# Lesson 7 Trigonometric Functions

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```					                               Lesson 7
Trigonometric Functions
PURPOSE
The purpose of this lesson is to introduce you to the six trigonometric functions. These functions
will be used with degree and radian-angle measurements.

OBJECTIVES
After completing this lesson, you should be able to:

find the measurement of an angle in either degrees or radians and find coterminal angles;

find the arc length and area of a sector of a circle and solve problems involving apparent size;

find values of the sine and cosine and solve simple trigonometric equations;

use reference angles to find the value of the sine and cosine and sketch graphs of these
functions;

find values of the tangent, cotangent, secant, and cosecant functions; and

find values of the inverse trigonometric functions.

Textbook, Chapter 7, Sections 7-1 through 7-6 (pages 257–293)

COMMENTARY
Section 7-1: Measurement of Angles pp. 257–262

Trigonometry means "triangle measurement." It deals with angle measures and side lengths of
triangles. A revolution is one complete turn, such as a complete turn of a Ferris wheel or a tire or a
1
steering wheel. There are 360° in one revolution. A degree is 360 of a revolution. There are 60
minutes in a degree and 60 seconds in a minute. The angle measure "30 degrees 40 minutes and 20
seconds" would be written as 30°40 '20".
Lesson 7

Another unit of angle measure other than the degree is the radian. The radian measure of the angle at
the left is s . The radian measure of a complete revolution is 2 ≺ . A complete revolution would
r
make s the circumference of the circle.

Circumference:

= π (diameter)
= π (2 r).

s π (2 r)
=      =       = 2 π.
r    r

Since a revolution is 2 π radians and a revolution is 360°, we can now have a conversion factor
between the two units of measurement:

2π
=   180°
π     ;

2π         π
1 degree =         360°
=   180°
.

Not only can we break the circle into units of 1 degree, but we can divide it into eighths, twelfths,
and sixths evenly. An eighth of a circle is 8, 45° . In radians that is 28π = π 45° is a multiple of π
360
4                      4
2π
radians. A sixth of a circle is     360
6, 60°
.   In radians that is        6
= π . Thus every multiple of 60° is a multiple
3
π                                                                                2π
of   3
radians. A twelfth of a circle is      360
12, 30°
.   In radians that is         12
= π . And—you guessed it—every
6
π
multiple of 30° is a multiple of      6

Now let's go over some angle terminology. An angle in standard position has its vertex at the origin
with its initial ray on the positive x-axis. The initial ray is the ray of the angle that does not rotate.
The terminal ray is the ray of the angle that does rotate. A quadrantal angle has its terminal ray on an
axis. It is always a multiple of 90° or π radians. Coterminal angles have the same terminal ray. An
2
angle of 30° and 390° degrees would be coterminal. The 390° angle is one complete revolution, 360°
, and an additional 30°. Positive angle measures have a terminal ray that rotates counterclockwise.
Negative angle measures have a terminal ray that rotates clockwise.

Example 1: Convert 240° to radians:

π         4π
180°        3

Example 2: Convert −315° to radians:

π         −7 π
180°        4
Lesson 7

4π
Example 3: Convert      3

4 π 180°
×     = 240°.
3   π

Example 4: Convert 80° to radians to the nearest hundredth.

π         4π
80°×          =
180°        9
= 1.396  or  1.4  radians .

Example 5: Convert −1.6 radians to the nearest 10 minutes:

−1.6× 180° = − 91.67°.
π

Convert .67° to minutes. .67×60 = 40.2. The measure is −91°40 '.

Example 6: Find two angles, one positive and one negative, that are coterminal with−270°35 '.

To find a positive coterminal angle, I need to reverse the rotation by adding 360°:

360° + − 270°35 ' = 89°25 '.

To find a negative coterminal angle, I need to continue with the negative rotation:

−360° + − 270°35 ' = − 630°35 '.

Example 7: Let's try problem 28 on page 262 of your text.

(a) 6.5 rpm means 6.5 revolutions per minute. A revolution is 360° so (6.5) (360°) = 2, 340° per
minute.
(b) Convert 2, 340° to radians:
π
2, 340°×           = 13 π
180°

Study Exercises
Complete odd-numbered problems 1–33 in the Written Exercises section on pages 261–262 of your

Section 7-2: Sectors of Circles pp. 263–267

A sector of a circle is like a slice of pie. It is bounded by a central angle, and the arc is intercepted
by that angle. The measurement of the central angle helps you determine the area of the sector. In
the diagram at the right, if the central angle has a measurement of 120°, then the sector has an area
of 1 because 120° is 1 of an entire circle, 360°.
3                   3
Lesson 7

So if the circle has a circumference of 6 cm, the area of the circle is π (6) 2 = 36 π cm, and 1 of that
3
is 12 π cm. The same is true of the length of the intercepted arc. It is     1
3
of the entire circumference.
The circumference of a circle with a radius of 6 cm is 2 π (6) = 12 π cm and 1 of that is 4 π cm. The
3
area of the sector and arc length can be found when the central angle is measured in radians also.
General formulas for both are below. They are also on page 263 of your text. The variable K stands
for the area of the sector, and the variable s stands for the arc length. θ is the measurement of the
central angle.

θ                    θ
Degrees s =
360°
×2 π r K =   360°
×π r 2

Radians s = r θ                K = 1 r2 θ
2

If θ is in radians and the arc length is known, the area of the sector can be found using K = 1 r s.
2

Example 1: Let's try problem 6 on page 265 of your text.

The book gives an area of 90 cm 2 and central angle in radians, 0.2.

1 2
K=        r θ
2
1 2
90 =      r (0.2)
2
1 2
450 =      r
2
900 = r 2
30  cm = r

Now find the arc length:

s = rθ
s = (30) (0.2)
s = 6  cm;

or

1
K=       rs
2
1
90 =      (30) s
2
90 = 15 s
6  cm = s.
Lesson 7

Example 2: Let's try problem 10 on page 265 of your text.

Solve given the perimeter of the sector is 12 cm and the area is 8 cm 2 .

r + r + s = 12
s = 12 − 2 r;
1
K=     rs
2
1
8=       r (12 − 2 r)
2
8 = 6 r − r2
0 = r2 − 6 r + 8
0 = (r − 4) (r − 2)
r = 4;   r = 2.

The possible radii are 4 cm and 2 cm.

An object looks smaller the farther away it is. The angle at which your eyes view it also affects the
appearance of its size. If you watch the sunrise, you will notice the sun looks a lot bigger as it comes
over the horizon than it does when it is up in the sky. The angle of the object above your eye's
horizontal line is called the apparent size. That angle represents the central angle, and the diameter
of the object represents the arc length.

Example 3: Let's try problem 12 on page 265 of your text.

The radius is 5.6×10 7 km. The central angle has a measure of 0 .00012 radians. Find the diameter
and arc length:

s = rθ
s = I5.6×10 7M (0.00012)

s = I5.6×10 7M I1.2×10 −4M

s = 6.72×10 3.

The diameter is approximately 6,720 km.

Example 4: Let's try problem 20 on page 266 of your text.

Given a 20 m arc length and a .005 radians central angle, find the radius and the distance from
lighthouse:

s = rθ
20 = r (.005)
4, 000 = r.
Lesson 7

Originally the ship was 4,000 m from the lighthouse. Now find the distance when the central angle is

20 = r (0.01)
2, 000 = r.

Ten minutes later the ship is 2,000 m from the lighthouse. In 10 minutes the ship traveled 2,000
meters. That is 12,000 meters per hour, or 12 km/h.

Example 5: Let's try problem 24 on page 267 of your text.

(a)
The perimeter of the sector is 20 cm.

s = θ r (arc length equation) and θ = s are used together.
r

2 r + s = 20
s = 20 − 2 r
20 − 2 r
θ=
r

The equation for area using radius and arc length is K = 1 s r.
2

1
K=     (20 − 2 r) r
2
K = 10 r − r 2

(b)
K is the area so K = 10 r − r 2 will give the maximum possible area. Find the vertex of
K (r) = 10 r − r 2:

y = − r 2 + 10 r
y − 25 = − Ir 2 − 10 r + 25M

y − 25 = − (r − 5) 2.

The vertex is (5, 25).

A radius of 5 cm will give a maximum area of 25 cm 2 .

(c)
θ
Use K = 1 r 2 θ for radian measure and K =
2                                      360°
×π r 2 for degree measure.

1 2
25 =     (5) θ
2
2=θ

Lesson 7

θ
25 =          ×π (5) 2
360°
360°
=θ
π
114.5°

Study Exercises
Complete odd-numbered problems 1–25 in the Written Exercises section on pages 264–267 of your

Section 7-3: The Sine and Cosine Functions pp. 268–274

The sine and cosine are two trigonometric functions based on a circle with a radius of 1 and a center
at the origin. The sine function is a ratio of the y-value to the radius. The cosine function is a ratio of

y
sin θ =
r
x
osθ =
r

Because the sine function uses the y -value, the sine function will have negative values in quadrants
III and IV. The cosine function will have negative values in quadrants II and III because it uses the
x-values.

At 90° or π , the radius lies on the y-axis. So the radius and y -coordinate are both r. So sin 90° =
2
1
1
or
π
1. On the other hand, at 90° or 2 , the x-coordinate is zero. So cos 90° = or 0. When would the sine
0
1
have a value of zero? It would have a value of 0 when the y-coordinate is 0. This happens when the
radius lies on the x-axis. At 0° and 180°, the sine is 0. The cos 0° = 1 because the x-coordinate is 1.
The cos 180° = − 1 because the x-coordinate is −1.
Lesson 7

Example 1: Find the value of sin (−90°).

−90° is a clockwise angle of 90°.

1
sin − 90° = −     = −1
1

Example 2: Find the value of cos 3 π .

Remember, 2 π is a full revolution so 3 π would be an extra 180°.

Example 3: Find the quadrant in which the sin θ < 0 and cos θ < 0.

The sine function uses y, and the cosine function uses x. Both are negative in quadrant III.

Example 4: Would sin 74π be positive, negative, or zero?

7π 7
= ×180° = 315°
4  4

315° is in quadrant IV. The y-value in quadrant IV is negative.

The sin 74π is negative.
Lesson 7

Example 5: Given cos θ = − 3 in quadrant II, what is the sin θ ?
5

3 x
cos θ = −       =
5 r

Using the Pythagorean Theorem, a 2 + b 2 = c 2:

9 + b 2 = 25
b 2 = 16
b = 4.

So the y-coordinate is 4. sin θ = 4 .
5

Example 6: Given the following diagram, find sin θ and cos θ .

The value of x is 2 and the value of y is −3. Use the Pythagorean Theorem to find r:

a2 + b2 = c2
x2 + y2 = r2
4 + 9 = r2
√
13 = r.
√         √
y −3         13 −3 13
sin θ = = √ × √ =
r     13     13    13
√      √
x    2       13 2 13
cos θ = = √ × √ =
r    13      13  13

Study Exercises
Complete odd-numbered problems 1–43 in the Written Exercises section on pages 272–274 of your
Lesson 7

Section 7-4: Evaluating and Graphing Sine and Cosine pp. 275–282

A reference angle is the angle measure from the terminal ray to the closest x-axis. The sine and
cosine of the reference angle will have the same absolute value as the given angle.

Example 1: Find the reference angle for 310°. Give sin 310°.

The reference angle is 50°. Use your calculator to find sin 310° and sin 50°. (Make sure your
calculator is set for degrees.) They will have the same absolute values. It makes sense that the actual
value of sin 310° is negative because the terminal ray is in quadrant IV where y is negative.

Example 2: Find the reference angle for −153.2°. Give sin − 153.2° and cos − 153.2°.

180° − 153.2° = 26.8°

The reference angle is 28.6°. Use your calculator to find sin − 153.2° and sin 26.8°. Again their
absolute values are both .4509. The sin − 153.2° is negative because it is in quadrant III. The
absolute values of cos − 153.2° and cos 26.8° are both .8926 and are negative because x is negative

Example 3: Use your calculator to find sin 6.

sin 6 is in radians. Make the appropriate changes in your calculator's mode.

sin 6 = − .2794

From geometry, we know that a right triangle with acute angles 30° and 60° have sides in the ratio of
√
1 : 3 : 2. We also know that a right triangle with acute angles 45° and 45° have sides in the ratio of
√
1 : 1 : 2. So when finding the sine and cosine of angles whose reference angles are 30°, 60°, or 45°
, we can use these geometric facts to find the function values quickly.
Lesson 7

Example 4: Find the exact value of cos π :
6

π    180°
=     = 30°
6      6
√
π     3
cos =       .
6    2

Example 5: Find sin 11 π :
4

11 π 11
= ×180° = 495°.
4   4

The reference angle is 45°.

√   √
11 π    1    2   2
sin      = √ ×√ =
4       2   2  2

Example 6: Let's try problem 26 on page 281 of your text.

To find the distance from Durham, North Carolina, to the North Pole you need to find the length of
the intercepted arc:

θ = 90 ◦ − 36 ◦ = 54 ◦
θ
s=             ×2 π r
360°
54°
s=          ×2 π (3963)
360°
s = 3, 735.

Durham is approximately 3,735 miles from the North Pole.
Lesson 7

Example 7: Let's try problem 30a on page 281 of your text.

We must start by finding the radius of the circle that goes through 45° N latitude. We are still using
3,963 as the radius of the earth:

x
cos 45° =
3963
√               x
2=
3, 963
2, 802 = x.

Now find the circumference of the circle through 45° N latitude :

C = 2π r
C = 2 π (2, 802).

The circumference is 17,605.49. Now divide that by 24 hours. The speed of the earth at 45° N
latitude is 734 mph.

Study Exercises
Complete odd-numbered problems 1–33 in the Written Exercises section on pages 279–281 of your

Section 7-5: The Other Trigonometric Functions pp. 282–286
y
Another trigonometric function is the tangent function. The tangent function is a ratio of x , x ≠ 0. All
three functions have reciprocals.

sine =
y           reciprocal cosecant            r       y≠0
r                                          y

cosine =     x       reciprocal secant =            r       x≠0
r                                      x

tangent =
y   reciprocal cotangent =             x   y≠0
x                                      y

Working with the ratios:
y
sine r ÷ cosine x ;
r

y r         y
r ×x   =    x   = tangent;

y
cosine x ÷ sine r ; and x × y =
r                r
r                x
y   = cotangent.
Lesson 7

This means tangent = sine/cosine and cotangent = cosine/sine. Because the tangent, cosecant, secant,
and cotangent have x or y in the denominator, their graphs will have asymptotes when
¡                     ¢
y = 0 (0, π , 2 π , etc .) or x = 0 π , 32π , 52π , etc . .
2

Example 1: Find the value of tan 175° using its reference angle.

The reference angle of 175° is 5° and it is in quadrant II where x is negative. Since the tangent
y
function is the ratio of x , it will be negative.

tan 175° = − .0875

Example 2: Give the value in radians of x for which sec x is (a) undefined, (b) 0, (c) 1, and (d) −1.

(a) The ratio for secant is r , so when x is 0 it is undefined. The value of x is zero when the
x
π         3π
terminal ray lies on the y-axis. This is at                     2
and    2

(b) The value of r is never zero, so the ratio of r will never be zero.
x

(c) The value of r will be 1 when the terminal ray lies on the x-axis and in quadrant I or IV. This
x
occurs at 0 and 2 π radians
(d) For r to have a value of −1, the terminal ray must lie on the x -axis and be in quadrant II or
x
III. This would occur at π radians.

Example 3: Let's try problem 18 on page 286 of your text.
r                         −5
Keeping in mind that the cosecant is a ratio of y ,                        1
, make a rough graph. Stay within the
π          3π
boundaries of            2
<x<     2
.

Use the Pythagorean Theorem to find the value of x:

a2 + 12 = 52
a 2 + 1 = 25
a 2 = 24
√
a = 2 6;
√
x  is 2 6.
y
sine =   r   = −1
5
√
−2 6
cosine =     x
r   =     5
√              √        √
y             −2 6            2 6        6
tangent =        x   =    −1
√ ×   √         =    24
=   12
−2 6 −2 6
√      √
−2 6
cotangent =          x
y   =    −1
=2 6
√                     √           √
−2 6                −10 6         −5 6
secant =     r
x   =     5√
× √              =     24
=    12
−2 6 −2 6
Lesson 7

Example 4: Let's try problem 22a on page 286 of your text. Use the special angle side ratios you
learned about in Section 7-4 for this problem. You can leave the ratios as is and solve, or you can do
as I did and first convert the ratios to the unit circle, x 2 + y 2 = 1.

1 + cot 2 π , csc 2 π , at π / 2  or  90°r = 1, y = 1, x = 0
2         2

1 + 02 = 12
1=1

1 + cot 2 34π , csc 2 34π , at   3π
4
or 135°
√           √
2        − 2
r = 1,   y =     ,  x =
2          2
⎛  √ ⎞ ⎛    ⎞2
− 2 2
⎝ √ ⎠ ⎝√ 1⎠
1+     2
2      2
2      2

1+1 = 2
2=2

1 + cot 2 76π , csc 2 76π at    7π
6
or 210°
√
−1            3
r = 1,   y =    ,  x = −
2           2
⎛     √   ⎞ Ã             !2
3 2
1+   ⎝−   2 ⎠        1
−1
2
−1
2

1 + 3 = (−2) 2
4=4

Example 5: Let's try problem 28 on page 286 of your text.
Lesson 7

(a)
tan π at   π
radians y = 1,   r = 1, and x = 0. tangent ratio is x , so tan π is undefined.
y
2      2                                                                  2

√                  √
(b) cot 7 π at       7π
radians r = 1,   y =
− 2
,    x =
2
.   cotangent ratio is x , so cot 74π is −1.
4            4                             2                      2                         y

(c) sec (−3 π ) at −3 π radians, y = 0,   r = 1,   x = − 1. secant ratio is r , so sec (−3 π ) is −1.
x
√
(d) csc 7 π at       7π
radians, r = 1,   y =   −1
,     x =
− 3
.    Cosecant ratio is y , so csc 76π = − 2.
r
6            6                             2              2

Study Exercises
Complete odd-numbered problems 1–27 in the Written Exercises section on pages 285–286 of your

Section 7-6: The Inverse Trigonometric Functions pp. 286–291

In this last section, we will review the horizontal-line test for inverse functions. When a horizontal
line crosses the graph more than once, the function does not have an inverse. The graphs of the
trigonometric functions are periodic and continue to infinity. So a horizontal line will cross the graph
several times. In order to work with the inverses of trigonometric functions, a range must be set so
that the inverse will pass the horizontal-line test.

The tangent function has an inverse when the domain is limited to − π < x < π . The sine function has
2       2
an inverse when the domain is limited to − π ≤ x ≤ π . The cosine function has an inverse when the
2       2
domain is limited to 0 ≤ x ≤ π . The inverse functions are used to find an angle measure when the
function value is known. So are you ready to think backwards?
³ √ ´
− 2
Example 1: Find the value of sin −1 2        without using a calculator.

Since we are finding the inverse of the sine, we are looking for an angle measure that gives a sine
√
Keep in mind we are limited to the region between − π and π . The sine ratio
− 2
function value of           2
.                                                  2     2
√                     ³ √ ´
y             − 2                   −1 − 2             7π
is r , so y =    2
and r = 1. Sin     2

¡             ¢
Example 2: Find the value of cos Sin −1 (−0.2) .

Sin −1 (−0.2) is an angle measure that gives the sine function a value of −0.2 and stays in the range
− π < x < π.
2       2

1
−.2 = −
5
Lesson 7

y
sine = r so y = − 1 and r = 5.

√                              √
2 6
The Pythagorean Theorem gives x a value of 2 6. The cosine of this angle is 5 . Find the value of
√                                           ¡             ¢
2 6
5
on your calculator and compare it to cos Sin −1 (−0.2) on your calculator. They are very
close.

Example 3: Let's try problem 18 on page 290 of your text.

Pick some ordered pairs.

x        cos −1 x         x           cos −1( − x)
π                            π
0        2                0           2
√                             √
2      π                 − 2        3π
2       4                  2          4

1        0                −1          π

In each case when the value of the functions are added, the sum is π . The domain is {x| − 1 ≤ x ≤ 1}.
The range is y = π .

Example 4: Let's try problem 28 on page 291 of your text.

The sin −1 x has a domain of − π ≤ x ≤ π . Both will have the same reference angle, but sin −1 x will
2       2
be positive and sin −1 − x will be negative. Their sum will be zero.
Lesson 7

Study Exercises
Complete odd-numbered problems 1–27 in the Written Exercises section on pages 289–291 of your