadrant, where both coordinates
are positive.
The 30
◦
,60
◦
Right Triangle
Draw an equilateral triangle. All its sides are of equal length, so all of its angles are of equal
measure; each angle is π/3 radians, or 60
◦
. This time let’s call the length of each side 2.
(We’ll chop one side in half right away.) Drop a perpendicular to create a right triangle
with acute angles of π/3 and π/6. The perpendicular cuts one of the sides of the original

equilateral triangle in two. We now have a right triangle in which the only missing piece
of information is the length of the side opposite π/3. We find this using the Pythagorean
Theorem.
1
2
+ x
2
= 2
2
x
2
= 3
x =±
√
3
x =
√
3 because x ≥ 0.
638 CHAPTER 20 Trigonometry—Circles and Triangles
π
3
π
3
π
3
π
3
π
6
x

222
11 1
√3
Figure 20.13
Notice that the side opposite the smallest angle is smallest in length; the side opposite
the largest angle is largest in length. From the triangle we see that
sin(π/3) =
√
3
2
cos(π/3) =
1
2
tan(π/3) =
√
3
sin(π/6) =
1
2
cos(π/6) =
√
3
2
tan(π/6) =
1
√
3
π
3
π

3
π
6
2
1
1
1
√3
2
1
2
1
2
1
2
1
2
√3
2
√3
2
√3
v
uu
v
π
3
2
1
2

1
(
(
,
P(
) =
2
√3
π
6
π
6
(
(
,
2
√3
P(
) =
Figure 20.14
We can transport all this information to the unit circle. Scale down the triangle so its
hypotenuse is 1 by cutting all sides in half. We use the symmetry of the circle to obtain
the coordinates of eight points on the circle, as shown in Figure 20.15. On the diagram we
have listed only x-values between 0 and 2π, but infinitely many others can be obtained by
adding multiples of 2π.
20.2 Triangles We Know and Love, and the Information They Give Us 639
2
√3
π
3

2
1
2
1
(
(
,
P(
) =
2
√3
π
6
(
(
,
P(
) =
v
u
1
2
2
√3
2π
3
1
(
(
,

P(
) =
–
2
2
√3
5π
3
1
(
(
,
P(
) =
–
2
2
√3
4π
3
1
(
(
,
P(
) =
––
2
2
√3

5π
6
1
(
(
,
P(
) =
–
2
2
√3
7π
6
1
(
(
,
P(
) =
––
2
2
√3
11π
6
1
(
(
,

P(
) =
–
Figure 20.15
We can determine the coordinates of P(x) for x any integer multiple of
π
6
, x =
nπ
6
where n is an integer. Now we have a host of angles for which we can evaluate trigonometric
functions exactly.
◆
EXAMPLE 20.4 (a) Evaluate cos(−5π/6). (b) Evaluate sin 3π/4.
SOLUTIONS (a) First locate P(−5π/6) on the unit circle. We want the coordinates of P = P(−5π/6).
Approach 1
Find the coordinates of P by using the symmetry of the circle to bounce into the first
quadrant where we have an acute angle and can use the π/6 − π/3 right triangle.
Then return to P .
π
6
π
6
1
2
1
2
√3
v
uu

v
2
1
2
√3
(
(
,
R =
2
1
2
√3
(
(
,
R =
2
1
2
√3
(
(
,
P =
––
(b) (c)
v
u
R

P
(a)
Figure 20.16
640 CHAPTER 20 Trigonometry—Circles and Triangles
Shortcut
We’re looking for the coordinates of P(x) where x is an integer multiple of
π
6
.We
know that a 30
◦
,60
◦
right triangle is involved and, due to the position of P , that
both coordinates are negative. Therefore, the coordinates of P are either

−1
2
,
−
√
3
2

or

−
√
3
2

,
−1
2

. We see from the position of P that the magnitude of the v-coordinate
is less than that of the u-coordinate.
10
Therefore, the coordinates of P are

−
√
3
2
,
−1
2

.
cos

−5π
6

=
−
√
3
2
.
v

u
2
2
√3
–5π
6
–5π
6
1
(
(
,
P(
) =
––
Figure 20.17
(b) To evaluate sin(3π/4), first locate P(3π/4) on the unit circle, then find its coordinates.
Due to its position, we know the coordinates of P(3π/4) are (negative, positive).
We know that a 45
◦
,45
◦
right triangle is involved, so the coordinates are

−1
√
2
,
1
√

2

.
Therefore, sin

3π
4

=
1
√
2
.
v
u
3π
4
3π
4
P(
)
Figure 20.18 ◆
EXERCISE 20.2 Evaluate, giving an exact answer. Check your work by comparing your answer with the
numerical approximation provided by a calculator.
(a) sin(3π/4) (b) sin(−3π/4) (c) tan(7π/6)
(d) cos(−π/3) (e) tan(−π/4)
10
We know that
√
3 > 1.