6
Trigonometric
Functions
Length of Day Revisited
The length of a day depends upon the day of the year as well as the latitude
of the location. Latitude gives the location of a point on Earth north or
south of the equator. In Chapter 4, we found a model that describes the
relation between the length of day and latitude for a specific day of the
year. In the Internet Project at the end of this chapter, we will find a model
that describes the relation between the length of day and day of the year
for a specific latitude.
—See the Internet-based Chapter Project I—
A Look Back
In Chapter 2, we began our discussion of functions. We defined domain and
range and independent and dependent variables; we found the value of a
function and graphed functions. We continued our study of functions by listing
properties that a function might have, like being even or odd, and we created a
library of functions, naming key functions and listing their properties, including
the graph.
A Look Ahead
In this chapter we define the trigonometric functions, six functions that have
wide application. We shall talk about their domain and range, see how to find
values, graph them, and develop a list of their properties.
There are two widely accepted approaches to the development of the
trigonometric functions: one uses right triangles; the other uses circles,
especially the unit circle. In this book, we develop the trigonometric functions
using the unit circle. In Chapter 8, we present right triangle trigonometry.
Outline
6.1
6.2
6.3
6.4
6.5
6.6
Angles and Their Measure
Trigonometric Functions: Unit Circle
Approach
Properties of the Trigonometric
Functions
Graphs of the Sine and Cosine
Functions
Graphs of the Tangent, Cotangent,
Cosecant, and Secant Functions
Phase Shift; Sinusoidal Curve Fitting
Chapter Review
Chapter Test
Cumulative Review
Chapter Projects
385
M06_SULL1772_10_GE_C06.indd 385
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386
CHAPTER 6 Trigonometric Functions
6.1 Angles and Their Measure
PREPARING FOR THIS SECTION Before getting started, review the following:
• Area and Circumference of a Circle (Appendix A,
Section A.2, p. 960)
• Uniform Motion (Appendix A, Section A.8,
pp. 1009–1011)
Now Work the ‘Are You Prepared?’ problems on page 395.
Objectives 1 Convert between Decimals and Degrees, Minutes, Seconds
Measures for Angles (p. 388)
2 Find the Length of an Arc of a Circle (p. 389)
3 Convert from Degrees to Radians and from Radians to Degrees (p. 390)
4 Find the Area of a Sector of a Circle (p. 393)
5 Find the Linear Speed of an Object Traveling in Circular Motion (p. 394)
V
Ray
Line
Figure 1 A ray or half-line
A ray, or half-line, is that portion of a line that starts at a point V on the line and
extends indefinitely in one direction. The starting point V of a ray is called its vertex.
See Figure 1.
If two rays are drawn with a common vertex, they form an angle. We call one
ray of an angle the initial side and the other the terminal side. The angle formed is
identified by showing the direction and amount of rotation from the initial side to
the terminal side. If the rotation is in the counterclockwise direction, the angle is
positive; if the rotation is clockwise, the angle is negative. See Figure 2.
de
i
al s
min
e
min
Ter a
id
al s
Figure 2
Ter
b
Vertex Initial side
min
Ter
Vertex
(a) Counterclockwise
rotation
Positive angle
Initial side
g
(b) Clockwise rotation
Negative angle
Vertex
ide
al s
Initial side
(c) Counterclockwise
rotation
Positive angle
Lowercase Greek letters, such as a (alpha), b (beta), g (gamma), and u (theta),
will often be used to denote angles. Notice in Figure 2(a) that the angle a is
positive because the direction of the rotation from the initial side to the terminal side
is counterclockwise. The angle b in Figure 2(b) is negative because the rotation is
clockwise. The angle g in Figure 2(c) is positive. Notice that the angle a in Figure 2(a)
and the angle g in Figure 2(c) have the same initial side and the same terminal side.
However, a and g are unequal, because the amount of rotation required to go from
the initial side to the terminal side is greater for angle g than for angle a.
An angle u is said to be in standard position if its vertex is at the origin of a rectangular
coordinate system and its initial side coincides with the positive x-axis. See Figure 3.
y
Terminal side
Vertex
y
u
Initial side
x
Terminal side
Figure 3
Standard position of an angle
M06_SULL1772_10_GE_C06.indd 386
(a) u is in standard position;
u is positive
Vertex
Initial side x
u
(b) u is in standard position;
u is negative
11/03/16 1:45 PM
Section 6.1 Angles and Their Measure
387
When an angle u is in standard position, the terminal side will lie either in a
quadrant, in which case we say that u lies in that quadrant, or the terminal side will
lie on the x-axis or the y-axis, in which case we say that u is a quadrantal angle. For
example, the angle u in Figure 4(a) lies in quadrant II, the angle u in Figure 4(b) lies
in quadrant IV, and the angle u in Figure 4(c) is a quadrantal angle.
II
I
y
II
I
y
y
u
u
x
Figure 4
III
IV
(a) u lies in quadrant II
x
x
III
IV
(b) u lies in quadrant IV
(c) u is a quadrantal angle
u
Angles are measured by determining the amount of rotation needed for the
initial side to become coincident with the terminal side. The two commonly used
measures for angles are degrees and radians.
Degrees
Historical Note One counterclockwise
rotation was said to measure 360°
because the Babylonian year had
360 days.
■
The angle formed by rotating the initial side exactly once in the counterclockwise
direction until it coincides with itself (1 revolution) is said to measure 360 degrees,
1
abbreviated 360°. One degree, 1°, is
revolution. A right angle is an angle that
360
1
measures 90°, or revolution; a straight angle is an angle that measures 180°, or
4
1
revolution. See Figure 5. As Figure 5(b) shows, it is customary to indicate a right
2
angle by using the symbol .
Terminal side
Figure 5
Initial side
Vertex
(a) 1 revolution
counterclockwise, 360°
Terminal
side
Vertex Initial side
1
(b) right angle, –4 revolution
counterclockwise, 90°
Terminal side Vertex Initial side
1
(c) straight angle, –2 revolution
counterclockwise, 180°
It is also customary to refer to an angle that measures u degrees as an angle of
u degrees.
Exampl e 1
Drawing an Angle
Draw each angle.
(a) 45° (b) - 90° (c) 225° (d) 405°
Solution
(a) An angle of 45° is
Te
rm
in
al
sid
e
See Figure 6.
45°
Vertex Initial side
Figure 6 45° angle
M06_SULL1772_10_GE_C06.indd 387
1
of a right angle.
2
1
revolution in
4
the clockwise direction. See Figure 7.
(b) An angle of - 90° is
Vertex
Initial side
Terminal
side
90°
Figure 7 −90° angle
11/03/16 1:45 PM
388
CHAPTER 6 Trigonometric Functions
(c) An angle of 225° consists of a
rotation through 180° followed by a
rotation through 45°. See Figure 8.
(d) An angle of 405° consists of
1 revolution (360°) followed by a
rotation through 45°. See Figure 9.
Te
rm
in
al
sid
e
225°
Initial side
Vertex
al
sid
e
405°
Initial side
Te
rm
in
Vertex
Figure 9 405° angle
Figure 8 225° angle
•
Now Work p r o b l e m 1 1
1 Convert between Decimals and Degrees, Minutes, Seconds
Measures for Angles
Although subdivisions of a degree may be obtained by using decimals, the terms
1
minute and second are also used. One minute, denoted by 1′, is defined as
degree.
60
1
One second, denoted by 1″, is defined as
minute, or equivalently as,
60
1
degree. An angle of, say, 30 degrees, 40 minutes, 10 seconds is written
3600
compactly as 30°40′10″. To summarize:
Comment Graphing calculators (and
some scientific calculators) have the
ability to convert from degrees, minutes,
seconds to decimal form, and vice versa.
Consult your owner’s manual.
■
Exampl e 2
1 counterclockwise revolution = 360°
1° = 60′
1′ = 60″
(1)
It is sometimes necessary to convert from the degrees, minutes, seconds notation
1D°M′S″2 to a decimal form, and vice versa.
Converting between Degrees, Minutes, Seconds,
and Decimal Forms
(a) Convert 50°6′21″ to a decimal in degrees. Round the answer to four decimal
places.
(b) Convert 21.256° to the D°M′S″ form. Round the answer to the nearest second.
Solution
(a) Because 1′ = a
1 5
1 =
1 1 5
b and 1″ = a b = a # b , convert as follows:
60
60
60 60
50°6′21″ = 50° + 6′ + 21″ = 50° + 6 # 1′ + 21 # 1″
= 50° + 6 # a
1 5
1 1 5
b + 21 # a # b
60
60 60
Convert minutes and
seconds to degrees.
≈ 50° + 0.1° + 0.0058°
= 50.1058°
M06_SULL1772_10_GE_C06.indd 388
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Section 6.1 Angles and Their Measure
389
(b) Because 1° = 60′ and 1′ = 60″, proceed as follows:
21.256° = 21° + 0.256°
= 21° + (0.256) # 1°
= 21° + 10.2562 160′2
Convert fraction of degree to
minutes; 1° = 60′.
= 21° + 15.36′
= 21° + 15′ + 0.36′
= 21° + 15′ + (0.36) # 1′
= 21° + 15′ + 10.362 160″2
Convert fraction of minute to
seconds; 1′ = 60″.
≈ 21°15′22″
Round to the nearest second.
= 21° + 15′ + 21.6″
•
Now Work p r o b l e m s 2 3 a n d 2 9
In many applications, such as describing the exact location of a star or the
precise position of a ship at sea, angles measured in degrees, minutes, and even seconds
are used. For calculation purposes, these are transformed to decimal form. In other
applications, especially those in calculus, angles are measured using radians.
Radians
Ter
mi
nal
si
de
Ter
mi
nal
si
de
A central angle is a positive angle whose vertex is at the center of a circle. The rays of
a central angle subtend (intersect) an arc on the circle. If the radius of the circle is r
and the length of the arc subtended by the central angle is also r, then the measure
of the angle is 1 radian. See Figure 10(a).
For a circle of radius 1, the rays of a central angle with measure 1 radian subtend
an arc of length 1. For a circle of radius 3, the rays of a central angle with measure
1 radian subtend an arc of length 3. See Figure 10(b).
r
1 radian
Figure 10
s
u
u1
r
s1
u
s
Figure 11
=
u1
s1
M06_SULL1772_10_GE_C06.indd 389
3
r
r
1
Initial side
Initial side
1
1 radian
(a)
3
(b)
2 Find the Length of an Arc of a Circle
Now consider a circle of radius r and two central angles, u and u1, measured in radians.
Suppose that these central angles subtend arcs of lengths s and s1, respectively, as
shown in Figure 11. From geometry, the ratio of the measures of the angles equals
the ratio of the corresponding lengths of the arcs subtended by these angles; that is,
u
s
=
s1
u1
(2)
11/03/16 1:45 PM
390
CHAPTER 6 Trigonometric Functions
Suppose that u1 = 1 radian. Refer again to Figure 10(a). The length s1 of the arc
subtended by the central angle u1 = 1 radian equals the radius r of the circle. Then
s1 = r, so equation (2) reduces to
u
s
=
r
1
Theorem
(3)
or s = ru
Arc Length
For a circle of radius r, a central angle of u radians subtends an arc whose
length s is
(4)
s = ru
NOTE Formulas must be consistent with regard to the units used. In equation (4), we write
s = ru
To see the units, however, we must go back to equation (3) and write
s length units
u radians
=
1 radian
r length units
u radians
s length units = r length units
1 radian
The radians divide out, leaving
s length units = (r length units)u
s = ru
where u appears to be “dimensionless” but, in fact, is measured in radians. So, in using the formula
s = r u, the dimension for u is radians, and any convenient unit of length (such as inches or meters)
may be used for s and r.
■
Exampl e 3
Finding the Length of an Arc of a Circle
Find the length of the arc of a circle of radius 2 meters subtended by a central angle
of 0.25 radian.
Solution
Use equation (4) with r = 2 meters and u = 0.25. The length s of the arc is
•
s = ru = 210.252 = 0.5 meter
Now Work p r o b l e m 7 1
3 Convert from Degrees to Radians and from Radians to Degrees
s 2 pr
1 revolution
r
With two ways to measure angles, it is important to be able to convert from one to
the other. Consider a circle of radius r. A central angle of 1 revolution will subtend
an arc equal to the circumference of the circle (Figure 12). Because the circumference
of a circle of radius r equals 2pr, we substitute 2pr for s in equation (4) to find that,
for an angle u of 1 revolution,
s = ru
2pr = ru
Figure 12
1 revolution = 2p radians
u = 2p radians
u = 1 revolution; s = 2pr
Solve for u.
From this, we have
1 revolution = 2p radians
M06_SULL1772_10_GE_C06.indd 390
(5)
11/03/16 1:45 PM
Section 6.1 Angles and Their Measure
391
Since 1 revolution = 360°, we have
360° = 2p radians
Dividing both sides by 2 yields
(6)
180° = p radians
Divide both sides of equation (6) by 180. Then
1 degree =
p
radian
180
Divide both sides of (6) by p. Then
180
degrees = 1 radian
p
We have the following two conversion formulas:*
1 degree =
Exampl e 4
p
radian
180
1 radian =
180
degrees
p
(7)
Converting from Degrees to Radians
Convert each angle in degrees to radians.
(a) 60° (b) 150° (c) - 45° (d) 90° (e) 107°
Solution
(a) 60° = 60 # 1 degree = 60 #
p
p
radian = radians
180
3
p
5p
(b) 150° = 150 # 1° = 150 #
radian =
radians
180
6
p
p
(c) - 45° = - 45 #
radian = - radian
180
4
p
p
(d) 90° = 90 #
radian = radians
180
2
p
(e) 107° = 107 #
radian ≈ 1.868 radians
180
•
Example 4, parts (a)–(d), illustrates that angles that are “nice” fractions of a
revolution are expressed in radian measure as fractional multiples of p, rather than as
p
decimals. For example, a right angle, as in Example 4(d), is left in the form radians,
2
p
3.1416
which is exact, rather than using the approximation
≈
= 1.5708 radians.
2
2
When the fractions are not “nice,” we use the decimal approximation of the angle,
as in Example 4(e).
Now Work p r o b l e m s 3 5 a n d 6 1
*Some students prefer instead to use the proportion
given, and solve for the measurement sought.
M06_SULL1772_10_GE_C06.indd 391
Degree
180°
=
Radian
, then substitute for what is
p
11/03/16 1:45 PM
392
CHAPTER 6 Trigonometric Functions
E xam pl e 5
Converting Radians to Degrees
Convert each angle in radians to degrees.
p
radian
6
7p
(d)
radians
3
(a)
Solution
(b)
3p
radians
2
(c) -
(e) 5 radians
(a)
p
p 180
p
degrees = 30°
radian = # 1 radian = #
6
6 p
6
(b)
3p # 180
3p
degrees = 270°
radians =
p
2
2
(c) (d)
3p
radians
4
3p
3p # 180
degrees = - 135°
radians = p
4
4
7p # 180
7p
degrees = 420°
radians =
p
3
3
(e) 5 radians = 5 #
180
degrees ≈ 286.48°
p
•
Now Work p r o b l e m 4 7
Table 1 lists the degree and radian measures of some commonly encountered
angles. You should learn to feel equally comfortable using either measure for these
angles.
Table 1
Example 6
Degrees
0°
30°
45°
60°
90°
120°
135°
150°
180°
Radians
0
p
6
p
4
p
3
p
2
2p
3
3p
4
5p
6
p
Degrees
210°
225°
240°
270°
300°
315°
330°
360°
Radians
7p
6
5p
4
4p
3
3p
2
5p
3
7p
4
11p
6
2p
Finding the Distance between Two Cities
The latitude of a location L is the measure of the angle formed by a ray drawn
from the center of Earth to the equator and a ray drawn from the center of Earth
to L. See Figure 13(a). Sioux Falls, South Dakota, is due north of Dallas, Texas. Find
the distance between Sioux Falls (43° 33′ north latitude) and Dallas (32° 46′ north
latitude). See Figure 13(b). Assume that the radius of Earth is 3960 miles.
North Pole
North Pole
L
Sioux Falls
43°33'
Dallas
u°
Equator
South Pole
Figure 13
M06_SULL1772_10_GE_C06.indd 392
(a )
Equator
32°46'
South Pole
(b)
11/03/16 1:45 PM
Section 6.1 Angles and Their Measure
Solution
393
The measure of the central angle between the two cities is 43° 33′ - 32° 46′ = 10° 47′.
Use equation (4), s = ru. But remember first to convert the angle of 10° 47′ to radians.
u = 10°47′ ≈ 10.7833° = 10.7833 #
c
47′ = 47 a
p
radian ≈ 0.188 radian
180
1 °
b
60
Use u = 0.188 radian and r = 3960 miles in equation (4). The distance between the
two cities is
s = ru = 3960 # 0.188 ≈ 744 miles
Note If the measure of an angle is given
as 5, it is understood to mean 5 radians;
if the measure of an angle is given as 5°,
it means 5 degrees.
■
•
When an angle is measured in degrees, the degree symbol will always be shown.
However, when an angle is measured in radians, the usual practice is to omit the word
p
p
radians. So if the measure of an angle is given as , it is understood to mean radian.
6
6
Now Work p r o b l e m 1 0 7
A
4 Find the Area of a Sector of a Circle
u
r
Figure 14 Sector of a Circle
A
u
Consider a circle of radius r. Suppose that u, measured in radians, is a central angle
of this circle. See Figure 14. We seek a formula for the area A of the sector (shown
in blue) formed by the angle u.
Now consider a circle of radius r and two central angles u and u1, both measured
in radians. See Figure 15. From geometry, we know that the ratio of the measures
of the angles equals the ratio of the corresponding areas of the sectors formed by
these angles. That is,
u
A
=
u1
A1
Suppose that u1 = 2p radians. Then A1 = area of the circle = pr 2. Solving
for A, we find
r
A = A1
u1
A1
Figure 15
u
u
1
= pr 2
= r 2u
u1
2p
2
c
A1 = pr 2
u1 = 2p
u
A
=
u1
A1
Theorem
Area of a Sector
The area A of the sector of a circle of radius r formed by a central angle
of u radians is
A =
Exampl e 7
Solution
1 2
r u
2
(8)
Finding the Area of a Sector of a Circle
Find the area of the sector of a circle of radius 2 feet formed by an angle of 30°.
Round the answer to two decimal places.
p
Use equation (8) with r = 2 feet and u = 30° = radian. [Remember, in equation (8),
6
u must be in radians.]
1 2
1
p
p
r u = 122 2 =
≈ 1.05
2
2
6
3
The area A of the sector is 1.05 square feet, rounded to two decimal places.
A =
Now Work p r o b l e m 7 9
M06_SULL1772_10_GE_C06.indd 393
•
11/03/16 1:45 PM
394
CHAPTER 6 Trigonometric Functions
5 Find the Linear Speed of an Object Traveling in Circular Motion
Earlier we defined the average speed of an object as the distance traveled divided
by the elapsed time. For motion along a circle, we distinguish between linear speed
and angular speed.
Definition
Time t
s
Suppose that an object moves around a circle of radius r at a constant speed.
If s is the distance traveled in time t around this circle, then the linear speed v
of the object is defined as
u
r
v =
t0
Figure 16 v =
s
t
(9)
As this object travels around the circle, suppose that u (measured in radians) is
the central angle swept out in time t. See Figure 16.
s
t
Definition
The angular speed v (the Greek letter omega) of this object is the angle u
(measured in radians) swept out, divided by the elapsed time t; that is,
v =
u
t
(10)
Angular speed is the way the turning rate of an engine is described. For example,
an engine idling at 900 rpm (revolutions per minute) is one that rotates at an angular
speed of
revolutions
revolutions #
radians
radians
= 900
2p
= 1800p
minute
minute
revolution
minute
There is an important relationship between linear speed and angular speed:
900
linear speed = v =
c
(9)
s
ru
u
=
= ra b = r # v
t
t
t
c
c
s = ru
(10)
v = rv
(11)
where v is measured in radians per unit time.
s
(the linear speed) has the
t
dimensions of length per unit of time (such as feet per second or miles per hour),
r (the radius of the circular motion) has the same length dimension as s, and v (the
angular speed) has the dimensions of radians per unit of time. If the angular speed is
given in terms of revolutions per unit of time (as is often the case), be sure to convert
it to radians per unit of time using the fact that 1 revolution = 2p radians before
attempting to use equation (11).
When using equation (11), remember that v =
Exampl e 8
Finding Linear Speed
A child is spinning a rock at the end of a 2-foot rope at a rate of 180 revolutions per
minute (rpm). Find the linear speed of the rock when it is released.
Solution
Look at Figure 17. The rock is moving around a circle of radius r = 2 feet. The
angular speed v of the rock is
v = 180
M06_SULL1772_10_GE_C06.indd 394
revolutions
revolutions #
radians
radians
= 180
2p
= 360p
minute
minute
revolution
minute
11/03/16 1:45 PM
Section 6.1 Angles and Their Measure
395
From equation (11), the linear speed v of the rock is
r=2
v = rv = 2 feet # 360p
radians
feet
feet
= 720p
≈ 2262
minute
minute
minute
The linear speed of the rock when it is released is 2262 ft/min ≈ 25.7 mi/h.
•
Now Work p r o b l e m 9 9
Figure 17
Historical Feature
T
rigonometry was developed by Greek astronomers, who regarded
the sky as the inside of a sphere, so it was natural that triangles
on a sphere were investigated early (by Menelaus of Alexandria
about ad 100) and that triangles in the plane were studied much later.
The first book containing a systematic treatment of plane and spherical
trigonometry was written by the Persian astronomer Nasir Eddin (about
ad 1250).
Regiomontanus (1436–1476) is the person most responsible for
moving trigonometry from astronomy into mathematics. His work
was improved by Copernicus (1473–1543) and Copernicus’s student
Rhaeticus (1514–1576). Rhaeticus’s book was the first to define the
six trigonometric functions as ratios of sides of triangles, although
he did not give the functions their present names. Credit for this is
due to Thomas Finck (1583), but Finck’s notation was by no means
universally accepted at the time. The notation was finally stabilized by the
textbooks of Leonhard Euler (1707–1783).
Trigonometry has since evolved from its use by surveyors, navigators,
and engineers to present applications involving ocean tides, the rise and
fall of food supplies in certain ecologies, brain wave patterns, and many
other phenomena.
6.1 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. What is the formula for the circumference C of a circle of
radius r? What is the formula for the area A of a circle of
radius r? (p. 960)
2. If a particle has a speed of r feet per second and travels a
distance d (in feet) in time t (in seconds), then d =
.
(pp. 1009–1011)
Concepts and Vocabulary
3. An angle u is in
if its vertex is at the
origin of a rectangular coordinate system and its initial side
coincides with the positive x-axis.
4. A
center of a circle.
is a positive angle whose vertex is at the
5. If the radius of a circle is r and the length of the arc
subtended by a central angle is also r, then the measure of
the angle is 1
.
(a) degree
7. 180° =
radians
p
3p
(a)
(b) p (c)
(d) 2p
2
2
8. An object travels around a circle of radius r with constant
speed. If s is the distance traveled in time t around the circle
and u is the central angle (in radians) swept out in time t,
then the linear speed of the object is v =
speed of the object is v =
(b) minute (c) second (d) radian
6. On a circle of radius r, a central angle of u radians subtends
an arc of length s =
; the area of the sector formed by
this angle u is A =
and the angular
.
9. True or False The angular speed v of an object traveling
around a circle of radius r is the angle u (measured in radians)
swept out, divided by the elapsed time t.
10. True or False For circular motion on a circle of radius r,
linear speed equals angular speed divided by r.
.
Skill Building
In Problems 11–22, draw each angle.
11. 30°
17.
4p
3
12. 60°
18.
3p
4
13. - 120°
19. -
2p
3
14. 135°
20. -
p
6
15. 540°
21.
21p
4
16. 450°
22.
16p
3
In Problems 23–28, convert each angle to a decimal in degrees. Round your answer to two decimal places.
23. 40°10′25″
M06_SULL1772_10_GE_C06.indd 395
24. 61°42′21″
25. 73°40′40″
26. 50°14′20″
27. 98°22′45″
28. 9°9′9″
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396
CHAPTER 6 Trigonometric Functions
In Problems 29–34, convert each angle to D°M′S″ form. Round your answer to the nearest second.
29. 40.32°
30. 61.24°
31. 29.411°
32. 18.255°
33. 44.01°
34. 19.99°
In Problems 35–46, convert each angle in degrees to radians. Express your answer as a multiple of p.
35. 30°
36. 120°
37. 330°
38. 240°
39. - 30°
40. - 60°
41. 270°
42. 180°
43. - 225°
44. - 135°
45. - 180°
46. - 90°
In Problems 47–58, convert each angle in radians to degrees.
p
5p
2p
5p
p
48.
49. 50. 51. 4p
52.
3
6
3
4
2
5p
p
p
3p
p
53.
54.
15°
55. - p
56. 57. 58. 6
12
12
2
4
In Problems 59–64, convert each angle in degrees to radians. Express your answer in decimal form, rounded to two decimal places.
47.
59. 73°
60. 17°
61. - 40°
62. - 51°
63. 350°
64. 125°
In Problems 65–70, convert each angle in radians to degrees. Express your answer in decimal form, rounded to two decimal places.
65. 0.75
66. 3.14
67. 3
68. 2
70. 6.32
69. 2 2
In Problems 71–78, s denotes the length of the arc of a circle of radius r subtended by the central angle u. Find the missing quantity. Round
answers to three decimal places.
71. r = 10 meters, u =
1
radian, s = ?
2
72. r = 6 feet, u = 2 radians, s = ?
1
radian, s = 6 centimeters, r = ?
4
75. r = 6 meters, s = 8 meters, u = ?
1
radian, s = 2 feet, r = ?
3
76. r = 5 miles, s = 3 miles, u = ?
73. u =
74. u =
77. r = 3 meters, u = 120°, s = ?
.
78. r = 2 inches, u = 30°, s = ?
In Problems 79–86, A denotes the area of the sector of a circle of radius r formed by the central angle u. Find the missing quantity. Round
answers to three decimal places.
79. r = 10 meters, u =
1
radian, A = ?
2
80. r = 6 feet, u = 2 radians, A = ?
1
radian, A = 6 square centimeters, r = ?
4
83. r = 6 meters, A = 8 square meters, u = ?
1
radian, A = 2 square feet, r = ?
3
84. r = 5 miles, A = 3 square miles, u = ?
81. u =
82. u =
85. r = 3 meters, u = 120°, A = ?
86. r = 2 inches, u = 30°, A = ?
In Problems 87–90, find the length s and area A. Round answers to three decimal places.
87.
88.
π
6
π A
A s
4m
s
3
2 ft
89.
A s
50˚
9 cm
90.
A s
70˚
12 yd
Applications and Extensions
91. Movement of a Minute Hand The minute hand of a clock is
4 inches long. How far does the tip of the minute hand move
in 20 minutes?
11
12
2
10
3
8
7
M06_SULL1772_10_GE_C06.indd 396
93. Area of a Sector Find the area of the sector of a circle of
radius 5 meters formed by an angle of 90°.
1
9
4
6
5
92. Movement of a Pendulum A pendulum swings through an
angle of 20° each second. If the pendulum is 40 inches long,
how far does its tip move each second? Round answers to
two decimal places.
94. Area of a Sector Find the area of the sector of a circle of
radius 3 centimeters formed by an angle of 60°. Round the
answer to two decimal places.
11/03/16 1:45 PM
Section 6.1 Angles and Their Measure
95. Watering a Lawn A
water sprinkler sprays
water over a distance
of 36 feet while rotating through an angle
of 150°. What area of
lawn receives water?
397
on this bicycle, through how many revolutions per minute
are the wheels turning?
150˚
36 ft
96. Designing a Water Sprinkler An engineer is asked to design
a water sprinkler that will cover a field of 100 square yards
that is in the shape of a sector of a circle of radius 15 yards.
Through what angle should the sprinkler rotate?
97. Windshield Wiper The windshield of a car has a total length
of arm and blade of 9 inches, and rotates back and forth
through an angle of 84°. What is the area of the portion of
the windshield cleaned by the 7-in wiper blade?
9 in.
7 in.
106. Car Wheels The radius of each wheel of a car is 15 inches.
If the wheels are turning at the rate of 3 revolutions per
second, how fast is the car moving? Express your answer in
inches per second and in miles per hour.
In Problems 107–110, the latitude of a location L is the angle
formed by a ray drawn from the center of Earth to the Equator
and a ray drawn from the center of Earth to L. See the figure.
North Pole
L
u
Equator
98. Windshield Wiper The arm and blade of a windshield wiper
have a total length of 30 inches. If the blade is 24 inches
long and the wiper sweeps out an angle of 125°, how much
window area can the blade clean?
99. Motion on a Circle An object is traveling around a circle
with a radius of 12 centimeters. If in 40 seconds a central
1
angle of radian is swept out, what are the linear and angu6
lar speeds of the object?
100. Motion on a Circle An object is traveling around a circle
with a radius of 2 meters. If in 20 seconds the object travels
5 meters, what is its angular speed? What is its linear speed?
101. Amusement Park Ride A gondola on an amusement park
ride spins at a speed of 11 revolutions per minute. If the gondola is 23 feet from the ride’s center, what is the linear speed
of the gondola in miles per hour?
102. Amusement Park Ride A centrifugal force ride, similar to
the Gravitron, spins at a speed of 22 revolutions per minute. If
the diameter of the ride is 13 meters, what is the linear speed
of the passengers in kilometers per hour?
103. Blu-ray Drive A drive has a maximum speed of 10,000 revolutions per minute. If a disc has a diameter of 14 cm, what is
the linear speed, in km/h, of a point 6 cm from the center if
the disc is spinning at a rate of 8000 revolutions per minute?
104. DVD Drive A DVD drive has a maximum speed of 7200
revolutions per minute. If a DVD has a diameter of 12 cm,
what is the linear speed, in km/h, of a point 5 cm from the
disc’s center if it is spinning at a rate of 5400 revolutions per
minute?
105. Bicycle Wheels The diameter of each wheel of a bicycle is
26 inches. If you are traveling at a speed of 40 miles per hour
M06_SULL1772_10_GE_C06.indd 397
South Pole
107. Distance between Cities City A is due north of City B. Find
the distance between City A (37°7′ north latitude) and City
B (21°55′ north latitude). Assume that the radius of Earth is
3960 miles.
108. Distance between Cities Charleston, West Virginia, is
due north of Jacksonville, Florida. Find the distance
between Charleston (38°21′ north latitude) and Jacksonville
(30°20′ north latitude). Assume that the radius of Earth is
3960 miles.
109. Linear Speed on a Planet A planet rotates on an axis
through its poles and 1 revolution takes 1 day (1 day is 18
hours). The distance from the axis to a location on the planet
30° north latitude is about 2166.5 miles. Therefore, a location
on the planet at 30° north latitude is spinning on a circle of
radius 2166.5 miles. Compute the linear speed on the surface
of the planet at 30° north latitude.
110. Linear Speed on Earth Earth rotates on an axis through its
poles. The distance from the axis to a location on Earth at 40°
north latitude is about 3033.5 miles. Therefore, a location on
Earth at 40° north latitude is spinning on a circle of radius
3033.5 miles. Compute the linear speed on the surface of
Earth at 40° north latitude.
111. Speed of a Planet’s Moon The mean distance of a moon
from a planet is 2.5 * 105 miles. Assuming that the orbit of
the moon around the planet is circular and that 1 revolution
takes 24.3 days (1 day is 26 hours), find the linear speed of the
moon. Express your answer in miles per hour.
3/22/16 10:56 AM
398
CHAPTER 6 Trigonometric Functions
112. Speed of Earth The mean distance of Earth from the Sun is
9.29 * 107 miles. Assuming that the orbit of Earth around
the Sun is circular and that 1 revolution takes 365 days, find
the linear speed of Earth. Express your answer in miles per
hour.
113. Pulleys Two pulleys, one with radius 4 inches and one
with radius 9 inches, are connected by a belt. (See the
figure.) If the 4-inch pulley is caused to rotate at 5 revolutions
per minute, determine the revolutions per minute of the
9-inch pulley.
[Hint: The linear speeds of the pulleys are the same; both
equal the speed of the belt.]
[Hint: Consult the figure. When a person at Q sees the first
rays of the Sun, a person at P is still in the dark. The person
at P sees the first rays after Earth has rotated so that P is
at the location Q. Now use the fact that at the latitude of
Ft. Lauderdale in 24 hours an arc of length 2p135592 miles
is subtended.]
90 miles
P Q
3559
miles
Rotation
of Earth
W
4 in.
9 in.
N
S
E
Naples, P
114. Ferris Wheels A neighborhood carnival has a Ferris wheel
whose radius is 30 feet. You measure the time it takes for
one revolution to be 70 seconds. What is the linear speed (in
feet per second) of this Ferris wheel? What is the angular
speed in radians per second?
115. Computing the Speed of a River Current To approximate
the speed of the current of a river, a circular paddle wheel
with radius 5 ft. is lowered into the water. If the current
causes the wheel to rotate at a speed of 15 revolutions per
minute, what is the speed of the current?
Sun
Fort
Lauderdale, Q
119. Let the Dog Roam A dog is attached to a 35-foot rope
fastened to the outside corner of a fenced-in garden that
measures 30 feet by 36 feet. Assuming that the dog cannot
enter the garden, compute the exact area that the dog can
wander.
120. Area of a Region The measure of arc ¬
BE is 2p. Find the exact
area of the portion of the rectangle ABCD that falls outside of
the circle whose center is at A.*
C
B
D
5 ft.
E
—
ED = 7
A
116. Spin Balancing Tires A spin balancer rotates the wheel of
a car at 480 revolutions per minute. If the diameter of the
wheel is 26 inches, what road speed is being tested? Express
your answer in miles per hour. At how many revolutions per
minute should the balancer be set to test a road speed of
80 miles per hour?
117. The Cable Cars of San Francisco At a museum you can see
the four cable lines that are used to pull cable cars up and
down a hill. Each cable travels at a speed of 8.65 miles per
hour, caused by a rotating wheel whose diameter is 9.5 feet.
How fast is the wheel rotating? Express your answer in revolutions per minute.
118. Difference in Time of Sunrise Naples, Florida, is about
90 miles due west of Ft. Lauderdale. How much sooner
would a person in Ft. Lauderdale first see the rising Sun than
a person in Naples? See the hint.
M06_SULL1772_10_GE_C06.indd 398
*Courtesy of the Joliet Junior College Mathematics Department
121. Keeping Up with the Sun How fast would an object have
to travel on the surface of Jupiter at the equator to keep up
with the Sun (that is, so the Sun would appear to remain in
the same position in the sky)? Use the facts that the radius
of Jupiter is approximately 44,360 miles and its revolution is
approximately 10 hours.
122. Nautical Miles A nautical mile equals the length of arc
subtended by a central angle of 1 minute on a great circle†
on the surface of Earth. See the figure on the next page. If
the radius of Earth is taken as 3960 miles, express 1 nautical
mile in terms of ordinary, or statute, miles.
†Any circle drawn on the surface of Earth that divides Earth into two
equal hemispheres.
3/22/16 10:56 AM
Section 6.1 Angles and Their Measure
1 nautical mile
1 minute
Equator
South Pole
123. Approximating the Circumference of Earth Eratosthenes
of Cyrene (276–195 bc) was a Greek scholar who lived and
worked in Cyrene and Alexandria. One day while visiting
in Syene he noticed that the Sun’s rays shone directly down
a well. On this date 1 year later, in Alexandria, which is
500 miles due north of Syene he measured the angle of the
Sun to be about 7.2 degrees. See the figure. Use this information
to approximate the radius and circumference of Earth.
minimum of 200 feet.The commissioner of parks and recreation
is making plans for a new 60-foot field. Because of limited
ground availability, he will use the minimum required
distance to the outfield fence. To increase safety, however,
he plans to include a 10-foot-wide warning track on the
inside of the fence. To further increase safety, the fence and
warning track will extend both directions into foul territory.
In total, the arc formed by the outfield fence (including the
extensions into the foul territories) will be subtended by a
central angle at home plate measuring 96°, as illustrated.
(a) Determine the length of the outfield fence.
(b) Determine the area of the warning track.
Outfield Fence
10' Warning
Track
200'
96°
200' Foul Line
7.2
Alexandria
200' Foul Line
North Pole
399
Source: www.littleleague.org
[Note: There is a 90° angle between the two foul lines. Then
there are two 3° angles between the foul lines and the dotted
lines shown. The angle between the two dotted lines outside
the 200-foot foul lines is 96°.]
500 miles
Syene
124. Designing a Little League Field For a 60-foot Little League
Baseball field, the distance from home base to the nearest
fence (or other obstruction) in fair territory should be a
125. Pulleys Two pulleys, one with radius r1 and the other with
radius r2, are connected by a belt. The pulley with radius r1
rotates at v1 revolutions per minute, whereas the pulley with
radius r2 rotates at v2 revolutions per minute. Show that
r1
v2
=
v1
r2
Explaining Concepts: Discussion and Writing
126. Do you prefer to measure angles using degrees or radians?
Provide justification and a rationale for your choice.
127. What is 1 radian? What is 1 degree?
128. Which angle has the larger measure: 1 degree or 1 radian?
Or are they equal?
129. Explain the difference between linear speed and angular speed.
130. For a circle of radius r, a central angle of u degrees subtends
p
ru. Discuss whether this
180
statement is true or false. Defend your position.
an arc whose length s is s =
131. Discuss why ships and airplanes use nautical miles to
measure distance. Explain the difference between a nautical
mile and a statute mile.
132. Investigate the way that speed bicycles work. In particular,
explain the differences and similarities between 5-speed and
9-speed derailleurs. Be sure to include a discussion of linear
speed and angular speed.
133. In Example 6, we found that the distance between Dallas,
Texas and Sioux Falls, South Dakota is approximately 744 miles.
According to mapquest.com, the distance is approximately
850 miles. What might account for the difference?
Retain Your Knowledge
Problems 134–137 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
134. Find the zero of f (x) = 3x + 7.
135. Find the domain of h 1x2 =
3x
x2 - 9
.
136. Write the function that is finally graphed if all of the following transformations are applied to the graph of y = x .
(a) Shift left 3 units. (b) Reflect about the x-axis. (c) Shift down 4 units.
137. Find the horizontal and vertical asymptotes of R 1x2 =
3x2 - 12
.
x2 - 5x - 14
‘Are You Prepared?’ Answers
1. C = 2pr; A = pr 2 2. r # t
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400
CHAPTER 6 Trigonometric Functions
6.2 Trigonometric Functions: Unit Circle Approach
Preparing for this section Before getting started, review the following:
• Geometry Essentials (Appendix A, Section A.2,
pp. 958–962)
• Unit Circle (Section 1.4, p. 69)
• Symmetry (Section 1.2, pp. 46–48)
• Functions (Section 2.1, pp. 79–90)
Now Work the ‘Are You Prepared?’ problems on page 412.
Objectives
1 Find the Exact Values of the Trigonometric Functions Using a Point on
the Unit Circle (p. 402)
2 Find the Exact Values of the Trigonometric Functions of Quadrantal
Angles (p. 403)
p
3 Find the Exact Values of the Trigonometric Functions of = 45° (p. 405)
4
p
4 Find the Exact Values of the Trigonometric Functions of = 30° and
6
p
= 60° (p. 406)
3
5 Find the Exact Values of the Trigonometric Functions for Integer Multiples
p
p
p
of = 30°, = 45°, and = 60° (p. 408)
6
4
3
6 Use a Calculator to Approximate the Value of a Trigonometric
Function (p. 410)
7 Use a Circle of Radius r to Evaluate the Trigonometric Functions (p. 411)
We now introduce the trigonometric functions using the unit circle.
The Unit Circle
Recall that the unit circle is a circle whose radius is 1 and whose center is at the
origin of a rectangular coordinate system. Also recall that any circle of radius r has
circumference of length 2pr. Therefore, the unit circle 1radius = 12 has a circumference
of length 2p. In other words, for 1 revolution around the unit circle the length of
the arc is 2p units.
The following discussion sets the stage for defining the trigonometric functions
using the unit circle.
Let t be any real number. Position the t-axis so that it is vertical with the
positive direction up. Place this t-axis in the xy-plane so that t = 0 is located at the
point 11, 02 in the xy-plane.
If t Ú 0, let s be the distance from the origin to t on the t-axis. See the red
portion of Figure 18(a). Beginning at the point 11, 02 on the unit circle, travel
P (x, y)
s t units
y
y
t
1
1
s t units
1
(1, 0) 0
x
(1, 0) 0
x
1
s |t | units
1
Figure 18
M06_SULL1772_10_GE_C06.indd 400
(a)
1
t
s |t | units
P (x, y)
(b)
11/03/16 1:46 PM
Section 6.2 Trigonometric Functions: Unit Circle Approach
401
s = t units in the counterclockwise direction along the circle, to arrive at the point
P = 1x, y2. In this sense, the length s = t units is being wrapped around the unit
circle.
If t 6 0, we begin at the point 11, 02 on the unit circle and travel s = 0 t 0 units
in the clockwise direction to arrive at the point P = 1x, y2. See Figure 18(b).
If t 7 2p or if t 6 - 2p, it will be necessary to travel around the unit circle
more than once before arriving at the point P. Do you see why?
Let’s describe this process another way. Picture a string of length s = 0 t 0 units
being wrapped around a circle of radius 1 unit. Start wrapping the string around the
circle at the point 11, 02. If t Ú 0, wrap the string in the counterclockwise direction;
if t 6 0, wrap the string in the clockwise direction. The point P = 1x, y2 is the point
where the string ends.
This discussion tells us that, for any real number t, we can locate a unique point
P = 1x, y2 on the unit circle. We call P the point on the unit circle that corresponds
to t. This is the important idea here. No matter what real number t is chosen, there is
a unique point P on the unit circle corresponding to it. The coordinates of the point
P = 1x, y2 on the unit circle corresponding to the real number t are used to define
the six trigonometric functions of t.
Definition
Let t be a real number and let P = 1x, y2 be the point on the unit circle that
corresponds to t.
The sine function associates with t the y-coordinate of P and is denoted by
In Words
The point P = (x, y) on the unit
circle corresponding to a real
number t is given by (cos t, sin t).
sin t = y
The cosine function associates with t the x-coordinate of P and is denoted by
cos t = x
If x ≠ 0, the tangent function associates with t the ratio of the y-coordinate to
the x-coordinate of P and is denoted by
tan t =
y
x
If y ≠ 0, the cosecant function is defined as
csc t =
1
y
If x ≠ 0, the secant function is defined as
sec t =
1
x
If y ≠ 0, the cotangent function is defined as
cot t =
x
y
Notice in these definitions that if x = 0, that is, if the point P is on the y-axis,
then the tangent function and the secant function are undefined. Also, if y = 0,
that is, if the point P is on the x-axis, then the cosecant function and the cotangent
function are undefined.
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402
CHAPTER 6 Trigonometric Functions
Because we use the unit circle in these definitions of the trigonometric
functions, they are sometimes referred to as circular functions.
1 Find the Exact Values of the Trigonometric Functions Using
a Point on the Unit Circle
Finding the Values of the Six Trigonometric Functions Using a Point
on the Unit Circle
Exampl e 1
1 23
Let t be a real number and let P = a - ,
b be the point on the unit circle that
2 2
corresponds to t. Find the values of sin t, cos t, tan t, csc t, sec t, and cot t.
y
P
Solution
1
3
–2 , ––
2
See Figure 19. We follow the definition of the six trigonometric functions, using
1 23
1
23
P = a- ,
b = 1x, y2. Then, with x = - and y =
, we have
2 2
2
2
t
(1, 0) x
23
2
sin t = y =
cos t = x = -
y
tan t = =
x
1
2
Figure 19
WARNING When writing the values of
the trigonometric functions, do not
forget the argument of the function.
sin t =
23
2
sin =
23
2
csc t =
1
223
1
=
=
y
3
23
sec t =
2
1
1
= -2
=
x
1
2
23
2
= - 23
1
2
-
cot t =
1
2
x
23
= =
y
3
23
2
Now Work p r o b l e m 1 3
■
•
Trigonometric Functions of Angles
Let P = 1x, y2 be the point on the unit circle corresponding to the real number t.
See Figure 20(a). Let u be the angle in standard position, measured in radians,
whose terminal side is the ray from the origin through P. Suppose u subtends an arc
of length s on the unit circle. See Figure 20(b). Since the unit circle has radius 1 unit,
if s = t units, then from the arc length formula s = r u , we have u = t radians.
See Figures 20(c) and (d).
y
y
1
1
t
P (x, y )
u
x
1
Figure 20
(a)
1
s
P (x, y )
(1, 0)
1
y
1
(b)
P (x, y )
s t units,
t0
u t radians
1
(1, 0) x
(1, 0)
x
1
y
1
1
u t radians
(1, 0)
x
s |t | units,
t0
1
1 P (x, y )
(c)
(d)
The point P = 1x, y2 on the unit circle that corresponds to the real number t
is also the point P on the terminal side of the angle u = t radians. As a result, we
can say that
sin t = sin u
c
Real number
c
u = t radians
and so on. We can now define the trigonometric functions of the angle u.
M06_SULL1772_10_GE_C06.indd 402
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Section 6.2 Trigonometric Functions: Unit Circle Approach
Definition
403
If u = t radians, the six trigonometric functions of the angle U are defined as
sin u = sin t
csc u = csc t
cos u = cos t
sec u = sec t
tan u = tan t
cot u = cot t
Even though the trigonometric functions can be viewed both as functions of
real numbers and as functions of angles, it is customary to refer to trigonometric
functions of real numbers and trigonometric functions of angles collectively as the
trigonometric functions. We shall follow this practice from now on.
If an angle u is measured in degrees, we shall use the degree symbol when writing
a trigonometric function of u, as, for example, in sin 30° and tan 45°. If an angle u is
measured in radians, then no symbol is used when writing a trigonometric function
p
of u, as, for example, in cos p and sec .
3
Finally, since the values of the trigonometric functions of an angle u are
determined by the coordinates of the point P = 1x, y2 on the unit circle
corresponding to u, the units used to measure the angle u are irrelevant. For example, it
p
does not matter whether we write u =
radians or u = 90°. The point on the unit
2
circle corresponding to this angle is P = 10, 12. As a result,
sin
p
p
= sin 90° = 1 and cos = cos 90° = 0
2
2
2 Find the Exact Values of the Trigonometric Functions
of Quadrantal Angles
To find the exact value of a trigonometric function of an angle u or a real number t
requires that we locate the point P = 1x, y2 on the unit circle that corresponds to t. This
is not always easy to do. In the examples that follow, we will evaluate the trigonometric
functions of certain angles or real numbers for which this process is relatively easy. A
calculator will be used to evaluate the trigonometric functions of most other angles.
Exampl e 2
Finding the Exact Values of the Six Trigonometric
Functions of Quadrantal Angles
Find the exact values of the six trigonometric functions of:
p
(a) u = 0 = 0°
(b) u =
= 90°
2
3p
(c) u = p = 180°
(d) u =
= 270°
2
Solution
(a) The point on the unit circle that corresponds to u = 0 = 0° is P = 11, 02. See
Figure 21(a). Then using x = 1 and y = 0
sin 0 = sin 0° = y = 0
tan 0 = tan 0° =
cos 0 = cos 0° = x = 1
y
= 0
x
sec 0 = sec 0° =
1
= 1
x
Since the y-coordinate of P is 0, csc 0 and cot 0 are not defined.
y
1
P 5 (1, 0)
21
u 5 0 5 0° 1
x
21
Figure 21 (a) u = 0 = 0°
M06_SULL1772_10_GE_C06.indd 403
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404
CHAPTER 6 Trigonometric Functions
p
(b) The point on the unit circle that corresponds to u =
= 90° is P = 10, 12. See
2
Figure 21(b). Then
y
1 P (0, 1)
π
u 2 90°
x
1
1
1
p
= 90°
2
Figure 21 (b) u =
cos
p
= cos 90° = x = 0
2
csc
p
1
= csc 90° = = 1
y
2
cot
p
x
= cot 90° = = 0
y
2
p
p
and sec are not defined.
2
2
(c) The point on the unit circle that corresponds to u = p = 180° is P = 1 - 1, 02 .
See Figure 21(c). Then
1
u π 180°
sin p = sin 180° = y = 0
x
1
1
p
= sin 90° = y = 1
2
Since the x-coordinate of P is 0, tan
y
P (1, 0)
sin
tan p = tan 180° =
1
Figure 21 (c) u = p = 180°
y
3π
1 u 2 270°
1
= -1
x
sin
3p
= sin 270° = y = - 1
2
cos
3p
= cos 270° = x = 0
2
csc
3p
1
= csc 270° = = - 1
y
2
cot
3p
x
= cot 270° = = 0
y
2
1 P (0, 1)
Figure 21 (d) u =
sec p = sec 180° =
Since the y-coordinate of P is 0, csc p and cot p are not defined.
3p
(d) The point on the unit circle that corresponds to u =
= 270° is P = 10, - 12 .
2
See Figure 21(d). Then
x
1
1
y
= 0
x
cos p = cos 180° = x = - 1
3p
= 270°
2
Since the x-coordinate of P is 0, tan
3p
3p
and sec
are not defined.
2
2
•
Table 2 summarizes the values of the trigonometric functions found in
Example 2.
Table 2
Quadrantal Angles
U (Radians)
U (Degrees)
sin U
cos U
tan U
csc U
0
p
2
0°
0
1
0
Not defined
90°
1
0
Not defined
p
3p
2
180°
0
-1
270°
-1
0
1
sec U
1
cot U
Not defined
Not defined
0
0
Not defined
-1
Not defined
Not defined
-1
Not defined
0
There is no need to memorize Table 2. To find the value of a trigonometric
function of a quadrantal angle, draw the angle and apply the definition, as we did
in Example 2.
E xam pl e 3
Finding Exact Values of the Trigonometric Functions of Angles
That Are Integer Multiples of Quadrantal Angles
Find the exact value of:
(a) sin 13p2
M06_SULL1772_10_GE_C06.indd 404
(b) cos 1 - 270°2
11/03/16 1:46 PM
Section 6.2 Trigonometric Functions: Unit Circle Approach
Solution
(a) See Figure 22(a). The point P on
the unit circle that corresponds
to u = 3p is P = 1 - 1, 02, so
sin 13p2 = y = 0.
(b) See Figure 22(b). The point P on
the unit circle that corresponds
to u = - 270° is P = 10, 12, so
cos 1 - 270°2 = x = 0.
y
y
1
1 P 5 (0, 1)
P (1, 0)
u 3
1
1
x
1
21
1
Figure 22
405
u 5 2270°
21
(a) u = 3p
x
•
(b) u = - 270°
Now Work p r o b l e m s 2 1 a n d 6 1
3 Find the Exact Values of the Trigonometric
P
Functions of = 45°
4
Exampl e 4
Finding the Exact Values of the Trigonometric
P
Functions of
= 45°
4
Find the exact values of the six trigonometric functions of
Solution
yx
y
We seek the coordinates of the point P = 1x, y2 on the unit circle that corresponds
p
to u =
= 45°. See Figure 23. First, observe that P lies on the line y = x. (Do you
4
1
see why? Since u = 45° = # 90°, P must lie on the line that bisects quadrant I.)
2
Since P = 1x, y2 also lies on the unit circle, x2 + y2 = 1, it follows that
x2 + y 2 = 1
1
P (x, y )
1
1
y = x, x 7 0, y 7 0
x2 + x2 = 1
45°
1
p
= 45°.
4
x
2x2 = 1
x2 y2 1
x =
Figure 23 u = 45°
Then
sin
p
22
= sin 45° =
4
2
cos
1
22
=
22
2
y =
p
22
= cos 45° =
4
2
22
2
tan
22
2
p
= tan 45° =
= 1
4
22
2
csc
p
1
= csc 45° =
= 22
4
22
2
M06_SULL1772_10_GE_C06.indd 405
sec
p
1
= sec 45° =
= 22
4
22
2
cot
22
2
p
= cot 45° =
= 1
4
22
2
•
11/03/16 1:46 PM
406
CHAPTER 6 Trigonometric Functions
Exampl e 5
Finding the Exact Value of a Trigonometric Expression
Find the exact value of each expression.
p
3p
p 2
p
(a) sin 45° cos 180° (b) tan - sin
(c) asec b + csc
4
2
4
2
Solution
(a) sin 45° cos 180° =
c
(b) tan
22 #
22
1 - 12 = 2
2
c
From Table 2
From Example 4
p
3p
- sin
= 1 - 1 - 12 = 2
4
2
c
c
From Table 2
From Example 4
(c) asec
p 2
p
2
b + csc = 1 222 + 1 = 2 + 1 = 3
4
2
•
Now Work p r o b l e m 3 5
4 Find the Exact Values of the Trigonometric
P
P
Functions of = 30° and = 60°
6
3
p
Consider a right triangle in which one of the angles is
= 30°. It then follows that
6
p
the third angle is
= 60°. Figure 24(a) illustrates such a triangle with hypotenuse
3
of length 1. Our problem is to determine a and b.
Begin by placing next to the triangle in Figure 24(a) another triangle congruent
to the first, as shown in Figure 24(b). Notice that we now have a triangle whose three
angles each equal 60°. This triangle is therefore equilateral, so each side is of length 1.
30°
c1
30° 30°
b
60°
c1
60°
a
Figure 24 30° - 60° - 90° triangle
30°
b
a
(a)
1
c1
60°
60°
b 23
a1
2
a
(b)
(c)
This means the base is 2a = 1, and so a =
b satisfies the equation a2 + b2 = c 2, so we have
1
. By the Pythagorean Theorem,
2
a 2 + b2 = c 2
1
+ b2 = 1
4
b2 = 1 b =
23
2
a =
1
,c = 1
2
1
3
=
4
4
b 7 0 because b
is the length of the
side of a triangle.
This results in Figure 24(c).
M06_SULL1772_10_GE_C06.indd 406
11/03/16 1:46 PM
Section 6.2 Trigonometric Functions: Unit Circle Approach
Finding the Exact Values of the Trigonometric
P
Functions of
= 60°
3
Exampl e 6
Find the exact values of the six trigonometric functions of
Solution
y
1
(
P (x, y ) 12, 3
2
)
3
60° 2
1
2
1
csc
p
23
= sin 60° =
3
2
p
1
2
223
= csc 60° =
=
=
3
3
23
23
cos
sec
2
x2y21
1
Figure 25 u =
x
1
23
p
2
= 23
tan = tan 60° =
3
1
2
p
= 60°
3
Exampl e 7
cot
p
1
= cos 60° =
3
2
p
1
= sec 60° =
= 2
3
1
2
y
1
P (x, y ) 1
30°
1
3
2
1
1
2
1
( 23 , 12)
1
2
p
1
23
= cot 60° =
=
=
3
3
23
23
2
•
Finding the Exact Values of the Trigonometric
P
Functions of
= 30°
6
Find the exact values of the trigonometric functions of
Solution
p
= 60°.
3
Position the triangle in Figure 24(c) so that the 60° angle is in standard position.
p
See Figure 25. The point on the unit circle that corresponds to u =
= 60° is
3
1 23
P = a ,
b . Then
2 2
sin
1
407
p
= 30°.
6
Position the triangle in Figure 24(c) so that the 30° angle is in standard position.
p
See Figure 26. The point on the unit circle that corresponds to u =
= 30° is
6
23 1
P = a
, b . Then
2 2
sin
x
csc
x2 y2 1
p
Figure 26 u =
= 30°
6
tan
p
1
= sin 30° =
6
2
cos
p
1
= csc 30° =
= 2
6
1
2
1
2
p
1
23
= tan 30° =
=
=
6
3
23
23
2
sec
p
23
= cos 30° =
6
2
p
1
2
223
= sec 30° =
=
=
6
3
23
23
2
23
p
2
cot = cot 30° =
= 23
6
1
2
•
Table 3 on the next page summarizes the information just derived for
p
p
p
= 30°, = 45°, and
= 60°. Until you memorize the entries in Table 3, you
6
4
3
should draw an appropriate diagram to determine the values given in the table.
M06_SULL1772_10_GE_C06.indd 407
11/03/16 1:46 PM
408
CHAPTER 6 Trigonometric Functions
Table 3
U (Radians)
U (Degrees)
sin U
cos U
tan U
csc U
sec U
cot U
p
6
30°
1
2
23
2
23
3
2
223
3
23
p
4
45°
22
2
22
2
1
22
22
1
p
3
60°
23
2
1
2
23
223
3
2
23
3
Now Work p r o b l e m 4 1
Exampl e 8
A rain gutter is to be constructed of aluminum sheets 12 inches wide. After marking
off a length of 4 inches from each edge, this length is bent up at an angle u. See
Figure 27. The area A of the opening may be expressed as a function of u as
12 in.
4 in.
4 in.
Constructing a Rain Gutter
A 1u2 = 16 sin u1cos u + 12
4 in.
4 in
.
4 in
u
u
.
Find the area A of the opening for u = 30°, u = 45°, and u = 60°.
Solution
For u = 30°: A 130°2 = 16 sin 30°1cos 30° + 12
1
23
= 16a b a
+ 1b = 423 + 8 ≈ 14.93
2
2
Figure 27
The area of the opening for u = 30° is about 14.93 square inches.
For u = 45°: A 145°2 = 16 sin 45°1cos 45° + 12
= 16a
22
22
ba
+ 1b = 8 + 822 ≈ 19.31
2
2
The area of the opening for u = 45° is about 19.31 square inches.
For u = 60°: A 160°2 = 16 sin 60°1cos 60° + 12
= 16a
23 1
b a + 1b = 1223 ≈ 20.78
2
2
The area of the opening for u = 60° is about 20.78 square inches.
•
5 Find the Exact Values of the Trigonometric Functions
P
P
P
for Integer Multiples of = 30°, = 45°, and = 60°
6
4
3
( ––22 , ––22 )
3–––
4
( ––22 , ––22 )
––
4
1
5–––
4
7–––
4
( ––22 , ––22 )
p
= 45°. Using
4
symmetry, we can find the exact values of the trigonometric functions of
3p
5p
7p
= 135°,
= 225°, and
= 315°. The point on the unit circle corresponding
4
4
4
p
22 22
to
= 45° is a
,
b . See Figure 28. Using symmetry with respect to the
4
2
2
We know the exact values of the trigonometric functions of
y
1
1
Figure 28
M06_SULL1772_10_GE_C06.indd 408
1 x
( ––22 , ––22 )
y-axis, the point a the angle
22 22
,
b is the point on the unit circle that corresponds to
2
2
3p
= 135°. Similarly, using symmetry with respect to the origin, the
4
11/03/16 1:46 PM
Section 6.2 Trigonometric Functions: Unit Circle Approach
409
22
22
,b is the point on the unit circle that corresponds to the angle
2
2
5p
22
22
= 225°. Finally, using symmetry with respect to the x-axis, the point a
,b
4
2
2
7p
is the point on the unit circle that corresponds to the angle
= 315°.
4
point a -
Exampl e 9
Finding Exact Values for Multiples of
P
= 45°
4
Find the exact value of each expression.
(a) cos
Solution
5p
p
11p
(b) sin 135° (c) tan 315° (d) sin a - b (e) cos
4
4
4
(a) From Figure 28, we see the point a cos
5p
22
= x = .
4
2
(b) Since 135° =
sin 135° =
3p
, the point
4
22
.
2
(c) Since 315° =
tan 315° =
7p
, the point
4
22
2
a
22 22
,
b
2
2
corresponds to 135°, so
22
22
,b
2
2
corresponds to 315°, so
= - 1.
22
2
(d) The point a
a-
22
22
5p
,b corresponds to
, so
2
2
4
22
22
p
p
22
,b corresponds to - , so sin a - b = .
2
2
4
4
2
(e) The point a -
22 22
11p
11p
22
,
b corresponds to
, so cos
= .
2
2
4
4
2
•
Now Work p r o b l e m s 5 1 a n d 5 5
The use of symmetry also provides information about certain integer multiples
p
p
of the angles
= 30° and
= 60°. See Figures 29 and 30.
6
3
y
1
( ––23 , 1–2 )
(
5–––
6
( ––23 , 1–2 )
1
3
12 , ––
2
––
6
7–––
6
11
–––
6
Figure 29
M06_SULL1772_10_GE_C06.indd 409
)
1 x
1
4–––
3
( ––23 , 1–2 )
1–2
,
3
––
2
( 1–2 , ––23 )
––
2–––
3
( ––23 , 1–2 )
(
1
y
1
)
1
3
1 x
5–––
3
( 1–2 , ––23 )
Figure 30
11/03/16 1:46 PM
410
CHAPTER 6 Trigonometric Functions
Exampl e 10
Finding Exact Values for Multiples of
P
P
= 30° or
= 60°
6
3
Based on Figures 29 and 30, we see that
(a) cos 210° = cos
5p
(c) tan
=
3
-
23
7p
= 6
2
(b) sin 1 - 60°2 = sin a -
23
2
= - 23
1
2
(d) cos
p
23
b = 3
2
8p
2p
1
= cos
= 3
3
2
•
Now Work p r o b l e m 4 7
6 Use a Calculator to Approximate the Value
of a Trigonometric Function
WARNING On your calculator the
second functions sin–1, cos–1, and tan–1
do not represent the reciprocal of sin,
cos, and tan.
■
Exampl e 11
Before getting started, you must first decide whether to enter the angle in the
calculator using radians or degrees and then set the calculator to the correct MODE.
Check your instruction manual to find out how your calculator handles degrees and
radians. Your calculator has keys marked sin , cos , and tan . To find the values of
the remaining three trigonometric functions, secant, cosecant, and cotangent, use the
fact that, if P = 1x, y2 is a point on the unit circle on the terminal side of u, then
sec u =
1
1
=
x
cos u
csc u =
1
1
=
y
sin u
cot u =
x
1
1
=
=
y
y
tan u
x
Using a Calculator to Approximate the Value
of a Trigonometric Function
Use a calculator to find the approximate value of:
(a) cos 48° (b) csc 21° (c) tan
p
12
Express your answer rounded to two decimal places.
Solution
(a) First, set the MODE to receive degrees. Rounded to two decimal places,
cos 48° = 0.6691306 ≈ 0.67
(b) Most calculators do not have a csc key. The manufacturers assume that the
user knows some trigonometry. To find the value of csc 21°, use the fact that
1
csc 21° =
. Rounded to two decimal places,
sin 21°
csc 21° ≈ 2.79
(c) Set the MODE to receive radians. Figure 31 shows the solution using a TI-84
Plus C graphing calculator. Rounded to two decimal places,
tan
Figure 31
M06_SULL1772_10_GE_C06.indd 410
p
≈ 0.27
12
•
Now Work p r o b l e m 6 5
11/03/16 1:46 PM
Section 6.2 Trigonometric Functions: Unit Circle Approach
411
7 Use a Circle of Radius r to Evaluate
the Trigonometric Functions
y
P (x, y)
P* (x*, y*)
y
u
y*
A* x* O
A
1
x
r x
Until now, finding the exact value of a trigonometric function of an angle u required
that we locate the corresponding point P = 1x, y2 on the unit circle. In fact, though,
any circle whose center is at the origin can be used.
Let u be any nonquadrantal angle placed in standard position. Let P = 1x, y2
be the point on the circle x2 + y2 = r 2 that corresponds to u, and let P* = 1x*, y*2
be the point on the unit circle that corresponds to u. See Figure 32, where u is shown
in quadrant II.
Notice that the triangles OA*P* and OAP are similar; as a result, the ratios of
corresponding sides are equal.
x2 y2 1
x2 y2 r 2
Figure 32
y*
y
=
r
1
x*
x
=
r
1
y*
y
=
x
x*
1
r
=
y
y*
1
r
=
x
x*
x*
x
=
y
y*
These results lead us to formulate the following theorem:
Theorem
For an angle u in standard position, let P = 1x, y2 be the point on the terminal
side of u that is also on the circle x2 + y2 = r 2. Then
y
r
r
csc u =
y
sin u =
E xam pl e 12
Solution
y ≠ 0
x ≠ 0
x ≠ 0
y ≠ 0
Finding the Exact Values of the Six Trigonometric Functions
Find the exact values of each of the six trigonometric functions of an angle u if
14, - 32 is a point on its terminal side in standard position.
See Figure 33. The point 14, - 32 is on a circle that has a radius of
y
3
= r
5
r
5
csc u = = y
3
sin u =
u
r
5
6 x
(4, 3)
6
y
x
x
cot u =
y
tan u =
r = 242 + 1 - 32 2 = 216 + 9 = 225 = 5 with the center at the origin.
For the point 1x, y2 = 14, - 32, we have x = 4 and y = - 3. Since r = 5, we find
y
6
6
x
r
r
sec u =
x
cos u =
x 2 y 2 25
y
3
= x
4
x
4
cot u = = y
3
x
4
=
r
5
r
5
sec u = =
x
4
cos u =
tan u =
•
Now Work p r o b l e m 7 7
Figure 33
Historical Feature
T
he name sine for the sine function arose from a medieval
confusion. The name comes from the Sanskrit word jiva
(meaning chord), first used in India by Araybhata the Elder (ad 510).
He really meant half-chord, but abbreviated it. This was brought into
Arabic as jiba, which was meaningless. Because the proper Arabic
word jaib would be written the same way (short vowels are not written
out in Arabic), jiba was pronounced as jaib, which meant bosom or
hollow, and jiba remains as the Arabic word for sine to this day. Scholars
M06_SULL1772_10_GE_C06.indd 411
translating the Arabic works into Latin found that the word sinus also
meant bosom or hollow, and from sinus we get sine.
The name tangent, due to Thomas Finck (1583), can be understood
by looking at Figure 34 on page 412. The line segment DC is tangent
to the circle at C. If d(O, B) = d(O, C) = 1, then the length of the line
segment DC is
d(D, C) =
d(D, C)
1
=
d(D, C)
d(O, C)
= tan a
continued
11/03/16 1:46 PM
412
CHAPTER 6 Trigonometric Functions
The old name for the tangent is umbra versa (meaning turned
shadow), referring to the use of the tangent in solving height problems
with shadows.
The names of the remaining functions came about as follows. If a
and b are complementary angles, then cos a = sin b. Because b is the
complement of a, it was natural to write the cosine of a as sin co a.
Probably for reasons involving ease of pronunciation, the co migrated to
the front, and then cosine received a three-letter abbreviation to match
sin, sec, and tan. The two other cofunctions were similarly treated,
except that the long forms cotan and cosec survive to this day in some
countries.
D
B
a
O
A
C
Figure 34
6.2 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. In a right triangle, with legs a and b and hypotenuse c, the
Pythagorean Theorem states that
. (p. 958)
2. The value of the function f 1x2 = 3x - 7 at 5 is
(pp. 79–90)
3. True or False For a function y = f 1x2, for each x in
the domain, there is exactly one element y in the range.
(pp. 79–90)
5. What point is symmetric with respect to the y-axis to the
.
1 23
b? (pp. 46–48)
point a ,
2 2
6. If 1x, y2 is a point on the unit circle in quadrant IV and if
x =
4. If two triangles are similar, then corresponding angles
are
and the lengths of corresponding sides are
. (pp. 958–962)
23
, what is y? (p. 69)
2
Concepts and Vocabulary
7. Which function takes as input a real number t that
corresponds to a point P = (x, y) on the unit circle and
outputs the x-coordinate?
(a) sine (b) cosine (c) tangent (d) secant
p
8. The point on the unit circle that corresponds to u =
is
2
P =
.
p
9. The point on the unit circle that corresponds to u =
is
4
P =
10. The point on the unit circle that corresponds to u =
1 23
(a) a ,
b
2 2
(c) a
23 1
, b
2 2
(b) a
p
is
3
22 22
,
b
2
2
(d) a 23,
223
b
3
11. For any angle u in standard position, let P = (x, y) be the
point on the terminal side of u that is also on the circle
.
x2 + y2 = r 2. Then, sin u =
and cos u =
.
12. True or False Exact values can be found for the sine of any
angle.
Skill Building
In Problems 13–20, P = 1x, y2 is the point on the unit circle that corresponds to a real number t. Find the exact values of the six
trigonometric functions of t.
13. a
17. a
23 1
, b
2 2
22 22
,
b
2
2
M06_SULL1772_10_GE_C06.indd 412
1
23
14. a , b
2
2
1 226
15. a - ,
b
5 5
2 221
16. a - ,
b
5 5
18. a -
19. a -
20. a
22 22
,
b
2
2
25 2
,- b
3
3
222 1
,- b
3
3
11/03/16 1:46 PM
Section 6.2 Trigonometric Functions: Unit Circle Approach
413
In Problems 21–30, find the exact value. Do not use a calculator.
11p
2
11p
26. csc
2
21. sin
22. cos 17p2
23. cot
27. sin1 - 3p2
7p
2
28. cos a -
3p
b
2
24. tan16p2
25. sec18p2
29. tan1 - 3p2
30. sec1 - p2
In Problems 31–46, find the exact value of each expression. Do not use a calculator.
31. sin 30° - cos 45°
32. sin 45° + cos 60°
33. cos 180° - sin 180°
34. sin 90° + tan 45°
35. sin 45° cos 45°
36. tan 45° cos 30°
37. sec 30° cot 45°
38. csc 45° tan 60°
39. 5 cos 90° - 8 sin 270°
40. 4 sin 90° - 3 tan 180°
41. 2 sin
p
p
- 3 tan
3
6
p
45. sec p - csc
2
p
p
+ 3 tan
4
4
p
p
46. csc + cot
2
2
43. 3 csc
p
p
+ cot
3
4
44. 2 sec
p
p
+ 4 cot
4
3
42. 2 sin
In Problems 47–64, find the exact values of the six trigonometric functions of the given angle. If any are not defined, say “not defined.”
Do not use a calculator.
2p
5p
3p
11p
47.
48.
49. 240°
50. 210°
51.
52.
3
6
4
4
13p
8p
p
p
54.
55. 405°
56. 390°
57. 58. 53.
6
3
3
6
5p
13p
14p
59. - 240°
60. - 135°
61.
62. 5p
63. 64. 2
6
3
In Problems 65–76, use a calculator to find the approximate value of each expression rounded to two decimal places.
65. sin 28°
66. cos 14°
p
8
70. tan
73. tan 1
74. sin 1
69. sin
p
10
67. cot 70°
71. csc
5p
13
75. tan 1°
68. sec 21°
72. cot
p
12
76. sin 1°
In Problems 77–84, a point on the terminal side of an angle u in standard position is given. Find the exact value of each of the six
trigonometric functions of u.
77. 1 - 3, 42
78. 15, - 122
81. 1 - 1, 12
82. 1 - 2, - 22
85. Find the exact value of:
sin 45° + sin 135° + sin 225° + sin 315°
86. Find the exact value of:
tan 60° + tan 150°
87. Find the exact value of:
sin 40° + sin 130° + sin 220° + sin 310°
88. Find the exact value of:
tan 40° + tan 140°
79. 1 - 1, - 22
83. 10.3, 0.42
80. 12, - 32
1 1
84. a , b
3 4
89. If f 1u2 = sin u = 0.1, find f 1u + p2.
90. If f 1u2 = cos u = 0.3, find f 1u + p2.
91. If f 1u2 = tan u = 3, find f 1u + p2.
92. If f 1u2 = cot u = - 2, find f 1u + p2.
1
, find csc u.
5
2
94. If cos u = , find sec u.
3
93. If sin u =
In Problems 95–106, f 1u2 = sin u and g1u2 = cos u. Find the exact value of each function below if u = 60°. Do not use a calculator.
95. g1u2
99. 3g1u24 2
103. 2g1u2
M06_SULL1772_10_GE_C06.indd 413
96. f 1u2
100. 3f 1u24 2
104. 2f 1u2
u
97. ga b
2
101. g12u2
105. g1 - u2
u
98. f a b
2
102. f 12u2
106. f 1 - u2
11/03/16 1:46 PM
414
CHAPTER 6 Trigonometric Functions
Mixed Practice
In Problems 107–116, f(x) = sin x, g(x) = cos x, h(x) = 2x, and p(x) =
107. (f + g)(30°)
108. (f - g)(60°)
p
111. (f ∘ h) a b
6
112. (g ∘ p)(60°)
x
. Find the value of each of the following:
2
109. (f # g) a
p
115. (a) Find f a b . What point is on the graph of f ?
4
(b) Assuming f is one-to-one*, use the result of part (a) to
find a point on the graph of f -1.
p
(c) What point is on the graph of y = f ax + b - 3
4
p
if x = ?
4
3p
b
4
110. (f # g) a
113. (p ∘ g)(315°)
114. (h ∘ f ) a
4p
b
3
5p
b
6
p
116. (a) Find ga b . What point is on the graph of g?
6
(b) Assuming g is one-to-one*, use the result of part (a) to
find a point on the graph of g -1.
p
p
(c) What point is on the graph of y = 2gax - b if x = ?
6
6
Applications and Extensions
117. Find two negative and three positive angles, expressed in
radians, for which the point on the unit circle that
118. Find two negative and three positive angles, expressed in
radians, for which the point on the unit circle that
1 23
corresponds to each angle is a - ,
b.
2 2
corresponds to each angle is a -
119. Use a calculator in radian mode to complete the following table.
What can be concluded about the value of f 1u2 =
0.5
U
0.4
22 22
,
b.
2
2
tan u
as u approaches 0?
u
0.2
0.1
0.01
0.001
0.0001
0.00001
0.001
0.0001
0.00001
tan u
f 1u2 =
tan u
u
120. Use a calculator in radian mode to complete the following table.
cos u - 1
What can you conclude about the value of g1u2 =
as u approaches 0?
u
0.5
U
0.4
0.2
0.1
0.01
cos u - 1
g1u2 =
cos u - 1
u
For Problems 121–124, use the following discussion.
Projectile Motion The path of a projectile fired at an inclination u
to the horizontal with initial speed v0 is a parabola (see the figure).
The range R of the projectile, that is, the horizontal distance that
the projectile travels, is found by using the function
R 1u2 =
v0 = Initial speed
Height, H
θ
v20 sin12u2
g
where g ≈ 32.2 feet per second per second ≈ 9.8 meters
per second per second is the acceleration due to gravity. The
maximum height H of the projectile is given by the function
Range, R
H1u2 =
v20 1sin u2 2
2g
*In Section 7.1, we discuss the necessary domain restriction so that the function is one-to-one.
M06_SULL1772_10_GE_C06.indd 414
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415
Section 6.2 Trigonometric Functions: Unit Circle Approach
In Problems 121–124, find the range R and maximum height H.
121. The projectile is fired at an angle of 45° to the horizontal
with an initial speed of 150 feet per second.
122. The projectile is fired at an angle of 30° to the horizontal
with an initial speed of 150 meters per second.
123. The projectile is fired at an angle of 50° to the horizontal
with an initial speed of 951 feet per second.
127. Calculating the Time of a Trip Two homes are located 4
miles apart, each 1 mile from a road that parallels the ocean.
Sally can jog 4 mph along the road, but only 2 mph in the
sand. Because of a river between the two houses, it is necessary to jog on the sand to the road, continue on the road, and
then jog on the sand to get from one house to the other. For
0° 6 u 6 90°, the time T to get from one house to the other
is a function of u, as shown.
124. The projectile is fired at an angle of 50° to the horizontal
with an initial speed of 200 feet per second.
T 1u2 = 1 +
125. Inclined Plane See the figure.
2
1
,
2 sin u
2 tan u
0° 6 u 6 90°
Ocean
2 mi
2 mi
Beach
Paved path
1 mi
u
x
u
River
u
a
If friction is ignored, the time t (in seconds) required for
a block to slide down an inclined plane (see the figure) is
given by the formula t
t1u2 =
2a
,
A g sin u cos u
where a is the length (in feet) of the base and g ≈ 32 feet
per second per second is the acceleration due to gravity.
How long does it take a block to slide down an inclined
plane with base a = 70 feet when
(a) u = 30°
(b) u = 45°
(c) u = 60°
126. Piston Engines In a certain piston engine, the distance x (in
centimeters) from the center of the drive shaft to the head of
a piston is given by the function below, where u is the angle
between the crank and the path of the piston head. See the
figure. Find x when u = 30° and u = 45°.
x1u2 = sin u + 249 + 0.6 sin12u2
(a) Calculate the time T for u = 30°. How long is Sally on
the paved road?
(b) Calculate the time T for u = 45°. How long is Sally on
the paved road?
(c) Calculate the time T for u = 60°. How long is Sally on
the paved road?
(d) Calculate the time T for u = 90°. Describe the path
taken. Why can’t the formula for T be used?
128. Designing Fine Decorative Pieces
Spheres of fixed radius R will be
enclosed in a cone of height h and
radius r. The volume V of the cone
can be expressed as a function of
the slant angle u of the cone by the
formula below. What volume V is
required to enclose a sphere of
radius 2 centimeters in a cone whose
slant angle u is 30°? 45°? 60°?
h
R
O
r
θ
3
V 1u2 =
11 + sec u2
1
pR3
,
3
1tan u2 2
0° 6 u 6 90°
u
x
M06_SULL1772_10_GE_C06.indd 415
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416
CHAPTER 6 Trigonometric Functions
Use the following to answer Problems 129–132. The viewing
angle, u, of an object is the angle the object forms at the lens of the
viewer’s eye.This is also known as the perceived or angular size of the
object. The viewing angle is related to the object’s height, H, and
u
H
distance from the viewer, D, through the formula tan =
.
2
2D
(a) Find the distance R that the object travels along the
inclined plane if the initial velocity is 32 feet per second
and u = 60°.
(b) Graph R = R 1u2 if the initial velocity is 32 feet per
second.
(c) What value of u makes R largest?
129. Tailgating While driving, Arletha observes the car in front
of her with a viewing angle of 26°. If the car is 5.5 feet wide,
how close is Arletha to the car in front of her?
136. If u, 0 6 u 6 p, is the angle between the positive x-axis and
a nonhorizontal, nonvertical line L, show that the slope m
of L equals tan u. The angle u is called the inclination of L.
[Hint: See the illustration, where we have drawn the line M
parallel to L and passing through the origin. Use the fact
that M intersects the unit circle at the point 1cos u, sin u2.]
130. Viewing Distance The Washington Monument in Washington,
D.C. is 555 feet tall. If a tourist sees the monument with a
viewing angle of 8°, how far away, to the nearest foot, is she
from the monument?
131. Tree Height A forest ranger views a tree that is 190 feet
away with a viewing angle of 21°. How tall is the tree?
y
1
132. Radius of the Moon An astronomer observes the moon
with a viewing angle of 0.52°. If the moon’s average distance
from Earth is 384,400 km, what is its radius to the nearest
kilometer?
133. Let u be the measure of an angle, in radians, in standard
3p
. Find the exact y-coordinate of the
position with p 6 u 6
2
intersection of the terminal side of u with the unit circle, given
19
cos u + sin2 u =
. State the answer as a single fraction.
25
134. Let u be the measure of an angle, in radians, in standard
p
6 u 6 p. Find the exact x-coordinate of the
position with
2
intersection of the terminal side of u with the unit circle, given
1
cos2 u - sin u = - . State the answer as a single fraction,
9
completely simplified, with rationalized denominator.
135. Projectile Motion An object is
propelled upward at an angle
u, 45° 6 u 6 90°, to the horizontal with an initial velocity
of v0 feet per second from the
base of an inclined plane that
makes an angle of 45° with the
horizontal. See the illustration.
u
O
1
L
u
1
x
1
In Problems 137 and 138, use the figure to approximate the value of
the six trigonometric functions at t to the nearest tenth. Then use a
calculator to approximate each of the six trigonometric functions at t.
2
b
1
Unit Circle
0.5
3
0.5
0.5
a
6
0.5
R
4
θ
5
45°
If air resistance is ignored, the distance R that it travels up
the inclined plane as a function of u is given by
R 1u2 =
(cos u, sin u)
M
137. (a) t = 1
(b) t = 5.1
138. (a) t = 2
(b) t = 4
v20 22
3sin12u2 - cos 12u2 - 14
32
Explaining Concepts: Discussion and Writing
139. Write a brief paragraph that explains how to quickly
compute the trigonometric functions of 30°, 45°, and 60°.
140. Write a brief paragraph that explains how to quickly
compute the trigonometric functions of 0°, 90°, 180°, and 270°.
141. How would you explain the meaning of the sine function to
a fellow student who has just completed college algebra?
M06_SULL1772_10_GE_C06.indd 416
p p p
7p
142. Draw a unit circle. Label the angles 0, , , , . . . ,
,
6 4 3
4
11p
, 2p and the coordinates of the points on the unit circle
6
that correspond to each of these angles. Explain how
symmetry can be used to find the coordinates of points
on the unit circle for angles whose terminal sides are in
quadrants II, III, and IV.
3/22/16 2:58 PM
417
Section 6.3 Properties of the Trigonometric Functions
Retain Your Knowledge
Problems 143–146 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
143. State the domain of f 1x2 = ln15x + 22.
144. Given that the polynomial function P 1x2 = x4 - 5x3 - 9x2 + 155x - 250 has zeros of 4 + 3i and 2, find the remaining
zeros of the function.
145. Find the remainder when P 1x2 = 8x4 - 2x3 + x - 8 is divided by x + 2.
146. Sidewalk Area A sidewalk with a uniform width of 3 feet is to be placed around a circular garden with a diameter
of 24 feet. Find the exact area of the sidewalk.
‘Are You Prepared?’ Answers
1. c 2 = a2 + b2
2. 8
3. True
1 23
b
5. a - ,
2 2
4. equal; proportional
6. -
1
2
6.3 Properties of the Trigonometric Functions
Preparing for this section Before getting started, review the following:
• Functions (Section 2.1, pp. 79–90)
• Identity (Appendix A, Section A.6, p. 987)
• Even and Odd Functions (Section 2.3, pp. 103–105)
Now Work the ‘Are You Prepared?’ problems on page 428.
Objectives
1 Determine the Domain and the Range of the Trigonometric Functions (p. 417)
2 Determine the Period of the Trigonometric Functions (p. 419)
3 Determine the Signs of the Trigonometric Functions in a Given
Quadrant (p. 421)
4 Find the Values of the Trigonometric Functions Using Fundamental
Identities (p. 422)
5 Find the Exact Values of the Trigonometric Functions of an Angle Given
One of the Functions and the Quadrant of the Angle (p. 424)
6 Use Even–Odd Properties to Find the Exact Values of the Trigonometric
Functions (p. 427)
1 Determine the Domain and the Range
of the Trigonometric Functions
Let u be an angle in standard position, and let P = 1x, y2 be the point on the unit
circle that corresponds to u. See Figure 35. Then, by the definition given earlier,
y
(0, 1)
P (x, y )
sin u = y
u
(1, 0)
O
(1, 0)
(0, 1)
Figure 35
M06_SULL1772_10_GE_C06.indd 417
x
csc u =
1
y
cos u = x
y ≠ 0
sec u =
1
x
x ≠ 0
tan u =
y
x
x ≠ 0
cot u =
x
y
y ≠ 0
For sin u and cos u, there is no concern about dividing by 0, so u can be any
angle. It follows that the domain of the sine function and cosine function is the set
of all real numbers.
11/03/16 1:46 PM
418
CHAPTER 6 Trigonometric Functions
The domain of the sine function is the set of all real numbers.
The domain of the cosine function is the set of all real numbers.
For the tangent function and secant function, the x-coordinate of P = 1x, y2
cannot be 0 since this results in division by 0. See Figure 35. On the unit circle, there
are two such points, 10, 12 and 10, - 12. These two points correspond to the angles
p
3p
190°2 and
1270°2 or, more generally, to any angle that is an odd integer multiple
2
2
p
p
3p
5p
of 190°2, such as { 1 {90°2, {
1 {270°2, {
1 {450°2, and so on. Such
2
2
2
2
angles must be excluded from the domain of the tangent function and secant function.
The domain of the tangent function is the set of all real numbers, except
p
odd integer multiples of 190°2.
2
The domain of the secant function is the set of all real numbers, except
p
odd integer multiples of 190°2.
2
For the cotangent function and cosecant function, the y-coordinate of P = 1x, y2
cannot be 0 since this results in division by 0. See Figure 35. On the unit circle, there
are two such points, 11, 02 and 1 - 1, 02. These two points correspond to the angles
010°2 and p1180°2 or, more generally, to any angle that is an integer multiple of
p1180°2, such as 010°2, {p1 {180°2, {2p1 {360°2, {3p1 {540°2, and so on.
Such angles must therefore be excluded from the domain of the cotangent function
and cosecant function.
The domain of the cotangent function is the set of all real numbers,
except integer multiples of p1180°2.
The domain of the cosecant function is the set of all real numbers, except
integer multiples of p1180°2.
Next we determine the range of each of the six trigonometric functions. Refer
again to Figure 35. Let P = 1x, y2 be the point on the unit circle that corresponds
to the angle u. It follows that - 1 … x … 1 and - 1 … y … 1. Since sin u = y and
cos u = x, we have
- 1 … sin u … 1
- 1 … cos u … 1
The range of both the sine function and the cosine function consists of all real
numbers between - 1 and 1, inclusive. Using absolute value notation, we have
0 sin u 0 … 1 and 0 cos u 0 … 1.
1
If u is not an integer multiple of p1180°2, then csc u = . Since y = sin u and
y
1
1
1
1
0 y 0 = 0 sin u 0 … 1, it follows that 0 csc u 0 =
=
Ú 1 a … - 1 or Ú 1b .
y
y
0 sin u 0
0y0
1
Since csc u = , the range of the cosecant function consists of all real numbers less
y
than or equal to - 1 or greater than or equal to 1. That is,
csc u … - 1 or csc u Ú 1
M06_SULL1772_10_GE_C06.indd 418
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Section 6.3 Properties of the Trigonometric Functions
419
p
1
190°2, then sec u = . Since x = cos u and
x
2
1
1
1
1
0 x 0 = 0 cos u 0 … 1, it follows that 0 sec u 0 =
=
Ú 1 a … - 1 or Ú 1b .
x
x
0 cos u 0
0x0
1
Since sec u = , the range of the secant function consists of all real numbers less
x
than or equal to - 1 or greater than or equal to 1. That is,
If u is not an odd integer multiple of
sec u … - 1 or sec u Ú 1
The range of both the tangent function and the cotangent function is the set of
all real numbers.
- q 6 tan u 6 q
- q 6 cot u 6 q
You are asked to prove this in Problems 121 and 122.
Table 4 summarizes these results.
Table 4
Function
Symbol
Domain
Range
sine
f(u) = sin u
All real numbers
All real numbers from - 1 to 1, inclusive
cosine
f(u) = cos u
All real numbers
All real numbers from - 1 to 1, inclusive
tangent
f(u) = tan u
p
All real numbers, except odd integer multiples of (90°)
2
All real numbers
cosecant
f(u) = csc u
All real numbers, except integer multiples of p (180°)
All real numbers greater than or equal to
1 or less than or equal to - 1
secant
f(u) = sec u
All real numbers, except odd integer multiples of
cotangent
f(u) = cot u
All real numbers, except integer multiples of p (180°)
y
1
1
p
(90°)
2
All real numbers greater than or equal to
1 or less than or equal to - 1
All real numbers
Now Work p r o b l e m 9 7
3
P –2 , ––
2
–
3
1
1
2 Determine the Period of the Trigonometric Functions
x
– 2
3
1
Figure 36 sina
p
p
+ 2pb = sin ;
3
3
cos a
p
p
+ 2pb = cos
3
3
y
1
P (x, y )
u
1
1
x
u 2p
1
Figure 37 sin (u + 2pk) = sin u;
cos (u + 2pk) = cos u
M06_SULL1772_10_GE_C06.indd 419
p
Look at Figure 36. This figure shows that for an angle of radians the corresponding
3
1 23
p
point P on the unit circle is a ,
b . Notice that, for an angle of
+ 2p radians,
2 2
3
1 23
the corresponding point P on the unit circle is also a ,
b . Then
2 2
sin
p
23
=
3
2
cos
p
1
=
3
2
and
sin a
and cos a
p
23
+ 2pb =
3
2
p
1
+ 2pb =
3
2
This example illustrates a more general situation. For a given angle u, measured
in radians, suppose that we know the corresponding point P = 1x, y2 on the unit
circle. Now add 2p to u. The point on the unit circle corresponding to u + 2p is
identical to the point P on the unit circle corresponding to u. See Figure 37. The
values of the trigonometric functions of u + 2p are equal to the values of the
corresponding trigonometric functions of u.
If we add (or subtract) integer multiples of 2p to u, the values of the sine and
cosine function remain unchanged. That is, for all u
11/03/16 1:46 PM
420
CHAPTER 6 Trigonometric Functions
sin 1u + 2pk2 = sin u
cos 1u + 2pk2 = cos u
where k is any integer.
(1)
Functions that exhibit this kind of behavior are called periodic functions.
Definition
A function f is called periodic if there is a positive number p such that,
whenever u is in the domain of f, so is u + p, and
f1u + p2 = f1u2
If there is a smallest such number p, this smallest value is called the
(fundamental) period of f.
Based on equation (1), the sine and cosine functions are periodic. In fact,
the sine and cosine functions have period 2p. You are asked to prove this fact in
Problems 123 and 124. The secant and cosecant functions are also periodic with
period 2p, and the tangent and cotangent functions are periodic with period p. You
are asked to prove these statements in Problems 125 through 128.
These facts are summarized as follows:
Periodic Properties
In Words
sin 1u + 2p2 = sin u
Tangent and cotangent have
period p; the others have
period 2p.
csc1u + 2p2 = csc u
cos 1u + 2p2 = cos u
sec1u + 2p2 = sec u
tan 1u + p2 = tan u
cot1u + p2 = cot u
Because the sine, cosine, secant, and cosecant functions have period 2p, once we
know their values over an interval of length 2p, we know all their values; similarly,
since the tangent and cotangent functions have period p, once we know their values
over an interval of length p, we know all their values.
E xam pl e 1
Finding Exact Values Using Periodic Properties
Find the exact value of:
(a) sin
Solution
17p
5p
(b) cos 15p2 (c) tan
4
4
(a) It is best to sketch the angle first, as shown in Figure 38(a). Since the period of
p
the sine function is 2p, each full revolution can be ignored leaving the angle .
4
Then
sin
17p
p
p
22
= sin a + 4pb = sin =
4
4
4
2
(b) See Figure 38(b). Since the period of the cosine function is 2p, each full revolution
can be ignored leaving the angle p. Then
cos 15p2 = cos 1p + 4p2 = cos p = - 1
(c) See Figure 38(c). Since the period of the tangent function is p, each
p
half-revolution can be ignored leaving the angle . Then
4
5p
p
p
tan
= tan a + pb = tan = 1
4
4
4
M06_SULL1772_10_GE_C06.indd 420
11/03/16 1:47 PM
Section 6.3 Properties of the Trigonometric Functions
y
y
y
5
u –––
4
u 5
–––
u 17
4
x
x
x
(a)
Figure 38
421
(c)
(b)
•
The periodic properties of the trigonometric functions will be very helpful to us
when we study their graphs later in the chapter.
Now Work p r o b l e m 1 1
3 Determine the Signs of the Trigonometric Functions
in a Given Quadrant
Let P = 1x, y2 be the point on the unit circle that corresponds to the angle u. If
we know in which quadrant the point P lies, then we can determine the signs of
the trigonometric functions of u. For example, if P = 1x, y2 lies in quadrant IV, as
shown in Figure 39, then we know that x 7 0 and y 6 0. Consequently,
y
1
u
1
1
1
sin u = y 6 0
x
P (x, y ), x 0, y 0
Figure 39
u in quadrant IV, x 7 0, y 6 0
csc u =
y
6 0
x
x
cot u = 6 0
y
cos u = x 7 0
1
6 0
y
sec u =
tan u =
1
7 0
x
Table 5 lists the signs of the six trigonometric functions for each quadrant. See
also Figure 40.
Table 5
Quadrant of P
sin U, csc U
cos U, sec U
tan U, cot U
I
Positive
Positive
Positive
II
Positive
Negative
Negative
III
Negative
Negative
Positive
IV
Negative
Positive
Negative
+
y
II (–, +)
I (+, +)
sin u 0, csc u 0
others negative
All positive
y
–
–
x
–
III (–, –)
IV (+, –)
tan u 0, cot u 0
others negative
cos u 0, sec u 0
others negative
–
–
y
y
x
sine
cosecant
x
cosine
secant
x
tangent
cotangent
+
+
+
(a)
+
+
–
(b)
Figure 40 Signs of the trigonometric functions
M06_SULL1772_10_GE_C06.indd 421
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422
CHAPTER 6 Trigonometric Functions
Exampl e 2
Finding the Quadrant in Which an Angle U Lies
If sin u 6 0 and cos u 6 0, name the quadrant in which the angle u lies.
Solution
Let P = 1x, y2 be the point on the unit circle corresponding to u. Then
sin u = y 6 0 and cos u = x 6 0. Because points in quadrant III have x 6 0
and y 6 0, u lies in quadrant III.
•
Now Work p r o b l e m 2 7
4 Find the Values of the Trigonometric Functions Using
Fundamental Identities
If P = 1x, y2 is the point on the unit circle corresponding to u, then
sin u = y
csc u =
1
y
cos u = x
if y ≠ 0
sec u =
1
x
if x ≠ 0
tan u =
y
x
if x ≠ 0
cot u =
x
y
if y ≠ 0
Based on these definitions, we have the reciprocal identities:
Reciprocal Identities
csc u =
1
sin u
sec u =
1
cos u
cot u =
1
tan u
(2)
Two other fundamental identities are the quotient identities.
Quotient Identities
tan u =
sin u
cos u
cot u =
cos u
sin u
(3)
The proofs of identities (2) and (3) follow from the definitions of the
trigonometric functions. (See Problems 129 and 130.)
If sin u and cos u are known, identities (2) and (3) make it easy to find the values
of the remaining trigonometric functions.
E xam pl e 3
Finding Exact Values Using Identities When Sine
and Cosine Are Given
25
225
and cos u =
, find the exact values of the four remaining
5
5
trigonometric functions of u using identities.
Given sin u =
Solution
Based on a quotient identity from (3), we have
25
sin u
5
1
tan u =
=
=
cos u
2
225
5
M06_SULL1772_10_GE_C06.indd 422
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Section 6.3 Properties of the Trigonometric Functions
423
Then we use the reciprocal identities from (2) to get
csc u =
1
5
1
=
=
= 25
sin u
25
25
sec u =
1
1
5
25
=
=
=
cos u
2
225
225
5
cot u =
5
1
1
=
= 2
tan u
1
2
•
Now Work p r o b l e m 3 5
The equation of the unit circle is x2 + y2 = 1 or, equivalently,
y 2 + x2 = 1
If P = 1x, y2 is the point on the unit circle that corresponds to the angle u, then
y = sin u and x = cos u, so we have
1sin u2 2 + 1cos u2 2 = 1
(4)
sin2 u + cos2 u = 1
(5)
It is customary to write sin2 u instead of 1sin u2 2, cos2 u instead of 1cos u2 2, and
so on. With this notation, we can rewrite equation (4) as
If cos u ≠ 0, we can divide each side of equation (5) by cos2 u.
sin2 u
cos2 u
1
+
=
2
cos u
cos2 u
cos2 u
2
sin u 2
1
b + 1 = a
b
cos u
cos u
Now use identities (2) and (3) to get
a
tan2 u + 1 = sec2 u
(6)
Similarly, if sin u ≠ 0, we can divide equation (5) by sin2 u and use identities (2)
and (3) to get 1 + cot 2 u = csc2 u, which we write as
cot 2 u + 1 = csc2 u
(7)
Collectively, the identities in (5), (6), and (7) are referred to as the Pythagorean
identities.
Fundamental Identities
sin u
cos u
1
1
csc u =
sec u =
sin u
cos u
sin2 u + cos2 u = 1
tan2 u + 1 = sec2 u
tan u =
E xam pl e 4
cos u
sin u
1
cot u =
tan u
cot 2 u + 1 = csc2 u
cot u =
Finding the Exact Value of a Trigonometric Expression
Using Identities
Find the exact value of each expression. Do not use a calculator.
(a) tan 20° -
M06_SULL1772_10_GE_C06.indd 423
sin 20°
cos 20°
(b) sin2
p
+
12
1
sec2
p
12
11/03/16 1:47 PM
424
CHAPTER 6 Trigonometric Functions
Solution
(a) tan 20° -
sin 20°
= tan 20° - tan 20° = 0
cos 20° c
sin u
= tan u
cos u
(b) sin2
p
+
12
1
p
sec2
12
p
p
+ cos2
= 1
12
12
= sin2
æ
cos u =
1
sec u
æ
sin u + cos2 u = 1
2
•
Now Work p r o b l e m 7 9
5 Find the Exact Values of the Trigonometric Functions
of an Angle Given One of the Functions and the Quadrant
of the Angle
Many problems require finding the exact values of the remaining trigonometric
functions when the value of one of them is known and the quadrant in which u lies
can be found. There are two approaches to solving such problems. One approach
uses a circle of radius r; the other uses identities.
When using identities, sometimes a rearrangement is required. For example, the
Pythagorean identity
sin2 u + cos2 u = 1
can be solved for sin u in terms of cos u (or vice versa) as follows:
sin2 u = 1 - cos2 u
sin u = { 21 - cos2 u
where the + sign is used if sin u 7 0 and the - sign is used if sin u 6 0. Similarly,
in tan2 u + 1 = sec2 u, we can solve for tan u (or sec u), and in cot 2 u + 1 = csc2 u,
we can solve for cot u (or csc u).
Exampl e 5
Finding Exact Values Given One Value and the Sign of Another
1
and cos u 6 0, find the exact value of each of the remaining
3
five trigonometric functions.
Given that sin u =
Option 1
Using a Circle
y
3
P (x, 1)
u
3
O
3
x
Suppose that P = 1x, y2 is the point on a circle that corresponds to u. Since
1
sin u =
7 0 and cos u 6 0, the point P = 1x, y2 is in quadrant II. Because
3
y
1
sin u = = , we let y = 1 and r = 3. The point P = 1x, y2 = 1x, 12 that
r
3
corresponds to u lies on the circle of radius 3 centered at the origin: x2 + y2 = 9.
See Figure 41.
To find x, we use the fact that x2 + y2 = 9, y = 1, and P is in quadrant II
1so x 6 02.
x2 + y 2 = 9
x
3
Figure 41
M06_SULL1772_10_GE_C06.indd 424
2 y2 9
x2 + 12 = 9
y = 1
x2 = 8
x = - 222
x 6 0
11/03/16 1:47 PM
Section 6.3 Properties of the Trigonometric Functions
425
Since x = - 222 , y = 1, and r = 3, we find that
cos u =
x
212
= r
3
tan u =
y
22
1
= =
x
4
- 222
csc u =
r
3
= = 3
y
1
sec u =
r
322
3
= =
x
4
- 222
Option 2
Using Identities
cot u =
x
- 222
= - 222
=
y
1
First, solve the identity sin2 u + cos2 u = 1 for cos u.
sin2 u + cos2 u = 1
cos2 u = 1 - sin2 u
cos u = { 21 - sin2 u
Because cos u 6 0, choose the minus sign and use the fact that sin u =
cos u = - 21 - sin2 u = c
sin u =
1
3
A
1 -
1
.
3
1
8
222
= = 9
A9
3
Now we know the values of sin u and cos u, so we can use quotient and reciprocal
identities to get
tan u =
1
3
sin u
1
22
=
=
= cos u
4
- 222
- 222
cot u =
1
= - 222
tan u
3
sec u =
1
1
-3
322
=
=
= cos u
4
222
- 222
3
csc u =
1
1
=
= 3
sin u
1
3
•
Finding the Values of the Trigonometric Functions of U When the
Value of One Function Is Known and the Quadrant of U Is Known
Given the value of one trigonometric function and the quadrant in which u lies,
the exact value of each of the remaining five trigonometric functions can be
found in either of two ways.
Option 1
Step 1:
Step 2:
Step 3:
Step 4:
Using a Circle of Radius r
Draw a circle centered at the origin showing the location of the
angle u and the point P = 1x, y2 that corresponds to u. The radius of
the circle that contains P = 1x, y2 is r = 2x2 + y2 .
Assign a value to two of the three variables x, y, r based on the value
of the given trigonometric function and the location of P.
Use the fact that P lies on the circle x2 + y2 = r 2 to find the value of
the missing variable.
Apply the theorem on page 411 to find the values of the remaining
trigonometric functions.
Option 2
Using Identities
Use appropriately selected identities to find the value of each remaining
trigonometric function.
M06_SULL1772_10_GE_C06.indd 425
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426
CHAPTER 6 Trigonometric Functions
Exampl e 6
Given the Value of One Trigonometric Function and the Sign
of Another, Find the Values of the Remaining Ones
1
and sin u 6 0, find the exact value of each of the remaining
2
five trigonometric functions of u.
Given that tan u =
Option 1
Using a Circle
y
3
x 2 y2 5
u
3 x
3
P (2, 1)
1
7 0 and sin u 6 0, the point P = 1x, y2 that corresponds
2
to u lies in quadrant III. See Figure 42.
y
1
Step 2: Since tan u = = and u lies in quadrant III, let x = - 2 and y = - 1.
x
2
Step 1:Since tan u =
Step 3:With x = - 2 and y = - 1, then r = 2x2 + y2 = 21 - 22 2 + 1 - 12 2 = 25,
so P lies on the circle x2 + y2 = 5.
Step 4: Apply the theorem on page 411 using x = - 2, y = - 1, and r = 25.
sin u =
3
1
Figure 42 tan u = ; u in quadrant III
2
Option 2
Using Identities
csc u =
y
-1
25
=
= r
5
25
cos u =
r
25
= - 25
=
y
-1
sec u =
-2
x
225
=
= r
5
25
r
25
25
= =
x
-2
2
cot u =
x
-2
= 2
=
y
-1
•
Because the value of tan u is known, use the Pythagorean identity that involves
1
tan u, that is, tan2 u + 1 = sec2 u. Since tan u = 7 0 and sin u 6 0, then u lies in
2
quadrant III, where sec u 6 0.
tan2 u + 1 = sec2 u
Pythagorean identity
2
1
a b + 1 = sec2 u
2
sec2 u =
1
2
1
5
+ 1 =
Proceed to solve for sec u.
4
4
sec u = Now we know tan u =
tan u =
25
2
sec u 6 0
1
15
and sec u = . Using reciprocal identities, we find
2
2
cos u =
cot u =
1
=
sec u
1
-
25
2
= -
2
25
= -
225
5
1
1
=
= 2
tan u
1
2
To find sin u, use the following reasoning:
tan u =
sin u
cos u
csc u =
1
=
sin u
1
225
25
so sin u = (tan u)(cos u) = a b # a b = 2
5
5
1
25
5
= -
5
25
= - 25
•
Now Work p r o b l e m 4 3
M06_SULL1772_10_GE_C06.indd 426
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Section 6.3 Properties of the Trigonometric Functions
427
6 Use Even–Odd Properties to Find the Exact Values
of the Trigonometric Functions
Recall that a function f is even if f1 - u2 = f1u2 for all u in the domain of f ; a
function f is odd if f1 - u2 = - f 1u2 for all u in the domain of f. We will now show
that the trigonometric functions sine, tangent, cotangent, and cosecant are odd
functions and the functions cosine and secant are even functions.
Even–Odd Properties
In Words
sin 1 - u2 = - sin u
Cosine and secant are even
functions; the others are odd
functions.
csc1 - u2 = - csc u
O
1
P (x, y )
u
A (1, 0)
u
1
sin u = y
x
so
Q (x, y )
1
tan 1 - u2 = - tan u
sec1 - u2 = sec u
cot1 - u2 = - cot u
Proof Let P = 1x, y2 be the point on the unit circle that corresponds to the angle u.
See Figure 43. Using symmetry, the point Q on the unit circle that corresponds to the
angle - u will have coordinates 1x, - y2. Using the definition of the trigonometric
functions, we have
y
1
cos 1 - u2 = cos u
sin 1 - u2 = - y
cos u = x
sin 1 - u2 = - y = - sin u
cos 1 - u2 = x
cos 1 - u2 = x = cos u
Now, using these results and some of the fundamental identities, we have
Figure 43
tan 1 - u2 =
sec1 - u2 =
E xam pl e 7
sin 1 - u2
- sin u
=
= - tan u
cos 1 - u2
cos u
1
1
=
= sec u
cos 1 - u2
cos u
1
1
=
= - cot u
tan 1 - u2
- tan u
cot1 - u2 =
1
1
=
= - csc u
sin 1 - u2
- sin u
csc1 - u2 =
■
Finding Exact Values Using Even–Odd Properties
Find the exact value of:
(a) sin 1 - 45°2 (b) cos 1 - p2 (c) cota -
Solution
(a) sin 1 - 45°2 = - sin 45° = c
Odd function
(c) cota -
22
2
3p
37p
b (d) tan a b
2
4
(b) cos 1 - p2 = cos p = - 1
c
Even function
3p
3p
b = - cot
= 0
2
2
c
Odd function
(d) tan a -
37p
37p
p
p
b = - tan
= - tan a + 9pb = - tan = - 1
4
4
4
4
c
c
Odd function
Period is p.
•
Now Work p r o b l e m 5 9
M06_SULL1772_10_GE_C06.indd 427
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428
CHAPTER 6 Trigonometric Functions
6.3 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
x + 1
1. The domain of the function f 1x2 =
is
2x + 1
(pp. 79–90)
.
2. A function for which f 1x2 = f 1 - x2 for all x in the domain
of f is called a(n)
function. (pp. 103–105)
3. True or False The function f 1x2 = 1x is even.
(pp. 103–105)
4. True or False The equation x2 + 2x = 1x + 12 2 - 1 is an
identity. (p. 987)
Concepts and Vocabulary
5. The sine, cosine, cosecant, and secant functions have period
; the tangent and cotangent functions have period .
6. The domain of the tangent function is
.
7. Which of the following is not in the range of the sine
function?
p
3
(a) (b) (c) - 0.37 (d) - 1
4
2
8. Which of the following functions is even?
(a) cosine (b) sine (c) tangent (d) cosecant
9. sin2 u + cos2 u =
.
10. True or False sec u =
1
sin u
Skill Building
In Problems 11–26, use the fact that the trigonometric functions are periodic to find the exact value of each expression. Do not use a
calculator.
11. sin 405°
12. cos 420°
13. sin 390°
17. sec 420°
18. cot 390°
19. sin
9p
4
20. cos
33p
4
17p
4
25. sec
25p
6
26. tan
19p
6
23. cot
17p
4
24. sec
14. tan 405°
15. sec 540°
21. csc
16. csc 450°
9p
2
22. tan121p2
In Problems 27–34, name the quadrant in which the angle u lies.
27. sin u 7 0,
cos u 6 0
28. sin u 6 0,
cos u 7 0
29. cos u 7 0,
tan u 7 0
30. sin u 6 0,
tan u 6 0
31. cos u 6 0,
tan u 7 0
32. cos u 7 0,
tan u 6 0
33. csc u 7 0,
cos u 6 0
34. sec u 6 0,
sin u 7 0
In Problems 35–42, sin u and cos u are given. Find the exact value of each of the four remaining trigonometric functions.
3
4
35. sin u = - , cos u =
5
5
36. sin u =
4
3
, cos u = 5
5
37. sin u = -
1
23
, cos u =
2
2
40. sin u =
38. sin u =
225
25
, cos u =
5
5
39. sin u =
41. sin u =
222
1
, cos u = 3
3
222
1
42. sin u = - , cos u =
3
3
25
225
, cos u = 5
5
1
23
, cos u =
2
2
In Problems 43–58, find the exact value of each of the remaining trigonometric functions of u.
12
, u in quadrant II
13
4
46. cos u = - , u in quadrant III
5
2
3p
49. sin u = - , p 6 u 6
3
2
2
52. sin u = , tan u 6 0
3
4
55. cot u = , cos u 6 0
3
1
58. tan u = - , sin u 7 0
3
43. sin u =
M06_SULL1772_10_GE_C06.indd 428
3
, u in quadrant IV
5
4
47. cos u = , 270° 6 u 6 360°
5
1 p
50. cos u = - ,
6 u 6 p
3 2
5
, 90° 6 u 6 180°
13
1
51. cos u = - , tan u 7 0
4
53. csc u = 3,
54. sec u = 2,
44. cos u =
56. tan u =
cot u 6 0
3
, sin u 6 0
4
45. sin u = -
5
, u in quadrant III
13
48. sin u =
57. sec u = - 2,
sin u 6 0
tan u 7 0
11/03/16 1:47 PM
Section 6.3 Properties of the Trigonometric Functions
429
In Problems 59–76, use the even–odd properties to find the exact value of each expression. Do not use a calculator.
60. cos 1 - 30°2
59. sin1 - 60°2
64. sec1 - 60°2
69. sina 74. csca -
p
b
3
p
b
4
61. sin1 - 135°2
62. tan1 - 30°2
63. csc1 - 30°2
65. cos 1 - 270°2
66. sin1 - 90°2
67. sin1 - p2
68. tana -
70. cos a -
71. sina -
72. tan1 - p2
73. sec1 - p2
75. csca -
p
b
4
p
b
3
76. seca -
3p
b
2
p
b
4
p
b
6
In Problems 77–88, use properties of the trigonometric functions to find the exact value of each expression. Do not use a calculator.
77. sec2 18° - tan2 18°
81. cot 20° 85. seca -
cos 20°
sin 20°
78. sin2 40° + cos2 40°
82. tan 40° -
p #
37p
b cos
18
18
86. sina -
89. If sin u = 0.3, find the value of:
sin 40°
cos 40°
p
25p
b csc
12
12
sin u + sin1u + 2p2 + sin1u + 4p2
90. If cos u = 0.2, find the value of:
79. sin 80° csc 80°
80. tan 10° cot 10°
83. tan 200° # cot 20°
84. cos 400° # sec 40°
87.
sin 70°
+ tan1- 70°2
cos 1- 430°2
88.
sin1 - 20°2
cos 380°
+ tan 200°
100. For what numbers u is f 1u2 = csc u not defined?
101. What is the range of the sine function?
102. What is the range of the cosine function?
cos u + cos 1u + 2p2 + cos 1u + 4p2
103. What is the range of the tangent function?
tan u + tan1u + p2 + tan1u + 2p2
105. What is the range of the secant function?
91. If tan u = 3, find the value of:
92. If cot u = - 2, find the value of:
104. What is the range of the cotangent function?
106. What is the range of the cosecant function?
cot u + cot 1u - p2 + cot 1u - 2p2
107. Is the sine function even, odd, or neither? Is its graph
symmetric? With respect to what?
sin 1° + sin 2° + sin 3° + g + sin 358° + sin 359°
108. Is the cosine function even, odd, or neither? Is its graph
symmetric? With respect to what?
93. Find the exact value of:
94. Find the exact value of:
cos 1° + cos 2° + cos 3° + g + cos 358° + cos 359°
95. What is the domain of the sine function?
96. What is the domain of the cosine function?
97. For what numbers u is f 1u2 = tan u not defined?
98. For what numbers u is f 1u2 = cot u not defined?
99. For what numbers u is f 1u2 = sec u not defined?
109. Is the tangent function even, odd, or neither? Is its graph
symmetric? With respect to what?
110. Is the cotangent function even, odd, or neither? Is its graph
symmetric? With respect to what?
111. Is the secant function even, odd, or neither? Is its graph
symmetric? With respect to what?
112. Is the cosecant function even, odd, or neither? Is its graph
symmetric? With respect to what?
Applications and Extensions
In Problems 113–118, use the periodic and even–odd properties.
1
113. If f 1x2 = cos 1x2 and f 1a2 = , find the exact value of the
2
following.
(a) f 1 - a2
(b) f 1a2 + f 1a + 2p2 + f 1a + 4p2
114. If f 1u2 = cos u and f 1a2 =
(a) f 1 - a2
M06_SULL1772_10_GE_C06.indd 429
1
, find the exact value of:
4
(b) f 1a2 + f 1a + 2p2 + f 1a - 2p2
115. If f 1x2 = cot 1x2 and f 1a2 = 2, find the exact value of the
following.
(a) f 1 - a2
(b) f 1a2 + f 1a + p2 + f 1a + 2p2
116. If f 1u2 = cot u and f 1a2 = - 3, find the exact value of:
(a) f 1 - a2
(b) f 1a2 + f 1a + p2 + f 1a + 4p2
117. If f 1x2 = sec1x2 and f 1a2 = - 11, find the exact value of
the following.
(a) f 1 - a2
(b) f 1a2 + f 1a + 2p2 + f 1a + 4p2
3/22/16 10:56 AM
430
CHAPTER 6 Trigonometric Functions
118. If f 1u2 = csc u and f 1a2 = 2, find the exact value of:
(a) f 1 - a2
(b) f 1a2 + f 1a + 2p2 + f 1a + 4p2
119. Calculating the Time of a Trip From a parking lot you
want to walk to a house on the ocean. The house is located
1480 ft down a paved path that parallels the beach, which
is 740 ft wide. Along the path, you can walk 330 ft/min,
but on the beach you can only walk 130 ft/min. Calculate
the time T if you walk directly from the parking lot to the
house.
Ocean
4 mi
4 mi
Beach
Paved path
1 mi
u
u
x
River
Ocean
740 ft
Beach
u
x
Forest
121. Show that the range of the tangent function is the set of all
real numbers.
Paved path
122. Show that the range of the cotangent function is the set of all
real numbers.
Parking lot
123. Show that the period of f 1u2 = sin u is 2p.
[Hint: Assume that 0 6 p 6 2p exists so that
sin1u + p2 = sin u for all u. Let u = 0 to find p. Then let
1480 ft
120. Calculating the Time of a Trip Two oceanfront homes are
located 8 miles apart on a straight stretch of beach, each a
distance of 1 mile from a paved path that parallels the ocean.
Sally can jog 8 miles per hour on the paved path, but only
3 miles per hour in the sand on the beach. Because a river
flows directly between the two houses, it is necessary to jog
in the sand to the road, continue on the path, and then jog
directly back in the sand to get from one house to the other.
See the illustration. The time T to get from one house to the
other as a function of the angle u shown in the illustration is
T 1u2 = 1 +
2
1
3 sin u
4 tan u
(a) Calculate the time T for tan u =
0 6 u 6
p
2
1
.
4
(b) Describe the path taken.
(c) Explain why u must be larger than 14°.
p
to obtain a contradiction.]
2
124. Show that the period of f 1u2 = cos u is 2p.
u =
125. Show that the period of f 1u2 = sec u is 2p.
126. Show that the period of f 1u2 = csc u is 2p.
127. Show that the period of f 1u2 = tan u is p.
128. Show that the period of f 1u2 = cot u is p.
1
1
129. Prove the reciprocal identities csc u =
, sec u =
,
sin u
cos u
1
and cot u =
.
tan u
130. Prove the quotient identities given in formula (3).
131. Establish the identity,
1sin u cos f2 2 + 1sin u sin f2 2 + cos2 u = 1.
Explaining Concepts: Discussion and Writing
132. Write down five properties of the tangent function. Explain
the meaning of each.
133. Describe your understanding of the meaning of a periodic
function.
135. Explain how to find the value of cos 1 - 45°2 using even–odd
properties.
136. Explain how to find the value of sin 390° and cos ( - 45°)
using the unit circle.
134. Explain how to find the value of sin 390° using periodic
properties.
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Section 6.4 Graphs of the Sine and Cosine Functions
431
Retain Your Knowledge
Problems 137–140 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
137. Given: f 1x2 = x2 - 3 and g1x2 = x - 7, find 1f ∘ g2 1x2 .
138. Graph f(x) = - 2x2 + 12x - 13 using transformations. Find the vertex and the axis of symmetry.
139. Solve exactly: e x - 4 = 6
140. Find the real zeros of f(x) = x3 - 9x2 + 3x - 27.
‘Are You Prepared?’ Answers
1. e x x ≠ -
1
f
2
2. even
3. False
4. True
6.4 Graphs of the Sine and Cosine Functions*
Preparing for this section Before getting started, review the following:
• Graphing Techniques: Transformations (Section 2.5, pp. 127–136)
Now Work the ‘Are You Prepared?’ problems on page 441.
Objectives 1 Graph Functions of the Form y = A sin 1vx2 Using Transformations (p. 432)
2 Graph Functions of the Form y = A cos 1vx2 Using Transformations (p. 434)
3 Determine the Amplitude and Period of Sinusoidal Functions (p. 435)
4 Graph Sinusoidal Functions Using Key Points (p. 437)
5 Find an Equation for a Sinusoidal Graph (p. 440)
Since we want to graph the trigonometric functions in the xy-plane, we shall use
the traditional symbols x for the independent variable (or argument) and y for the
dependent variable (or value at x) for each function. So the six trigonometric functions
can be written as
y = f 1x2 = sin x
y = f1x2 = csc x
y = f1x2 = cos x
y = f1x2 = tan x
y = f1x2 = sec x
y = f1x2 = cot x
Here the independent variable x represents an angle, measured in radians. In calculus,
x will usually be treated as a real number. As noted earlier, these are equivalent
ways of viewing x.
The Graph of the Sine Function y = sin x
Because the sine function has period 2p, it is only necessary to graph y = sin x on
the interval 3 0, 2p4 . The remainder of the graph will consist of repetitions of this
portion of the graph.
To begin, consider Table 6 on page 432, which lists some points on the graph
of y = sin x, 0 … x … 2p. As the table shows, the graph of y = sin x, 0 … x … 2p,
p
begins at the origin. As x increases from 0 to , the value of y = sin x increases
2
p
3p
from 0 to 1; as x increases from to p to , the value of y decreases from 1 to 0 to - 1;
2
2
*For those who wish to include phase shifts here, Section 6.6 can be covered immediately after Section 6.4
without loss of continuity.
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