Trigonometric Functions

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5
Trigonometric Functions
Highway transportation is critical to the economy of the United States.
In 1970 there were 1150 billion miles traveled, and by the year 2000 this
increased to approximately 2500 billion miles. When an automobile
travels around a curve, objects like trees, buildings, and fences situated on
the curve may obstruct a driver’s vision. Trigonometry is used to determine how far inside the curve land must be cleared to provide visibility
for a safe stopping distance. (Source: Mannering, F. and W. Kilareski,
Principles of Highway Engineering and Traffic Analysis, 2nd Edition,
John Wiley & Sons, 1998.)
A problem like this is presented in Exercise 61 of Section 5.4.
5.1 Angles
5.2 Trigonometric Functions
5.3 Evaluating Trigonometric Functions
5.4 Solving Right Triangles
473
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474 CHAPTER 5 Trigonometric Functions
5.1 Angles
Basic Terminology
в– Degree Measure
Line AB
A
B
A
B
A
B
Segment AB
Ray AB
Figure 1
Terminal side
в– Standard Position
в– Coterminal Angles
Basic Terminology Two distinct points A and B determine a line called
line AB. The portion of the line between A and B, including points A and B
themselves, is line segment AB, or simply segment AB. The portion of line AB
that starts at A and continues through B, and on past B, is called ray AB. Point A
is the endpoint of the ray. (See Figure 1.)
An angle is formed by rotating a ray around its endpoint. The ray in its initial position is called the initial side of the angle, while the ray in its location after the rotation is the terminal side of the angle. The endpoint of the ray is the
vertex of the angle. Figure 2 shows the initial and terminal sides of an angle
with vertex A. If the rotation of the terminal side is counterclockwise, the angle
is positive. If the rotation is clockwise, the angle is negative. Figure 3 shows
two angles, one positive and one negative.
Vertex A
Initial side
A
C
Figure 2
B
Positive angle
Negative angle
Figure 3
An angle can be named by using the name of its vertex. For example, the
angle on the right in Figure 3 can be called angle C. Alternatively, an angle can
be named using three letters, with the vertex letter in the middle. Thus, the angle
on the right also could be named angle ACB or angle BCA.
A complete rotation of a ray
gives an angle whose measure
is 360В°.
Figure 4
Degree Measure The most common unit for measuring angles is the
degree. (The other common unit of measure, called the radian, is discussed in
Section 6.1.) Degree measure was developed by the Babylonians, 4000 years
ago. To use degree measure, we assign 360 degrees to a complete rotation of a
ray.* In Figure 4, notice that the terminal side of the angle corresponds to its initial side when it makes a complete rotation. One degree, written 1В°, represents
90
1
1
360 of a rotation. Therefore, 90В° represents 360 а·‡ 4 of a complete rotation, and
180
1
180В° represents 360 а·‡ 2 of a complete rotation. An angle measuring between 0В°
and 90В° is called an acute angle. An angle measuring exactly 90В° is a right
angle. An angle measuring more than 90В° but less than 180В° is an obtuse angle,
and an angle of exactly 180В° is a straight angle. See Figure 5, where we use the
Greek letter вђЄ (theta)** to name each angle.
*The Babylonians were the first to subdivide the circumference of a circle into 360 parts. There are various theories as to why the number 360 was chosen. One is that it is approximately the number of days in
a year, and it has many divisors, which makes it convenient to work with. Another involves a roundabout
theory dealing with the length of a Babylonian mile.
**In addition to вђЄ (theta), other Greek letters such as вђЈ (alpha) and вђ¤ (beta) are sometimes used to name
angles.
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5.1 Angles 475
вђЄ
TEACHING TIP Point out that two
angles can be complementary or
supplementary without having a
side in common. For instance, the
two acute angles of a right triangle
are complementary.
Acute angle
0В° < вђЄ < 90В°
вђЄ
вђЄ
вђЄ
Right angle
вђЄ = 90В°
Obtuse angle
90В° < вђЄ < 180В°
Straight angle
вђЄ = 180В°
Figure 5
If the sum of the measures of two positive angles is 90В°, the angles are
called complementary. Two positive angles with measures whose sum is 180В°
are supplementary.
EXAMPLE 1 Finding Measures of Complementary and Supplementary Angles
Find the measure of each angle in Figure 6.
Solution
(a) In Figure 6(a), since the two angles form a right angle (as indicated by
the symbol), they are complementary angles. Thus,
6m П© 3m а·‡ 90
(6m)В°
(3m)В°
9m а·‡ 90
m а·‡ 10.
(a)
The two angles have measures of 6Н‘10Н’ а·‡ 60РЉ and 3Н‘10Н’ а·‡ 30РЉ.
(b) The angles in Figure 6(b) are supplementary, so
(4k)В°
4k П© 6k а·‡ 180
(6k)В°
10k а·‡ 180
(b)
k а·‡ 18.
Figure 6
These angle measures are 4Н‘18Н’ а·‡ 72РЉ and 6Н‘18Н’ а·‡ 108РЉ.
Now try Exercises 13 and 15.
y
A = 35В°
O
x
Do not confuse an angle with its measure. Angle A of Figure 7 is a rotation;
the measure of the rotation is 35В°. This measure is often expressed by saying
that m͑angle A͒ is 35°, where m͑angle A͒ is read “the measure of angle A.” It is
convenient, however, to abbreviate mН‘angle AН’ а·‡ 35РЉ as A а·‡ 35РЉ.
Traditionally, portions of a degree have been measured with minutes and
1
seconds. One minute, written 1Р€, is 60 of a degree.
1Р€ а·‡
1РЉ
60
60Р€ а·‡ 1РЉ
or
1
Figure 7
One second, 1Р‰, is 60 of a minute.
1Р‰ а·‡
1 РЉ
1Р€
а·‡
60
3600
or
60Р‰ а·‡ 1Р€
The measure 12РЉ 42Р€ 38Р‰ represents 12 degrees, 42 minutes, 38 seconds.
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476 CHAPTER 5 Trigonometric Functions
EXAMPLE 2 Calculating with Degrees, Minutes, and Seconds
Perform each calculation.
(a) 51РЉ 29Р€ П© 32РЉ 46Р€
(b) 90РЉ ПЄ 73РЉ 12Р€
Solution
(a) Add the degrees and the minutes separately.
51РЉ 29Р€
П© 32РЉ 46Р€
83РЉ 75Р€
Since 75Р€ а·‡ 60Р€ П© 15Р€ а·‡ 1РЉ 15Р€, the sum is written
83РЉ
П© 1РЉ 15Р€
84РЉ 15Р€.
(b)
89РЉ 60Р€ Write 90В° as 89РЉ 60Р€.
ПЄ 73РЉ 12Р€
16РЉ 48Р€
Now try Exercises 23 and 27.
Because calculators are now so prevalent, angles are commonly measured
in decimal degrees. For example, 12.4238В° represents
12.4238РЉ а·‡ 12
4238 РЉ
.
10,000
EXAMPLE 3 Converting Between Decimal Degrees and Degrees, Minutes,
and Seconds
(a) Convert 74РЉ 8Р€ 14Р‰ to decimal degrees.
(b) Convert 34.817В° to degrees, minutes, and seconds.
Solution
(a) 74РЉ 8Р€ 14Р‰ а·‡ 74РЉ П©
8РЉ
14 РЉ
П©
60
3600
1РЉ
1 РЉ
and 1Р‰ а·‡ 3600
1Р€ а·‡ 60
П· 74РЉ П© .1333РЉ П© .0039РЉ
П· 74.137РЉ
Add; round to the nearest thousandth.
(b) 34.817РЉ а·‡ 34РЉ П© .817РЉ
A graphing calculator performs the conversions in Example 3 as shown above.
а·‡ 34РЉ П© .817Н‘60Р€Н’
1РЉ а·‡ 60Р€
а·‡ 34РЉ П© 49.02Р€
а·‡ 34РЉ П© 49Р€ П© .02Р€
а·‡ 34РЉ П© 49Р€ П© .02Н‘60Р‰Н’ 1Р€ а·‡ 60Р‰
а·‡ 34РЉ П© 49Р€ П© 1.2Р‰
а·‡ 34РЉ 49Р€ 1.2Р‰
Now try Exercises 33 and 37.
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5.1 Angles 477
Standard Position An angle is in standard position if its vertex is at the
origin and its initial side is along the positive x-axis. The angles in Figures 8(a)
and 8(b) are in standard position. An angle in standard position is said to lie in
the quadrant in which its terminal side lies. An acute angle is in quadrant I
(Figure 8(a)) and an obtuse angle is in quadrant II (Figure 8(b)). Figure 8(c)
shows ranges of angle measures for each quadrant when 0РЉ ПЅ вђЄ ПЅ 360РЉ. Angles
in standard position having their terminal sides along the x-axis or y-axis, such as
angles with measures 90В°, 180В°, 270В°, and so on, are called quadrantal angles.
90В°
y
y
Q II
QI
Q II
90В° < вђЄ < 180В°
0В°
360В°
180В°
Terminal side
x
Vertex 0
QI
0В° < вђЄ < 90В°
x
0
Initial side
Q III
Q IV
180В° < вђЄ < 270В° 270В° < вђЄ < 360В°
270В°
(a)
(b)
(c)
Figure 8
Coterminal Angles A complete rotation of a ray results in an angle measuring 360В°. By continuing the rotation, angles of measure larger than 360В° can
be produced. The angles in Figure 9 with measures 60В° and 420В° have the same
initial side and the same terminal side, but different amounts of rotation. Such
angles are called coterminal angles; their measures differ by a multiple of 360В°.
As shown in Figure 10, angles with measures 110В° and 830В° are coterminal.
y
y
y
830В°
420В°
60В°
x
0
188В°
908В°
110В°
Coterminal
angles
Coterminal
angles
x
0
x
0
Figure 9
Figure 10
EXAMPLE 4 Finding Measures of Coterminal Angles
Find the angles of smallest possible positive measure coterminal with each angle.
Figure 11
(a) 908В°
y
(b) ПЄ75РЉ
Solution
0
285В°
x
–75°
(a) Add or subtract 360В° as many times as needed to obtain an angle with measure greater than 0В° but less than 360В°. Since 908РЉ ПЄ 2 Рё 360РЉ а·‡ 908РЉ ПЄ
720РЉ а·‡ 188РЉ, an angle of 188В° is coterminal with an angle of 908В°. See
Figure 11.
(b) Use a rotation of 360РЉ П© Н‘ПЄ75РЉН’ а·‡ 285РЉ. See Figure 12.
Figure 12
Now try Exercises 45 and 49.
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478 CHAPTER 5 Trigonometric Functions
Sometimes it is necessary to find an expression that will generate all angles
coterminal with a given angle. For example, we can obtain any angle coterminal
with 60В° by adding an appropriate integer multiple of 360В° to 60В°. Let n represent any integer; then the expression
60РЉ П© n Рё 360РЉ
represents all such coterminal angles. The table shows a few possibilities.
Value of n
Angle Coterminal with 60Ш‡
2
60РЉ П© 2 Рё 360РЉ а·‡ 780РЉ
1
60РЉ П© 1 Рё 360РЉ а·‡ 420РЉ
0
60РЉ П© 0 Рё 360РЉ а·‡ 60РЉ (the angle itself)
ПЄ1
60РЉ П© Н‘ПЄ1Н’ Рё 360РЉ а·‡ ПЄ300РЉ
EXAMPLE 5 Analyzing the Revolutions of a CD Player
CAV (Constant Angular Velocity) CD players always spin at the same speed.
Suppose a CAV player makes 480 revolutions per min. Through how many degrees will a point on the edge of a CD move in 2 sec?
480
The player revolves 480 times in 1 min or 60 times а·‡ 8 times per
sec (since 60 sec а·‡ 1 min). In 2 sec, the player will revolve 2 Рё 8 а·‡ 16 times.
Each revolution is 360В°, so a point on the edge of the CD will revolve
16 Рё 360РЉ а·‡ 5760РЉ in 2 sec.
Solution
Now try Exercise 75.
5.1 Exercises
1
3.
8
5. (a) 60РЉ
6. (a) 30РЉ
7. (a) 45РЉ
8. (a) 72РЉ
9. (a) 36РЉ
10. (a) 1РЉ
11. 150РЉ
2.
45РЉ 4. 90РЉ
9
1. Explain the difference between a segment and a ray.
2. What part of a complete revolution is an angle of 45РЉ?
(b) 150РЉ
(b) 120РЉ
(b) 135РЉ
(b) 162РЉ
(b) 126РЉ
(b) 91РЉ
12. 142.5РЉ
3. Concept Check
What angle is its own complement?
4. Concept Check
What angle is its own supplement?
Find (a) the complement and (b) the supplement of each angle.
5. 30РЉ
6. 60РЉ
7. 45РЉ
8. 18РЉ
9. 54РЉ
10. 89РЉ
Find the measure of the smaller angle formed by the hands of a clock at the following
times.
11.
12.
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5.1 Angles 479
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39.
41.
45.
48.
51.
53.
54.
55.
56.
57.
58.
60.
70РЉ; 110РЉ 14. 30РЉ; 60РЉ
55РЉ; 35РЉ 16. 107РЉ; 73РЉ
80РЉ; 100РЉ 18. 69РЉ; 21РЉ
Н‘90 ПЄ xН’РЉ 20. Н‘180 ПЄ xН’РЉ
Н‘x ПЄ 360Н’РЉ 22. Н‘360 П© xН’РЉ
83РЉ 59Р€ 24. 158РЉ 47Р€
23РЉ 49Р€ 26. ПЄ26РЉ 25Р€
38РЉ 32Р€ 28. 55РЉ 9Р€
17РЉ 1Р€ 49Р‰ 30. 53РЉ 41Р€ 13Р‰
20.900РЉ 32. 38.700РЉ
91.598РЉ 34. 34.860РЉ
274.316РЉ 36. 165.853РЉ
31РЉ 25Р€ 47Р‰ 38. 59РЉ 5Р€ 7Р‰
89РЉ 54Р€ 1Р‰ 40. 102РЉ 22Р€ 38Р‰
178РЉ 35Р€ 58Р‰ 42. 122РЉ 41Р€ 7Р‰
320РЉ 46. 262РЉ 47. 235РЉ
157РЉ 49. 179РЉ 50. 339РЉ
130РЉ 52. 280РЉ
30РЉ П© n Ш’ 360РЉ
45РЉ П© n Ш’ 360РЉ
135РЉ П© n Ш’ 360РЉ
270РЉ П© n Ш’ 360РЉ
ПЄ90РЉ П© n Ш’ 360РЉ
ПЄ135РЉ П© n Ш’ 360РЉ
C and D
Find the measure of each angle in Exercises 13 –18. See Example 1.
13.
14.
15.
(5k + 5)В°
(2y)В°
(7x)В°
(11x)В°
(3k + 5)В°
(4y)В°
16. supplementary angles with measures 10m П© 7 and 7m П© 3 degrees
17. supplementary angles with measures 6x ПЄ 4 and 8x ПЄ 12 degrees
18. complementary angles with measures 9z П© 6 and 3z degrees
Concept Check
Answer each question.
19. If an angle measures xРЉ, how can we represent its complement?
20. If an angle measures xРЉ, how can we represent its supplement?
21. If a positive angle has measure xРЉ between 0РЉ and 60РЉ, how can we represent the first
negative angle coterminal with it?
22. If a negative angle has measure xРЉ between 0РЉ and ПЄ60РЉ, how can we represent the
first positive angle coterminal with it?
Perform each calculation. See Example 2.
23. 62РЉ 18Р€ П© 21РЉ 41Р€
24. 75РЉ 15Р€ П© 83РЉ 32Р€
25. 71РЉ 18Р€ ПЄ 47РЉ 29Р€
26. 47РЉ 23Р€ ПЄ 73РЉ 48Р€
27. 90РЉ ПЄ 51РЉ 28Р€
28. 180РЉ ПЄ 124РЉ 51Р€
29. 90РЉ ПЄ 72РЉ 58Р€ 11Р‰
30. 90РЉ ПЄ 36РЉ 18Р€ 47Р‰
Convert each angle measure to decimal degrees. Round to the nearest thousandth of a
degree. See Example 3.
31. 20РЉ 54Р€
32. 38РЉ 42Р€
33. 91РЉ 35Р€ 54Р‰
34. 34РЉ 51Р€ 35Р‰
35. 274РЉ 18Р€ 59Р‰
36. 165РЉ 51Р€ 9Р‰
Convert each angle measure to degrees, minutes, and seconds. See Example 3.
37. 31.4296РЉ
38. 59.0854РЉ
39. 89.9004РЉ
40. 102.3771РЉ
41. 178.5994РЉ
42. 122.6853РЉ
9 43.
9 44.
Read about the degree symbol (РЉ) in the manual for your graphing calculator. How
is it used?
Show that 1.21 hr is the same as 1 hr, 12 min, 36 sec. Discuss the similarity between
converting hours, minutes, and seconds to decimal hours and converting degrees,
minutes, and seconds to decimal degrees.
Find the angle of smallest positive measure coterminal with each angle. See Example 4.
45. ПЄ40РЉ
46. ПЄ98РЉ
47. ПЄ125РЉ
48. ПЄ203РЉ
49. 539РЉ
50. 699РЉ
51. 850РЉ
52. 1000РЉ
Give an expression that generates all angles coterminal with each angle. Let n represent
any integer.
53. 30РЉ
54. 45РЉ
55. 135РЉ
56. 270РЉ
57. ПЄ90РЉ
58. ПЄ135РЉ
9 59. Explain why the answers to Exercises 56 and 57 give the same set of angles.
60. Concept Check
A. 360РЉ П© rРЉ
Which two of the following are not coterminal with rРЉ?
B. rРЉ ПЄ 360РЉ
C. 360РЉ ПЄ rРЉ
D. rРЉ П© 180РЉ
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480 CHAPTER 5 Trigonometric Functions
Angles other than those given are
possible in Exercises 61– 68.
y
61. y
62.
75В°
0
89В°
x
0
435°; –285°;
quadrant I
63.
64.
234В°
x
0
x
0
534°; –186°;
quadrant II
594°; –126°;
quadrant III
x
0
68.
y
x
70. Н™29
y
y
(–5, 2)
(–3, –3)
x
2
x
72. 2
y
y
(в€љ3, 1)
1
0
68. ПЄ159РЉ
69. Н‘ПЄ3, ПЄ3Н’
72. Н‘ Н™3, 1 Н’
70. Н‘ПЄ5, 2Н’
73. Н‘ ПЄ2, 2Н™3 Н’
71. Н‘ПЄ3, ПЄ5Н’
74. Н‘ 4Н™3, ПЄ4 Н’
76. Revolutions of a Windmill A windmill makes 90 revolutions per min. How many
revolutions does it make per second?
78. Rotating Airplane Propeller An airplane propeller rotates 1000 times per min.
Find the number of degrees that a point on the edge of the propeller will rotate in
1 sec.
0
–5
71. Н™34
–3
(–3, –5)
67. ПЄ61РЉ
201°; –519°;
quadrant III
69. 3Н™2
0
–3
66. 512РЉ
–159°
299°; –421°;
quadrant IV
–3
65. 300РЉ
x
0
–61°
64. 234РЉ
77. Rotating Tire A tire is rotating 600 times per min.
Through how many degrees does a point on the edge of
the tire move in 12 sec?
152°; –208°;
quadrant II
y
0
x
0
660°; –60°;
quadrant IV
63. 174РЉ
75. Revolutions of a Turntable A turntable in a shop makes 45 revolutions per min.
How many revolutions does it make per second?
512В°
300В°
62. 89РЉ
Solve each problem. See Example 5.
y
66.
y
61. 75РЉ
Concept Check Locate each point in a coordinate system. Draw a ray from the origin
through the given point. Indicate with an arrow the angle in standard position having
smallest positive measure. Then find the distance r from the origin to the point, using the
distance formula of Section 2.1.
y
174В°
67.
x
449°; –271°;
quadrant I
y
65.
Concept Check Sketch each angle in standard position. Draw an arrow representing
the correct amount of rotation. Find the measure of two other angles, one positive and
one negative, that are coterminal with the given angle. Give the quadrant of each angle.
x
x
0
2
–5
73. 4
79. Rotating Pulley A pulley rotates through 75РЉ in 1 min. How many rotations does
the pulley make in an hour?
80. Surveying One student in a surveying
class measures an angle as 74.25РЉ, while
another student measures the same angle
as 74РЉ 20Р€. Find the difference between
these measurements, both to the nearest
minute and to the nearest hundredth of
a degree.
74.25В°
74. 8
y
y
4
(–2, 2√3)
–2 0
x
4
0
–4
(4√3, –4)
3
76. 1.5 77. 1800РЉ
4
78. 6000РЉ 79. 12.5 rotations per
hr 80. 5Р€ or .08РЉ 81. 4 sec
75.
x
81. Viewing Field of a Telescope Due to Earth’s rotation, celestial objects like the
moon and the stars appear to move across the sky, rising in the east and setting in
the west. As a result, if a telescope on Earth remains stationary while viewing a
celestial object, the object will slowly move outside the viewing field of the telescope. For this reason, a motor is often attached to telescopes so that the telescope
rotates at the same rate as Earth. Determine how long it should take the motor to turn
the telescope through an angle of 1 min in a direction perpendicular to Earth’s axis.
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5.2 Trigonometric Functions 481
82. Angle Measure of a Star on the
вђЄ
American Flag Determine the
вђЄ
measure of the angle in each point
of the five-pointed star appearing on
2вђЄ
the American flag. (Hint: Inscribe
the star in a circle, and use the following theorem from geometry: An
angle whose vertex lies on the circumference of a circle is equal to half the central angle that cuts off the same arc.
See the figure.)
82. 36РЉ
5.2 Trigonometric Functions
Trigonometric Functions в– Quadrantal Angles в– Reciprocal Identities
Function Values в– Pythagorean Identities в– Quotient Identities
y
P(x, y)
r
y
вђЄ
Q
x
O
Figure 13
x
в– Signs and Ranges of
Trigonometric Functions To define the six trigonometric functions, we
start with an angle вђЄ in standard position, and choose any point P having coordinates Н‘x, yН’ on the terminal side of angle вђЄ. (The point P must not be the vertex of
the angle.) See Figure 13. A perpendicular from P to the x-axis at point Q determines a right triangle, having vertices at O, P, and Q. We find the distance r
from PН‘x, yН’ to the origin, Н‘0, 0Н’, using the distance formula.
r а·‡ Н™Н‘x ПЄ 0Н’2 П© Н‘y ПЄ 0Н’2 а·‡ Н™x 2 П© y 2 (Section 2.1)
Notice that r Пѕ 0 since distance is never negative.
The six trigonometric functions of angle вђЄ are sine, cosine, tangent, cotangent, secant, and cosecant. In the following definitions, we use the customary
abbreviations for the names of these functions.
Trigonometric Functions
Let Н‘x, yН’ be a point other than the origin on the terminal side of an angle вђЄ
in standard position. The distance from the point to the origin is
r а·‡ Н™x 2 П© y 2. The six trigonometric functions of вђЄ are defined as follows.
sin ␪ ‫؍‬
y
r
csc ␪ ‫؍‬
r
y
Н‘y
0Н’
cos ␪ ‫؍‬
x
r
sec ␪ ‫؍‬
r
x
Н‘x
0Н’
tan ␪ ‫؍‬
y
x
Н‘x
0Н’
cot ␪ ‫؍‬
x
y
Н‘y
0Н’
Although Figure 13 shows a second quadrant angle, these definitions
apply to any angle вђЄ. Because of the restrictions on the denominators in the definitions of tangent, cotangent, secant, and cosecant, some angles will have undefined function values.
NOTE
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482 CHAPTER 5 Trigonometric Functions
EXAMPLE 1 Finding Function Values of an Angle
The terminal side of an angle вђЄ in standard position passes through the point
Н‘8, 15Н’. Find the values of the six trigonometric functions of angle вђЄ.
Figure 14 shows angle вђЄ and the triangle formed by dropping a perpendicular from the point Н‘8, 15Н’ to the x-axis. The point Н‘8, 15Н’ is 8 units to the
right of the y-axis and 15 units above the x-axis, so x а·‡ 8 and y а·‡ 15. Since
r а·‡ Н™x 2 П© y 2,
y
Solution
(8, 15)
17
15
x= 8
y = 15
r = 17
We can now find the values of the six trigonometric functions of angle вђЄ.
вђЄ
0
r а·‡ Н™82 П© 152 а·‡ Н™64 П© 225 а·‡ Н™289 а·‡ 17.
x
8
sin вђЄ а·‡
y
15
а·‡
r
17
cos вђЄ а·‡
x
8
а·‡
r
17
tan вђЄ а·‡
y
15
а·‡
x
8
csc вђЄ а·‡
r
17
а·‡
y
15
sec вђЄ а·‡
r
17
а·‡
x
8
cot вђЄ а·‡
x
8
а·‡
y
15
Figure 14
Now try Exercise 7.
EXAMPLE 2 Finding Function Values of an Angle
The terminal side of an angle вђЄ in standard position passes through the point
Н‘ПЄ3, ПЄ4Н’. Find the values of the six trigonometric functions of angle вђЄ.
y
Solution
вђЄ
x = –3
y = –4
r = 5
–3
–4
r а·‡ Н™Н‘ПЄ3Н’2 П© Н‘ПЄ4Н’2 а·‡ Н™25 а·‡ 5. Remember that r Пѕ 0.
Then by the definitions of the trigonometric functions,
x
0
As shown in Figure 15, x а·‡ ПЄ3 and y а·‡ ПЄ4. The value of r is
sin вђЄ а·‡
ПЄ4
4
а·‡ПЄ
5
5
cos вђЄ а·‡
ПЄ3
3
а·‡ПЄ
5
5
tan вђЄ а·‡
ПЄ4
4
а·‡
ПЄ3
3
csc вђЄ а·‡
5
5
а·‡ПЄ
ПЄ4
4
sec вђЄ а·‡
5
5
а·‡ПЄ
ПЄ3
3
cot вђЄ а·‡
ПЄ3
3
а·‡ .
ПЄ4
4
5
(–3, –4)
Figure 15
Now try Exercise 3.
We can find the six trigonometric functions using any point other than the
origin on the terminal side of an angle. To see why any point may be used, refer
to Figure 16, which shows an angle вђЄ and two distinct points on its terminal
side. Point P has coordinates ͑x, y͒, and point PЈ (read “P-prime”) has coordinates ͑xЈ, yЈ͒. Let r be the length of the hypotenuse of triangle OPQ, and let rЈ be
the length of the hypotenuse of triangle OPР€QР€. Since corresponding sides of
similar triangles are proportional,
y
(x ′, y′)
OP = r
OP′ = r′
P′
(x, y)
P
y
yР€
а·‡ ,
r
rР€
вђЄ
O
Q
Figure 16
Q′
x
y
so sin вђЄ а·‡ r is the same no matter which point is used to find it. A similar result
holds for the other five trigonometric functions.
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5.2 Trigonometric Functions 483
y
We can also find the trigonometric function values of an angle if we know
the equation of the line coinciding with the terminal ray. Recall from algebra
that the graph of the equation
Ax П© By а·‡ 0 (Section 2.3)
x
0
x + 2y = 0, x ≥ 0
is a line that passes through the origin. If we restrict x to have only nonpositive
or only nonnegative values, we obtain as the graph a ray with endpoint at the origin. For example, the graph of x П© 2y а·‡ 0, x Х† 0, shown in Figure 17, is a ray
that can serve as the terminal side of an angle in standard position. By choosing
a point on the ray, we can find the trigonometric function values of the angle.
Figure 17
EXAMPLE 3 Finding Function Values of an Angle
Find the six trigonometric function values of the angle вђЄ in standard position, if
the terminal side of вђЄ is defined by x П© 2y а·‡ 0, x Х† 0.
The angle is shown in Figure 18. We can use any point except Н‘0, 0Н’
on the terminal side of вђЄ to find the trigonometric function values. We choose
x а·‡ 2 and find the corresponding y-value.
Solution
y
x=2
y = –1
r = в€љ5
x П© 2y а·‡ 0, x Х† 0
вђЄ
2 П© 2y а·‡ 0
x
0
2y а·‡ ПЄ2
(2, –1)
x + 2y = 0, x ≥ 0
Figure 18
y а·‡ ПЄ1
Let x а·‡ 2.
Subtract 2.
Divide by 2.
The point Н‘2, ПЄ1Н’ lies on the terminal side, and the corresponding value of r is
r а·‡ Н™22 П© Н‘ПЄ1Н’2 а·‡ Н™5. Now we use the definitions of the trigonometric
functions.
y
ПЄ1
ПЄ1
а·‡
а·‡
r
Н™5 Н™5
x
2
2
а·‡
cos вђЄ а·‡ а·‡
r
Н™5 Н™5
y
1
tan вђЄ а·‡ а·‡ ПЄ
x
2
sin вђЄ а·‡
csc вђЄ а·‡
r
а·‡ ПЄН™5
y
Н™5
Н™5
а·‡ПЄ
5
Н™5
Н™5 2Н™5
Рё
а·‡
5
Н™5
Рё
sec вђЄ а·‡
r
Н™5
а·‡
x
2
Rationalize denominators.
(Section R.7)
cot вђЄ а·‡
x
а·‡ ПЄ2
y
Now try Exercise 17.
Recall that when the equation of a line is written in the form y а·‡ mx П© b,
the coefficient of x is the slope of the line. In Example 3, x П© 2y а·‡ 0 can be
1
1
1
written as y а·‡ ПЄ 2 x, so the slope is ПЄ 2 . Notice that tan вђЄ а·‡ ПЄ 2 . In general, it
is true that m ‫ ؍‬tan ␪.
NOTE
The trigonometric function values we found in Examples 1–3 are
exact. If we were to use a calculator to approximate these values, the decimal
results would not be acceptable if exact values were required.
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484 CHAPTER 5 Trigonometric Functions
Quadrantal Angles If the terminal side of an angle in standard position lies
along the y-axis, any point on this terminal side has x-coordinate 0. Similarly, an
angle with terminal side on the x-axis has y-coordinate 0 for any point on the terminal side. Since the values of x and y appear in the denominators of some
trigonometric functions, and since a fraction is undefined if its denominator is 0,
some trigonometric function values of quadrantal angles (i.e., those with terminal side on an axis) are undefined.
EXAMPLE 4 Finding Function Values of Quadrantal Angles
Find the values of the six trigonometric functions for each angle.
(a) an angle of 90В°
(b) an angle вђЄ in standard position with terminal side through Н‘ПЄ3, 0Н’
Solution
(a) First, we select any point on the terminal side of a 90В° angle. We choose the
point Н‘0, 1Н’, as shown in Figure 19. Here x а·‡ 0 and y а·‡ 1, so r а·‡ 1. Then,
sin 90РЉ а·‡
1
а·‡1
1
cos 90РЉ а·‡
0
а·‡0
1
tan 90РЉ а·‡
1
0
csc 90РЉ а·‡
1
а·‡1
1
sec 90РЉ а·‡
1
0
cot 90РЉ а·‡
0
а·‡ 0.
1
Н‘undefinedН’
y
y
A calculator in degree mode returns
the correct values for sin 90В° and
cos 90В°. The second screen shows an
ERROR message for tan 90В°, because
90В° is not in the domain of the tangent function.
Н‘undefinedН’
(0, 1)
Оё
90В°
0
x
(–3, 0)
Figure 19
x
0
Figure 20
(b) Figure 20 shows the angle. Here, x а·‡ ПЄ3, y а·‡ 0, and r а·‡ 3, so the trigonometric functions have the following values.
sin вђЄ а·‡
0
а·‡0
3
cos вђЄ а·‡
ПЄ3
а·‡ ПЄ1
3
tan вђЄ а·‡
0
а·‡0
ПЄ3
csc вђЄ а·‡
3
0
sec вђЄ а·‡
3
а·‡ ПЄ1
ПЄ3
cot вђЄ а·‡
ПЄ3
Н‘undefinedН’
0
Н‘undefinedН’
Now try Exercises 5 and 9.
The conditions under which the trigonometric function values of quadrantal
angles are undefined are summarized here.
Undefined Function Values
If the terminal side of a quadrantal angle lies along the y-axis, then the tangent and secant functions are undefined. If it lies along the x-axis, then the
cotangent and cosecant functions are undefined.
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5.2 Trigonometric Functions 485
The function values of the most commonly used quadrantal angles, 0В°, 90В°,
180В°, 270В°, and 360В°, are summarized in the following table.
вђЄ
sin вђЄ
cos вђЄ
tan вђЄ
cot вђЄ
sec вђЄ
1
csc вђЄ
0В°
0
1
0
Undefined
90В°
1
0
Undefined
0
Undefined
180В°
0
ПЄ1
0
Undefined
ПЄ1
Undefined
270В°
ПЄ1
0
Undefined
0
Undefined
ПЄ1
360В°
0
1
0
Undefined
1
Undefined
1
Undefined
The values given in this table can be found with a calculator that has
trigonometric function keys. Make sure the calculator is set in degree mode.
One of the most common errors involving calculators in
trigonometry occurs when the calculator is set for radian measure, rather than
degree measure. (Radian measure of angles is discussed in Chapter 6.) Be sure
you know how to set your calculator in degree mode.
CAUTION
Reciprocal Identities Identities are equations that are true for all values of
the variables for which all expressions are defined.
Н‘x П© yН’2 а·‡ x 2 П© 2xy П© y 2
2Н‘x П© 3Н’ а·‡ 2x П© 6 Identities
The definitions of the trigonometric functions at the beginning of this section were written so that functions in the same column are reciprocals of each
y
r
other. Since sin вђЄ а·‡ r and csc вђЄ а·‡ y ,
sin вђЄ а·‡
TEACHING TIP Students may be
tempted to associate secant with
sine and cosecant with cosine.
Note this common misconception.
1
csc вђЄ
and
csc вђЄ а·‡
1
,
sin вђЄ
provided sin вђЄ 0. Also, cos вђЄ and sec вђЄ are reciprocals, as are tan вђЄ and cot вђЄ.
In summary, we have the reciprocal identities that hold for any angle вђЄ that
does not lead to a 0 denominator.
Reciprocal Identities
(a)
(b)
Figure 21
sin ␪ ‫؍‬
1
csc вђЄ
cos ␪ ‫؍‬
1
sec вђЄ
tan ␪ ‫؍‬
1
cot вђЄ
csc ␪ ‫؍‬
1
sin вђЄ
sec ␪ ‫؍‬
1
cos вђЄ
cot ␪ ‫؍‬
1
tan вђЄ
The screen in Figure 21(a) shows how to find csc 90В°, sec 180В°, and
cscН‘ПЄ270РЉН’, using the appropriate reciprocal identities and the reciprocal key of
a graphing calculator in degree mode. Be sure not to use the inverse trigonometric function keys to find the reciprocal function values. Attempting to find
sec 90В° by entering 1Нћcos 90РЉ produces an ERROR message, indicating the reciprocal is undefined. See Figure 21(b). Compare these results with the ones
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486 CHAPTER 5 Trigonometric Functions
NOTE
Identities can be written in different forms. For example,
sin вђЄ а·‡
csc вђЄ а·‡
can be written
1
sin вђЄ
1
csc вђЄ
Н‘sin вђЄН’ Н‘csc вђЄН’ а·‡ 1.
and
EXAMPLE 5 Using the Reciprocal Identities
Find each function value.
(a) cos вђЄ, if sec вђЄ а·‡
5
3
(b) sin вђЄ, if csc вђЄ а·‡ ПЄ
Н™12
2
Solution
(a) Since cos вђЄ is the reciprocal of sec вђЄ,
cos вђЄ а·‡
1
1
3
а·‡ 5 а·‡ .
sec вђЄ
5
3
Simplify the complex
fraction. (Section R.5)
(b) sin вђЄ а·‡
1
12
ПЄ Н™2
sin вђЄ а·‡ csc1 вђЄ
2
Н™12
2
а·‡ПЄ
Н™12 а·‡ Н™4 Рё 3 а·‡ 2Н™3 (Section R.7)
2Н™3
а·‡ПЄ
а·‡ПЄ
1
Н™3
Simplify.
Н™3
3
Multiply by
а·‡ПЄ
Н™3
Н™3
to rationalize the denominator.
Now try Exercises 45 and 47.
TEACHING TIP Some students use
the sentence “All Students Take
Calculus” to remember which of
the three basic functions are positive in each quadrant. A indicates
“all” in quadrant I, S represents
“sine” in quadrant II, T represents
“tangent” in quadrant III, and C
stands for “cosine” in quadrant IV.
Signs and Ranges of Function Values In the definitions of the trigonometric functions, r is the distance from the origin to the point Н‘x, yН’. Distance is
never negative, so r Пѕ 0. If we choose a point Н‘x, yН’ in quadrant I, then both x
and y will be positive. Thus, the values of all six functions will be positive in
quadrant I.
A point Н‘x, yН’ in quadrant II has x ПЅ 0 and y Пѕ 0. This makes the values of
sine and cosecant positive for quadrant II angles, while the other four functions
take on negative values. Similar results can be obtained for the other quadrants,
as summarized on the next page.
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5.2 Trigonometric Functions 487
Signs of Function Values
y
вђЄ in Quadrant
sin вђЄ
cos вђЄ
tan вђЄ
cot вђЄ
sec вђЄ
csc вђЄ
I
П©
П©
П©
П©
П©
П©
II
П©
ПЄ
ПЄ
ПЄ
ПЄ
П©
III
ПЄ
ПЄ
П©
П©
ПЄ
ПЄ
IV
ПЄ
П©
ПЄ
ПЄ
П©
ПЄ
x < 0, y > 0, r > 0
x > 0, y > 0, r > 0
II
Sine and cosecant
positive
x < 0, y < 0, r > 0
III
Tangent and cotangent
positive
I
All functions
positive
0
x
x > 0, y < 0, r > 0
IV
Cosine and secant
positive
EXAMPLE 6 Identifying the Quadrant of an Angle
Identify the quadrant (or quadrants) of any angle вђЄ that satisfies sin вђЄ Пѕ 0,
tan вђЄ ПЅ 0.
Since sin вђЄ Пѕ 0 in quadrants I and II, while tan вђЄ ПЅ 0 in quadrants II
and IV, both conditions are met only in quadrant II.
Solution
Now try Exercise 57.
y
r
вђЄ
x
0
y
x
Figure 22 shows an angle вђЄ as it increases in measure from near 0В° toward
90В°. In each case, the value of r is the same. As the measure of the angle increases, y increases but never exceeds r, so y Х… r. Dividing both sides by the
y
positive number r gives r Х… 1.
In a similar way, angles in quadrant IV suggest that
ПЄ1 Х…
y
,
r
so
ПЄ1 Х…
y
Х…1
r
and
ПЄ1 Х… sin вђЄ Х… 1.
Similarly,
ПЄ1 Х… cos вђЄ Х… 1.
y
r
y
вђЄ
x
0
x
y
y
r
а·‡ sin вђЄ for any angle вђЄ.
y
r
y
вђЄ
x
x
0
The tangent of an angle is defined as x . It is possible that x ПЅ y, x а·‡ y, or
y
x Пѕ y. Thus, x can take any value, so tan вђЄ can be any real number, as can cot вђЄ.
The functions sec вђЄ and csc вђЄ are reciprocals of the functions cos вђЄ and
sin вђЄ, respectively, making
sec вђЄ Х… ПЄ1 or sec вђЄ Х† 1
y
csc вђЄ Х… ПЄ1 or csc вђЄ Х† 1.
In summary, the ranges of the trigonometric functions are as follows.
r
Ranges of Trigonometric Functions
y
вђЄ
0
and
x
x
Figure 22
For any angle вђЄ for which the indicated functions exist:
ПЄ1 Х… cos вђЄ Х… 1;
1. ПЄ1 Х… sin вђЄ Х… 1
and
2. tan вђЄ and cot вђЄ can equal any real number;
csc вђЄ Х… ПЄ1 or csc вђЄ Х† 1.
3. sec вђЄ Х… ПЄ1 or sec вђЄ Х† 1
and
(Notice that sec вђЄ and csc вђЄ are never between ПЄ1 and 1.)
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488 CHAPTER 5 Trigonometric Functions
EXAMPLE 7 Deciding Whether a Value Is in the Range of a Trigonometric
Function
Decide whether each statement is possible or impossible.
(a) sin вђЄ а·‡ Н™8
(b) tan вђЄ а·‡ 110.47
(c) sec вђЄ а·‡ .6
Solution
(a) For any value of вђЄ, ПЄ1 Х… sin вђЄ Х… 1. Since Н™8 Пѕ 1, it is impossible to find
a value of вђЄ with sin вђЄ а·‡ Н™8.
(b) Tangent can equal any value. Thus, tan вђЄ а·‡ 110.47 is possible.
(c) Since sec вђЄ Х… ПЄ1 or sec вђЄ Х† 1, the statement sec вђЄ а·‡ .6 is impossible.
Now try Exercises 71 and 73.
Pythagorean Identities We derive three new identities from the relationship x 2 П© y 2 а·‡ r 2.
x2
y2 r 2
Divide by r 2.
П©
а·‡
r2
r2 r2
Н©НЄ Н©НЄ
x
r
2
П©
y
r
2
а·‡1
Power rule for exponents (Section R.3)
Н‘cos вђЄН’2 П© Н‘sin вђЄН’2 а·‡ 1
or
cos вђЄ а·‡ xr , sin вђЄ а·‡
y
r
sin2 вђЄ П© cos2 вђЄ а·‡ 1.
Starting again with x 2 П© y 2 а·‡ r 2 and dividing through by x 2 gives
x2
y2 r 2
П© 2а·‡ 2
2
x
x
x
1П©
Divide by x 2.
Н©НЄ Н©НЄ
y
x
2
а·‡
r
x
2
1 П© Н‘tan вђЄН’2 а·‡ Н‘sec вђЄН’2
or
Power rule for exponents
y
tan вђЄ а·‡ x , sec вђЄ а·‡ xr
tan2 вђЄ П© 1 а·‡ sec2 вђЄ.
On the other hand, dividing through by y 2 leads to
1 П© cot 2 вђЄ а·‡ csc2 вђЄ.
These three identities are called the Pythagorean identities since the original
equation that led to them, x 2 П© y 2 а·‡ r 2, comes from the Pythagorean theorem.
Pythagorean Identities
sin2 ␪ ؉ cos2 ␪ ‫ ؍‬1
tan2 ␪ ؉ 1 ‫ ؍‬sec2 ␪
1 ؉ cot2 ␪ ‫ ؍‬csc2 ␪
Although we usually write sin2 вђЄ, for example, it should be entered
as Н‘sin вђЄН’2 in your calculator. To test this, verify that in degree mode,
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As before, we have given only one form of each identity. However, algebraic transformations produce equivalent identities. For example, by subtracting
sin2 вђЄ from both sides of sin2 вђЄ П© cos2 вђЄ а·‡ 1, we get the equivalent identity
cos2 вђЄ а·‡ 1 ПЄ sin2 вђЄ.
You should be able to transform these identities quickly and also recognize their
equivalent forms.
Looking Ahead to Calculus
The reciprocal, Pythagorean, and quotient identities are used in calculus to
find derivatives and integrals of
trigonometric functions. A standard
technique of integration called trigonometric substitution relies on the
Pythagorean identities.
y
Quotient Identities Recall that sin вђЄ а·‡ r and cos вђЄ а·‡ xr . Consider the quotient of sin вђЄ and cos вђЄ, where cos вђЄ 0.
sin вђЄ
а·‡
cos вђЄ
Similarly,
cos вђЄ
sin вђЄ
y
r
x
r
а·‡
y
x
y r
y
П¬ а·‡ Рё а·‡ а·‡ tan вђЄ
r
r
r x
x
а·‡ cot вђЄ, for sin вђЄ
0. Thus, we have the quotient identities.
Quotient Identities
sin вђЄ
‫ ؍‬tan ␪
cos вђЄ
cos вђЄ
‫ ؍‬cot ␪
sin вђЄ
EXAMPLE 8 Finding Other Function Values Given One Value and the Quadrant
4
Find sin вђЄ and cos вђЄ, if tan вђЄ а·‡ 3 and вђЄ is in quadrant III.
Solution Since вђЄ is in quadrant III, sin вђЄ and cos вђЄ will both be negative. It is
sin вђЄ
4
tempting to say that since tan вђЄ а·‡ cos вђЄ and tan вђЄ а·‡ 3 , then sin вђЄ а·‡ ПЄ4 and
cos вђЄ а·‡ ПЄ3. This is incorrect, however, since both sin вђЄ and cos вђЄ must be in
the interval Н“ПЄ1, 1Н”.
We use the Pythagorean identity tan2 вђЄ П© 1 а·‡ sec2 вђЄ to find sec вђЄ, and then
1
the reciprocal identity cos вђЄ а·‡ sec вђЄ to find cos вђЄ.
tan2 вђЄ П© 1 а·‡ sec2 вђЄ
Н©НЄ
4
3
2
П© 1 а·‡ sec2 вђЄ
tan вђЄ а·‡
4
3
16
П© 1 а·‡ sec2 вђЄ
9
25
а·‡ sec2 вђЄ
9
ПЄ
5
а·‡ sec вђЄ
3
Choose the negative square root since sec вђЄ is
negative when вђЄ is in quadrant III.
ПЄ
3
а·‡ cos вђЄ
5
Secant and cosine are reciprocals.
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490 CHAPTER 5 Trigonometric Functions
Since sin2 вђЄ а·‡ 1 ПЄ cos2 вђЄ,
Н© НЄ
sin2 вђЄ а·‡ 1 ПЄ ПЄ
sin2 вђЄ а·‡ 1 ПЄ
sin2 вђЄ а·‡
3
5
2
cos вђЄ а·‡ ПЄ 35
9
25
16
25
sin вђЄ а·‡ ПЄ
4
.
5
Choose the negative square root.
4
3
Therefore, we have sin вђЄ а·‡ ПЄ 5 and cos вђЄ а·‡ ПЄ 5 .
Now try Exercise 79.
NOTE
Example 8 can also be worked by drawing вђЄ in standard position in
quadrant III, finding r to be 5, and then using the definitions of sin вђЄ and cos вђЄ
in terms of x, y, and r.
5.2 Exercises
y
1.
Concept Check Sketch an angle вђЄ in standard position such that вђЄ has the smallest possible positive measure, and the given point is on the terminal side of вђЄ.
y
2.
вђЄ
0
x
вђЄ
5
0
–12
–12
(5, –12)
(–12, –5)
x
1. Н‘5, ПЄ12Н’
2. Н‘ПЄ12, ПЄ5Н’
–5
In Exercises 3– 10 and 17– 21, we
give, in order, sine, cosine,
tangent, cotangent, secant, and
cosecant.
5 5
4
3
4
3
;ПЄ ;ПЄ ;ПЄ ;ПЄ ;
3.
5
5
3
4
3 4
4 3 4
5
5
3
4. ПЄ ; ПЄ ; ; ; ПЄ ; ПЄ
5
5 4 3
4
3
5. 1; 0; undefined; 0; undefined; 1
6. 0; ПЄ1; 0; undefined; ПЄ1;
Н™3 1
undefined 7.
; ; Н™3;
2 2
1
2Н™3
Н™3
Н™3
; 2;
;
8. ПЄ ; ПЄ
3
3
2
2
2Н™3
Н™3
; Н™3; ПЄ
; ПЄ2
3
3
9. 0; ПЄ1; 0; undefined; ПЄ1;
4 3
undefined 10. ПЄ ; ;
5 5
3 5
5
4
ПЄ ;ПЄ ; ;ПЄ
3
4 3
4
Find the values of the six trigonometric functions for each angle in standard position
having the given point on its terminal side. Rationalize denominators when applicable.
See Examples 1, 2, and 4.
3. Н‘ПЄ3, 4Н’
4. Н‘ПЄ4, ПЄ3Н’
7. Н‘ 1, Н™3 Н’
9 11.
8. Н‘ ПЄ2Н™3, ПЄ2 Н’
5. Н‘0, 2Н’
9. Н‘ПЄ2, 0Н’
6. Н‘ПЄ4, 0Н’
10. Н‘3, ПЄ4Н’
For any nonquadrantal angle вђЄ, sin вђЄ and csc вђЄ will have the same sign. Explain why.
12. Concept Check If the terminal side of an angle вђЄ is in quadrant III, what is the sign
of each of the trigonometric function values of вђЄ ?
Concept Check Suppose that the point Н‘x, yН’ is in the indicated quadrant. Decide
whether the given ratio is positive or negative. (Hint: Drawing a sketch may help.)
13. II,
x
r
14. III,
y
r
15. IV,
y
x
16. IV,
x
y
In Exercises 17– 20, an equation of the terminal side of an angle ␪ in standard position
is given with a restriction on x. Sketch the smallest positive such angle вђЄ, and find the
values of the six trigonometric functions of вђЄ. See Example 3.
17. 2x П© y а·‡ 0, x Х† 0
18. 3x П© 5y а·‡ 0, x Х† 0
19. ПЄ6x ПЄ y а·‡ 0, x Х… 0
20. ПЄ5x ПЄ 3y а·‡ 0, x Х… 0
21. Find the six trigonometric function values of the quadrantal angle 450В°.
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5.2 Trigonometric Functions 491
12. tan вђЄ and cot вђЄ are positive; all
other function values are negative.
13. negative 14. negative
15. negative 16. negative
y
17.
Use the trigonometric function values of quadrantal angles given in this section to evaluate each expression. An expression such as cot 2 90В° means Н‘cot 90В°Н’2, which is equal to
0 2 а·‡ 0.
22. 3 sec 180В° ПЄ 5 tan 360В°
23. 4 csc 270В° П© 3 cos 180В°
24. tan 360В° П© 4 sin 180В° П© 5 cos 180В°
25. 2 sec 0В° П© 4 cot 2 90В° П© cos 360В°
26. sin2 180В° П© cos2 180В°
27. sin2 360В° П© cos2 360В°
2
2x + y = 0, x ≥ 0
вђЄ
0
x
1
(1, –2)
–2
1
2Н™5 Н™5
Н™5
;
; ПЄ2; ПЄ ; Н™5; ПЄ
5
5
2
2
y
18.
ПЄ
3x + 5y = 0, x ≥ 0
вђЄ
–3
5
(5, –3)
3Н™34 5Н™34
3
5
;
;ПЄ ;ПЄ ;
34
34
5
3
Н™34 Н™34
;ПЄ
5
3
y
19.
ПЄ
(–1, 6)
6
–6x – y = 0, x ≤ 0
вђЄ
–1 0
x
6Н™37 Н™37
1
;ПЄ
; ПЄ6; ПЄ ;
37
37
6
Н™37
ПЄН™37;
6
20. (–3, 5) y
5
–3
28. cosН“Н‘2n П© 1Н’ Рё 90В°Н”
29. sinН“n Рё 180В°Н”
30. tanН“n Рё 180В°Н”
31. tanН“Н‘2n П© 1Н’ Рё 90В°Н”
Provide conjectures in Exercises 32 – 35.
x
0
If n is an integer, n Рё 180В° represents an integer multiple of 180В°, and Н‘2n П© 1Н’ Рё 90В°
represents an odd integer multiple of 90В°. Decide whether each expression is equal to 0,
1, ПЄ1, or is undefined.
–5x – 3y = 0, x ≤ 0
вђЄ
x
0
5
3
5Н™34 3Н™34
;ПЄ
;ПЄ ;ПЄ ;
34
34
3
5
Н™34 Н™34
ПЄ
21. 1; 0;
;
3
5
undefined; 0; undefined; 1
22. ПЄ3 23. ПЄ7 24. 5
25. 3 26. 1 27. 1 28. 0
29. 0 30. 0 31. undefined
32. They are equal.
33. They are equal. 34. They
are negatives of each other.
35. They are equal. 36. about
.940; about .342 37. 40В°
38. 35В° 39. 45В°
40. decrease; increase
32. The angles 15В° and 75В° are complementary. With your calculator determine sin 15В°
and cos 75В°. Make a conjecture about the sines and cosines of complementary
angles, and test your hypothesis with other pairs of complementary angles. (Note:
This relationship will be discussed in detail in the next section.)
33. The angles 25В° and 65В° are complementary. With your calculator determine tan 25В°
and cot 65В°. Make a conjecture about the tangents and cotangents of complementary
angles, and test your hypothesis with other pairs of complementary angles. (Note:
This relationship will be discussed in detail in the next section.)
34. With your calculator determine sin 10В° and sinН‘ПЄ10В°Н’. Make a conjecture about the
sines of an angle and its negative, and test your hypothesis with other angles. Also,
use a geometry argument with the definition of sin вђЄ to justify your hypothesis.
(Note: This relationship will be discussed in detail in Section 7.1.)
35. With your calculator determine cos 20В° and cosН‘ПЄ20В°Н’. Make a conjecture about the
cosines of an angle and its negative, and test your hypothesis with other angles. Also,
use a geometry argument with the definition of cos вђЄ to justify your hypothesis.
(Note: This relationship will be discussed in detail in Section 7.1.)
In Exercises 36–41, set your graphing calculator in parametric and degree modes. Set
the window and functions (see the third screen) as shown here, and graph. A circle of
radius 1 will appear on the screen. Trace to move a short distance around the circle. In
the screen, the point on the circle corresponds to an angle T а·‡ 25В°. Since r а·‡ 1, cos 25В°
is X а·‡ .90630779, and sin 25В° is Y а·‡ .42261826.
1.2
–1.8
This screen is a continuation
of the previous one.
1.8
–1.2
36. Use the right- and left-arrow keys to move to the point corresponding to 20В°. What
are cos 20В° and sin 20В°?
37. For what angle T, 0В° Х… T Х… 90В°, is cos T П· .766?
38. For what angle T, 0В° Х… T Х… 90В°, is sin T П· .574?
39. For what angle T, 0В° Х… T Х… 90В°, does cos T а·‡ sin T?
40. As T increases from 0В° to 90В°, does the cosine increase or decrease? What about
the sine?
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492 CHAPTER 5 Trigonometric Functions
41.
42.
43.
44.
decrease; decrease
1; вђЄ а·‡ 90В°
ПЄ1; вђЄ а·‡ 180В°
ПЄ.4 45. ПЄ5
3Н™5
Н™15
46.
47. ПЄ
15
5
48. .70069071 49. .10199657
51. The range of the cosine
function is Н“ПЄ1, 1Н”, so cos вђЄ
3
cannot equal .
2
1
52. ПЄ
53. Н™3 54. 2.5
3
55. 2В° 56. 1В° 57. II
58. I 59. I or III 60. II or IV
61. П©; ПЄ; ПЄ 62. ПЄ; ПЄ; П©
63. ПЄ; П©; ПЄ 64. П©; П©; П©
65. ПЄ; П©; ПЄ 66. ПЄ; ПЄ; П©
67. tan 30В° 68. sin 21В°
69. sec 33В° 70. impossible
71. impossible 72. possible
73. possible 74. possible
75. possible 76. possible
77. impossible 78. ПЄ2Н™2
41. As T increases from 90В° to 180В°, does the cosine increase or decrease? What about
the sine?
42. Concept Check What positive number a is its own reciprocal? Find a value of вђЄ
for which sin вђЄ а·‡ csc вђЄ а·‡ a.
43. Concept Check What negative number a is its own reciprocal? Find a value of вђЄ
for which cos вђЄ а·‡ sec вђЄ а·‡ a.
Use the appropriate reciprocal identity to find each function value. Rationalize denominators when applicable. In Exercises 48 and 49, use a calculator. See Example 5.
1
5
44. cos вђЄ, if sec вђЄ а·‡ ПЄ2.5
45. cot вђЄ, if tan вђЄ а·‡ ПЄ
46. sin вђЄ, if csc вђЄ а·‡ Н™15
Н™5
47. tan вђЄ, if cot вђЄ а·‡ ПЄ
3
48. sin вђЄ, if csc вђЄ а·‡ 1.42716321
49. cos вђЄ, if sec вђЄ а·‡ 9.80425133
9 50.
Can a given angle вђЄ satisfy both sin вђЄ Пѕ 0 and csc вђЄ ПЅ 0? Explain.
51. Concept Check Explain what is wrong with the following item that appears on a
trigonometry test:
Find sec вђЄ, given that cos вђЄ а·‡
3
.
2
Find the tangent of each angle. See Example 5.
52. cot вђЄ а·‡ ПЄ3
53. cot вђЄ а·‡
Н™3
3
54. cot вђЄ а·‡ .4
Find a value of each variable.
55. tanН‘3вђЄ ПЄ 4В°Н’ а·‡
1
cotН‘5вђЄ ПЄ 8В°Н’
56. secН‘2вђЄ П© 6В°Н’ cosН‘5вђЄ П© 3В°Н’ а·‡ 1
Identify the quadrant or quadrants for the angle satisfying the given conditions. See
Example 6.
57. sin вђЄ Пѕ 0, cos вђЄ ПЅ 0
58. cos вђЄ Пѕ 0, tan вђЄ Пѕ 0
59. tan вђЄ Пѕ 0, cot вђЄ Пѕ 0
60. tan вђЄ ПЅ 0, cot вђЄ ПЅ 0
Concept Check Give the signs of the sine, cosine, and tangent functions for each angle.
61. 129В°
62. 183В°
63. 298В°
64. 412В°
65. ПЄ82В°
66. ПЄ121В°
Concept Check
Without using a calculator, decide which is greater.
67. sin 30В° or tan 30В°
68. sin 20В° or sin 21В°
69. sin 33В° or sec 33В°
Decide whether each statement is possible or impossible for an angle вђЄ. See Example 7.
70. sin вђЄ а·‡ 2
71. cos вђЄ а·‡ ПЄ1.001
72. tan вђЄ а·‡ .92
73. cot вђЄ а·‡ ПЄ12.1
74. sec вђЄ а·‡ 1
75. tan вђЄ а·‡ 1
76. sin вђЄ а·‡
1
and csc вђЄ а·‡ 2
2
77. tan вђЄ а·‡ 2 and cot вђЄ а·‡ ПЄ2
Use identities to find each function value. Use a calculator in Exercises 84 and 85. See
Example 8.
78. tan вђЄ, if sec вђЄ а·‡ 3, with вђЄ in quadrant IV
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5.2 Trigonometric Functions 493
Н™5
Н™15
80. ПЄ
4
2
4
Н™15
81. ПЄ
82. ПЄ
3
4
Н™3
83. ПЄ
84. 3.44701905
2
85. ПЄ.56616682 86. yes
In Exercises 87–92, we give, in
order, sine, cosine, tangent,
cotangent, secant, and cosecant.
17
15
8
15
8
87.
;ПЄ ;ПЄ ;ПЄ ;ПЄ ;
17 17
8
15
8
5
17
4
3 4 3
88. ПЄ ; ПЄ ; ; ; ПЄ ;
15
5
5 3 4
3
5
1
Н™3
ПЄ
89. ПЄ
; ПЄ ; Н™3 ;
4
2
2
2Н™3
Н™5
Н™3
; ПЄ2; ПЄ
;
90.
3
3
7
2Н™11 Н™55 2Н™55 7Н™11
;
;
;
;
7
22
5
22
7Н™5
91. ПЄ.555762; .831342;
5
ПЄ.668512; ПЄ1.49586; 1.20287;
ПЄ1.79933 92. .164215;
ПЄ.986425; ПЄ.166475; ПЄ6.00691;
ПЄ1.01376; 6.08958 95. false;
for example, sin 30В° П© cos 30В° П·
.5 П© .8660 а·‡ 1.3660 1.
96. false; sin вђЄ Х… 1 for all вђЄ.
97. 146 ft
79.
1
79. sin вђЄ, if cos вђЄ а·‡ ПЄ , with вђЄ in quadrant II
4
1
80. csc вђЄ, if cot вђЄ а·‡ ПЄ , with вђЄ in quadrant IV
2
81. sec вђЄ, if tan вђЄ а·‡
Н™7
, with вђЄ in quadrant III
3
82. cos вђЄ, if csc вђЄ а·‡ ПЄ4, with вђЄ in quadrant III
83. sin вђЄ, if sec вђЄ а·‡ 2, with вђЄ in quadrant IV
84. cot вђЄ, if csc вђЄ а·‡ ПЄ3.5891420, with вђЄ in quadrant III
85. tan вђЄ, if sin вђЄ а·‡ .49268329, with вђЄ in quadrant II
86. Concept Check
Does there exist an angle вђЄ with cos вђЄ а·‡ ПЄ.6 and sin вђЄ а·‡ .8?
Find all trigonometric function values for each angle. Use a calculator in Exercises 91
and 92. See Example 8.
87. tan вђЄ а·‡ ПЄ
15
, with вђЄ in quadrant II
8
89. tan вђЄ а·‡ Н™3, with вђЄ in quadrant III
88. cos вђЄ а·‡ ПЄ
90. sin вђЄ а·‡
3
, with вђЄ in quadrant III
5
Н™5
, with вђЄ in quadrant I
7
91. cot вђЄ а·‡ ПЄ1.49586, with вђЄ in quadrant IV
92. sin вђЄ а·‡ .164215, with вђЄ in quadrant II
Work each problem.
93. Derive the identity 1 П© cot 2 вђЄ а·‡ csc2 вђЄ by dividing x 2 П© y 2 а·‡ r 2 by y 2.
94. Using a method similar to the one given in this section showing that
вђЄ
show that cos
sin вђЄ а·‡ cot вђЄ.
sin вђЄ
cos вђЄ
а·‡ tan вђЄ,
95. Concept Check True or false: For all angles вђЄ, sin вђЄ П© cos вђЄ а·‡ 1. If false, give an
example showing why it is false.
96. Concept Check
вђЄ
1
True or false: Since cot вђЄ а·‡ cos
sin вђЄ , if cot вђЄ а·‡ 2 with вђЄ in quadrant I,
then cos вђЄ а·‡ 1 and sin вђЄ а·‡ 2. If false, explain why.
Use a trigonometric function ratio to solve each problem. (Source for Exercises 97– 98:
Parker, M., Editor, She Does Math, Mathematical Association of America, 1995.)
97. Height of a Tree A civil engineer must determine the height of the tree shown in
the figure. The given angle was measured with a clinometer. She knows that
sin 70В° П· .9397, cos 70В° П· .3420, and tan 70В° П· 2.747. Use the pertinent trigonometric function and the measurement given in the figure to find the height of the tree
to the nearest whole number.
70В°
50 ft
This is a picture of one type of
clinometer, called an Abney hand
level and clinometer. The picture is
courtesy of Keuffel & Esser Co.
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494 CHAPTER 5 Trigonometric Functions
98. (a) 13 prism diopters
5
(b) tan вђЄ а·‡
12
y
99. (a) tan вђЄ а·‡
x
y
(b) x а·‡
tan вђЄ
100. area а·‡ 3x 2 sin вђЄ
98. (Modeling) Double Vision To correct mild double vision, a small amount of
prism is added to a patient’s eyeglasses. The amount of light shift this causes is
measured in prism diopters. A patient needs 12 prism diopters horizontally and
5 prism diopters vertically. A prism that corrects for both requirements should have
length r and be set at angle вђЄ. See the figure.
r
5
вђЄ
12
(a) Use the Pythagorean theorem to find r.
(b) Write an equation involving a trigonometric function of вђЄ and the known
prism measurements 5 and 12.
99. (Modeling) Distance Between the Sun and a Star
Suppose that a star forms an angle вђЄ with respect to
Earth and the sun. Let the coordinates of Earth be
Н‘x, yН’, those of the star Н‘0, 0Н’, and those of the sun
Н‘x, 0Н’. See the figure. Find an equation for x, the distance between the sun and the star, as follows.
Earth
r
Оё
y
x
Star
Sun
Not to scale
(a) Write an equation involving a trigonometric
function that relates x, y, and вђЄ.
(b) Solve your equation for x.
100. Area of a Solar Cell A solar cell converts the energy of sunlight directly into
electrical energy. The amount of energy a cell produces depends on its area.
Suppose a solar cell is hexagonal, as shown in the figure. Express its area in terms
of sin вђЄ and any side x. (Hint: Consider one of the six equilateral triangles from the
hexagon. See the figure.) (Source: Kastner, B., Space Mathematics, NASA, 1985.)
вђЄ
h = в€љ3 x
2
x
x
вђЄ
9 101.
The straight line in the figure determines both angle вђЈ (alpha) and angle вђ¤ (beta)
with the positive x-axis. Explain why tan вђЈ а·‡ tan вђ¤.
y
ОІ
О±
0
x
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5.3 Evaluating Trigonometric Functions 495
5.3 Evaluating Trigonometric Functions
Definitions of the Trigonometric Functions в– Trigonometric Function Values of Special Angles
Angles в– Special Angles as Reference Angles в– Finding Function Values with a Calculator в– Measures
y
B
r
A
(x, y)
x
C
Figure 23
Right-Triangle-Based Definitions of Trigonometric Functions
TEACHING TIP Introduce the
mnemonic sohcahtoa to help students remember that “sine is
opposite over hypotenuse, cosine
is adjacent over hypotenuse,
and tangent is opposite over
adjacent.”
These definitions will be used
in applications of trigonometry in
Section 5.4.
B
For any acute angle A in standard position,
sin A ‫؍‬
y
side opposite
‫؍‬
r
hypotenuse
csc A ‫؍‬
hypotenuse
r
‫؍‬
y
side opposite
cos A ‫؍‬
x
side adjacent
‫؍‬
r
hypotenuse
sec A ‫؍‬
hypotenuse
r
‫؍‬
x
side adjacent
tan A ‫؍‬
y
side opposite
‫؍‬
x
side adjacent
cot A ‫؍‬
side adjacent
x
‫؍‬
.
y
side opposite
EXAMPLE 1 Finding Trigonometric Function Values of an Acute Angle
C
7
Reference
Finding Angle
Definitions of the Trigonometric Functions In Section 5.2 we used angles in standard position to define the trigonometric functions. There is another
way to approach them: as ratios of the lengths of the sides of right triangles.
Figure 23 shows an acute angle A in standard position. The definitions of the
trigonometric function values of angle A require x, y, and r. As drawn in
Figure 23, x and y are the lengths of the two legs of the right triangle ABC, and r
is the length of the hypotenuse.
The side of length y is called the side opposite angle A, and the side of
length x is called the side adjacent to angle A. We use the lengths of these sides
to replace x and y in the definitions of the trigonometric functions, and the
length of the hypotenuse to replace r, to get the following right-triangle-based
definitions.
y
x
в– Find the values of sin A, cos A, and tan A in the right triangle in Figure 24.
24
A
25
Figure 24
sin A а·‡
Solution The length of the side opposite angle A is 7, the length of the side adjacent to angle A is 24, and the length of the hypotenuse is 25. Use the relationships given in the box.
side opposite
7
а·‡
hypotenuse
25
cos A а·‡
side adjacent 24
а·‡
hypotenuse
25
tan A а·‡
side opposite
7
а·‡
side adjacent 24
Now try Exercise 1.
NOTE
Because the cosecant, secant, and cotangent ratios are the reciprocals
of the sine, cosine, and tangent values, respectively, in Example 1 we can con25
24
clude that csc A а·‡ 25
7 , sec A а·‡ 24 , and cot A а·‡ 7 .
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496 CHAPTER 5 Trigonometric Functions
60В°
2
2
60В°
60В°
2
Equilateral triangle
(a)
Trigonometric Function Values of Special Angles Certain special
angles, such as 30В°, 45В°, and 60В°, occur so often in trigonometry and in more advanced mathematics that they deserve special study. We start with an equilateral
triangle, a triangle with all sides of equal length. Each angle of such a triangle
measures 60В°. While the results we will obtain are independent of the length, for
convenience we choose the length of each side to be 2 units. See Figure 25(a).
Bisecting one angle of this equilateral triangle leads to two right triangles,
each of which has angles of 30В°, 60В°, and 90В°, as shown in Figure 25(b). Since
the hypotenuse of one of these right triangles has length 2, the shortest side will
have length 1. (Why?) If x represents the length of the medium side, then,
22 а·‡ 12 П© x 2 Pythagorean theorem (Section 1.5)
30В°
2
3 а·‡ x2
2
x
60В°
4 а·‡ 1 П© x2
30В°
x
90В°
Н™3 а·‡ x.
90В°
60В°
1
1
30°– 60° right triangle
(b)
Choose the positive root. (Section 1.4)
Figure 26 summarizes our results using a 30°– 60° right triangle. As shown in
the figure, the side opposite the 30В° angle has length 1; that is, for the 30В° angle,
hypotenuse а·‡ 2,
side opposite а·‡ 1,
side adjacent а·‡ Н™3.
Now we use the definitions of the trigonometric functions.
Figure 25
30В°
в€љ3
Subtract 1.
sin 30РЉ а·‡
side opposite
1
а·‡
hypotenuse
2
csc 30РЉ а·‡
2
а·‡2
1
cos 30РЉ а·‡
side adjacent Н™3
а·‡
hypotenuse
2
sec 30РЉ а·‡
2Н™3
2
а·‡
3
Н™3
tan 30РЉ а·‡
side opposite
1
Н™3
а·‡
а·‡
side adjacent Н™3
3
cot 30РЉ а·‡
Н™3
а·‡ Н™3
1
2
60В°
1
EXAMPLE 2 Finding Trigonometric Function Values for 60В°
Figure 26
Find the six trigonometric function values for a 60В° angle.
Solution
Refer to Figure 26 to find the following ratios.
sin 60РЉ а·‡
Н™3
2
cos 60РЉ а·‡
csc 60РЉ а·‡
2Н™3
3
sec 60РЉ а·‡ 2
1
2
tan 60РЉ а·‡ Н™3
cot 60РЉ а·‡
Н™3
3
Now try Exercises 11, 13, and 15.
45В°
1
r = в€љ2
45В°
1
45°– 45° right triangle
Figure 27
We find the values of the trigonometric functions for 45В° by starting with a
45°–45° right triangle, as shown in Figure 27. This triangle is isosceles; we
choose the lengths of the equal sides to be 1 unit. (As before, the results are independent of the length of the equal sides.) Since the shorter sides each have
length 1, if r represents the length of the hypotenuse, then
12 П© 12 а·‡ r 2 Pythagorean theorem
2 а·‡ r2
Н™2 а·‡ r.
Choose the positive square root.
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5.3 Evaluating Trigonometric Functions 497
Now we use the measures indicated on the 45°– 45° right triangle in Figure 27.
sin 45РЉ а·‡
1
Н™2
а·‡
2
Н™2
cos 45РЉ а·‡
1
Н™2
а·‡
2
Н™2
tan 45РЉ а·‡
1
а·‡1
1
csc 45РЉ а·‡
Н™2
а·‡ Н™2
1
sec 45РЉ а·‡
Н™2
а·‡ Н™2
1
cot 45РЉ а·‡
1
а·‡1
1
Function values for 30В°, 45В°, and 60В° are summarized in the table that follows.
Function Values of Special Angles
TEACHING TIP Tell students that a
good way to obtain one of these
function values is to draw the
appropriate “famous” right triangle
as shown in Figure 26 or 27 and
then use the right-triangle-based
definition of the function value.
вђЄ
sin вђЄ
cos вђЄ
tan вђЄ
cot вђЄ
sec вђЄ
csc вђЄ
30В°
1
2
Н™3
2
Н™3
3
Н™3
2Н™3
3
2
45В°
Н™2
2
Н™2
2
1
1
Н™2
Н™2
60В°
Н™3
2
1
2
Н™3
Н™3
3
2
2Н™3
3
Reference Angles Associated with every nonquadrantal angle in standard
position is a positive acute angle called its reference angle. A reference angle
for an angle вђЄ, written вђЄР€, is the positive acute angle made by the terminal side of
angle вђЄ and the x-axis. Figure 28 shows several angles вђЄ (each less than one
complete counterclockwise revolution) in quadrants II, III, and IV, respectively,
with the reference angle вђЄР€ also shown. In quadrant I, вђЄ and вђЄР€ are the same. If
an angle вђЄ is negative or has measure greater than 360В°, its reference angle is
found by first finding its coterminal angle that is between 0В° and 360В°, and then
using the diagrams in Figure 28.
y
вђЄ
␪′
y
y
O
вђЄ
x
␪′
вђЄ in quadrant II
вђЄ
O
вђЄ in quadrant III
x
O
x
␪′
вђЄ in quadrant IV
Figure 28
CAUTION
A common error is to find the reference angle by using the terminal side of вђЄ and the y-axis. The reference angle is always found with reference to the x-axis.
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498 CHAPTER 5 Trigonometric Functions
y
EXAMPLE 3 Finding Reference Angles
Find the reference angle for each angle.
218В°
(a) 218В°
x
0
38В°
(b) 1387В°
Solution
(a) As shown in Figure 29, the positive acute angle made by the terminal side of
this angle and the x-axis is 218РЉ ПЄ 180РЉ а·‡ 38РЉ. For вђЄ а·‡ 218РЉ, the reference
angle вђЄР€ а·‡ 38РЉ.
218° – 180° = 38°
(b) First find a coterminal angle between 0В° and 360В°. Divide 1387В° by 360В° to
get a quotient of about 3.9. Begin by subtracting 360В° three times (because
of the 3 in 3.9):
Figure 29
y
1387РЉ ПЄ 3 Рё 360РЉ а·‡ 307РЉ.
The reference angle for 307В° (and thus for 1387В°) is 360РЉ ПЄ 307РЉ а·‡ 53РЉ.
See Figure 30.
x
0
53В°
307В°
Now try Exercises 41 and 45.
360° – 307° = 53°
Special Angles as Reference Angles We can now find exact trigonometric function values of angles with reference angles of 30В°, 45В°, or 60В°.
Figure 30
EXAMPLE 4 Finding Trigonometric Function Values of a Quadrant III Angle
Find the values of the trigonometric functions for 210В°.
y
210В°
x
90В°
y
60В°
x
30В°
O
r
x = –√3
y = –1
r=2
P
Figure 31
Solution An angle of 210В° is shown in Figure 31. The reference angle is
210РЉ ПЄ 180РЉ а·‡ 30РЉ. To find the trigonometric function values of 210В°, choose
point P on the terminal side of the angle so that the distance from the origin O to
P is 2. By the results from 30°– 60° right triangles, the coordinates of point P
become Н‘ ПЄН™3, ПЄ1 Н’, with x а·‡ ПЄН™3, y а·‡ ПЄ1, and r а·‡ 2. Then, by the definitions of the trigonometric functions,
1
2
cos 210РЉ а·‡ ПЄ
Н™3
2
tan 210РЉ а·‡
csc 210РЉ а·‡ ПЄ2
sec 210РЉ а·‡ ПЄ
2Н™3
3
cot 210РЉ а·‡ Н™3.
sin 210РЉ а·‡ ПЄ
Н™3
3
Now try Exercise 59.
TEACHING TIP Review the signs of
each trigonometric function by
quadrant. As mentioned previously, the sentence “All Students
Take Calculus” may help some students remember signs of the basic
functions by quadrant.
Notice in Example 4 that the trigonometric function values of 210В° correspond in absolute value to those of its reference angle 30В°. The signs are different for the sine, cosine, secant, and cosecant functions because 210В° is a
quadrant III angle. These results suggest a shortcut for finding the trigonometric
function values of a nonacute angle, using the reference angle. In Example 4, the
reference angle for 210В° is 30В°. Using the trigonometric function values of 30В°,
and choosing the correct signs for a quadrant III angle, we obtain the results
found in Example 4.
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5.3 Evaluating Trigonometric Functions 499
Similarly, we determine the values of the trigonometric functions for any
nonquadrantal angle вђЄ by finding the function values for its reference angle
between 0В° and 90В°, and choosing the appropriate signs.
Finding Trigonometric Function Values for Any
Nonquadrantal Angle вђЄ
Step 1 If вђЄ Пѕ 360РЉ, or if вђЄ ПЅ 0РЉ, then find a coterminal angle by adding or
subtracting 360В° as many times as needed to get an angle greater
than 0В° but less than 360В°.
Step 2 Find the reference angle вђЄР€.
Step 3 Find the trigonometric function values for reference angle вђЄР€.
Step 4 Determine the correct signs for the values found in Step 3. (Use the
table of signs in Section 5.2, if necessary.) This gives the values of
the trigonometric functions for angle вђЄ.
y
EXAMPLE 5 Finding Trigonometric Function Values Using Reference Angles
Find the exact value of each expression.
␪′ = 60°
(a) cosН‘ПЄ240РЉН’
0
x
␪ = –240°
(b) tan 675РЉ
Solution
(a) Since an angle of ПЄ240РЉ is coterminal with an angle of
ПЄ240РЉ П© 360РЉ а·‡ 120В°,
(a)
the reference angle is 180РЉ ПЄ 120РЉ а·‡ 60РЉ, as shown in Figure 32(a). Since
the cosine is negative in quadrant II,
y
cosН‘ПЄ240РЉН’ а·‡ cos 120РЉ а·‡ ПЄcos 60РЉ а·‡ ПЄ
x
0
вђЄ = 675В°
␪′ = 45°
(b)
Figure 32
1
.
2
(b) Begin by subtracting 360В° to get a coterminal angle between 0В° and 360В°.
675РЉ ПЄ 360РЉ а·‡ 315РЉ
As shown in Figure 32(b), the reference angle is 360РЉ ПЄ 315РЉ а·‡ 45РЉ. An
angle of 315В° is in quadrant IV, so the tangent will be negative, and
tan 675РЉ а·‡ tan 315РЉ а·‡ ПЄtan 45РЉ а·‡ ПЄ1.
Now try Exercises 65 and 67.
Degree mode
Figure 33
Finding Function Values with a Calculator Calculators are capable of
finding trigonometric function values. For example, the values of cosН‘ПЄ240РЉН’
and tan 675РЉ, found in Example 5, are found with a calculator as shown in
Figure 33.
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500 CHAPTER 5 Trigonometric Functions
TEACHING TIP Caution students
not to use the sinПЄ1, cosПЄ1, and
tanПЄ1 keys when evaluating reciprocal functions with a calculator.
CAUTION
We have studied only degree measure of angles; radian measure
will be introduced in Chapter 6. When evaluating trigonometric functions of
angles given in degrees, remember that the calculator must be set in degree
mode. Get in the habit of always starting work by entering sin 90. If the displayed answer is 1, then the calculator is set for degree measure.
EXAMPLE 6 Finding Function Values with a Calculator
Approximate the value of each expression.
(a) sin 49РЉ 12Р€
(b) sec 97.977РЉ
(c) cot 51.4283РЉ
(d) sinН‘ПЄ246РЉН’
Solution
(a)
49РЉ 12Р€ а·‡ 49
12 В°
а·‡ 49.2РЉ
60
Convert 49РЉ 12Р€ to decimal
degrees. (Section 5.1)
sin 49РЉ 12Р€ а·‡ sin 49.2РЉ П· .75699506 To eight decimal places
These screens support the results of Example 6. We entered the angle measure in degrees and minutes for part (a).
In the fifth line of the first
screen, Ans–1 tells the calculator to find the reciprocal of
the answer given in the previous line.
(b) Calculators do not have secant keys. However, sec вђЄ а·‡ cos1 вђЄ for all angles вђЄ
where cos вђЄ 0. First find cos 97.977РЉ, and then take the reciprocal to get
sec 97.977РЉ П· ПЄ7.205879213.
(c) cot 51.4283РЉ П· .79748114 Use the identity cot вђЄ а·‡ tan1 вђЄ .
(d) sinН‘ПЄ246РЉН’ П· .91354546
Now try Exercises 75, 79, 81, and 85.
Finding Angle Measures Sometimes we need to find the measure of an angle having a certain trigonometric function value. Graphing calculators have
three inverse functions (denoted sinПЄ1, cosПЄ1, and tanПЄ1) that do just that. If x is
an appropriate number, then sinПЄ1x, cosПЄ1x, or tanПЄ1x give the measure of an angle whose sine, cosine, or tangent is x. For the applications in this section, these
functions will return values of x in quadrant I.
EXAMPLE 7 Using an Inverse Trigonometric Function to Find an Angle
Use a calculator to find an angle вђЄ in the interval Н“0В°, 90В°Н” that satisfies
sin вђЄ П· .9677091705.
Solution With the calculator in degree mode, we find that an angle вђЄ having
sine value .9677091705 is 75.4В°. (While there are infinitely many such
angles, the calculator gives only this one.) We write this result as
sinПЄ1 .9677091705 П· 75.4РЉ.
Degree mode
Figure 34
See Figure 34.
Now try Exercise 91.
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5.3 Evaluating Trigonometric Functions 501
EXAMPLE 8 Finding Angle Measures
Н™2
2 .
Find all values of вђЄ, if вђЄ is in the interval Н“0РЉ, 360РЉН’ and cos вђЄ а·‡ ПЄ
Algebraic Solution
Graphing Calculator Solution
Since cos вђЄ is negative, вђЄ must lie in quadrant II or
Н™2
III. Since the absolute value of cos вђЄ is 2 , the reference angle вђЄ Р€ must be 45В°. The two possible angles вђЄ
are sketched in Figure 35. The quadrant II angle вђЄ
equals 180РЉ ПЄ 45РЉ а·‡ 135В°, and the quadrant III angle
вђЄ equals 180РЉ П© 45РЉ а·‡ 225РЉ.
The screen in Figure 36 shows how the inverse cosine
function is used to find the two values in Н“0РЉ, 360РЉН’
y
for which cos вђЄ а·‡ ПЄ 2 . Notice that cosПЄ1Н‘ ПЄ 2 Н’
yields only one value, 135В°; to find the other value,
we use the reference angle.
Н™2
Н™2
y
вђЄ in quadrant II
вђЄ' = 45В°
0
вђЄ = 135В°
x
вђЄ' = 45В°
0 вђЄ = 225В°
x
Degree mode
вђЄ in quadrant III
Figure 36
Figure 35
Now try Exercise 95.
EXAMPLE 9 Finding Grade Resistance
вђЄ > 0В°
вђЄ < 0В°
When an automobile travels uphill or downhill on a highway, it experiences a
force due to gravity. This force F in pounds is called grade resistance and is
modeled by the equation F а·‡ W sin вђЄ, where вђЄ is the grade and W is the weight
of the automobile. If the automobile is moving uphill, then вђЄ Пѕ 0РЉ; if downhill,
then вђЄ ПЅ 0РЉ. See Figure 37. (Source: Mannering, F. and W. Kilareski, Principles
of Highway Engineering and Traffic Analysis, 2nd Edition, John Wiley & Sons,
1998.)
(a) Calculate F to the nearest 10 lb for a 2500-lb car traveling an uphill grade
with вђЄ а·‡ 2.5РЉ.
Figure 37
(b) Calculate F to the nearest 10 lb for a 5000-lb truck traveling a downhill
grade with вђЄ а·‡ ПЄ6.1РЉ.
(c) Calculate F for вђЄ а·‡ 0РЉ and вђЄ а·‡ 90РЉ. Do these answers agree with your
intuition?
Solution
(a) F а·‡ W sin вђЄ а·‡ 2500 sin 2.5РЉ П· 110 lb
(b) F а·‡ W sin вђЄ а·‡ 5000 sinН‘ПЄ6.1РЉН’ П· ПЄ530 lb
F is negative because the truck is moving downhill.
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502 CHAPTER 5 Trigonometric Functions
(c) F а·‡ W sin вђЄ а·‡ W sin 0РЉ а·‡ WН‘0Н’ а·‡ 0 lb
F а·‡ W sin вђЄ а·‡ W sin 90РЉ а·‡ WН‘1Н’ а·‡ W lb
This agrees with intuition because if вђЄ а·‡ 0РЉ, then there is level ground and gravity does not cause the vehicle to roll. If вђЄ а·‡ 90РЉ, the road would be vertical and
the full weight of the vehicle would be pulled downward by gravity, so F а·‡ W.
Now try Exercises 109 and 111.
5.3 Exercises
In Exercises 1– 4, we give, in
order, sine, cosine, and tangent.
21 20 21
1.
; ;
29 29 20
45 28 45
2.
; ;
53 53 28
n m n
3.
;
;
p p m
k y k
4.
; ;
z z y
5. C 6. H 7. B 8. G
Н™3
9. E 10. A 11.
12. Н™3
3
1
2Н™3
Н™3
13.
14.
15.
2
2
3
16. 2 17. Н™2 18. Н™2
Н™2
19.
20. 1
2
Find exact values or expressions for sin A, cos A, and tan A. See Example 1.
1.
2.
21
29
53
A
20
A
3.
4.
p
Concept Check
Column II.
28
z
A
y
n
m
45
k
A
For each trigonometric function in Column I, choose its value from
I
II
5. sin 30РЉ
6. cos 45РЉ
A. Н™3
7. tan 45РЉ
8. sec 60РЉ
D.
9. csc 60РЉ
10. cot 30РЉ
Н™3
2
G. 2
B. 1
C.
1
2
E.
2Н™3
3
F.
Н™3
3
H.
Н™2
2
I.
Н™2
For each expression, give the exact value. See Example 2.
11. tan 30РЉ
12. cot 30РЉ
13. sin 30РЉ
14. cos 30РЉ
15. sec 30РЉ
16. csc 30РЉ
17. csc 45РЉ
18. sec 45РЉ
19. cos 45РЉ
20. cot 45РЉ
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5.3 Evaluating Trigonometric Functions 503
21.
y
Relating Concepts
P
45В°
22.
P
2 в€љ2
23.
24.
25.
27.
2в€љ2
x
the legs; Н‘ 2Н™2, 2Н™2 Н’
Н‘ 1, Н™3 Н’
y
The figure shows a 45РЉ central angle in a circle with radius
4 units. To find the coordinates of point P on the circle, work
Exercises 21–24 in order.
y
4
45В°
For individual or collaborative investigation
(Exercises 21–24)
x
sin x; tan x 26. cos x; csc x
60РЉ
P
4
45В°
21. Add a line from point P perpendicular to the x-axis.
x
22. Use the trigonometric ratios for a 45РЉ angle to label the
sides of the right triangle you sketched in Exercise 21.
23. Which sides of the right triangle give the coordinates of
point P? What are the coordinates of P?
y
24. Follow the same procedure to find the coordinates of P
in the figure given here.
P
2
60В°
25. Refer to the table. What trigonometric functions are y1 and y2 ?
26. Refer to the table. What trigonometric functions are y1 and y2?
x؇
0
15
30
45
60
75
90
0
.25882
.5
.70711
.86603
.96593
1
.26795
.57735
1
1.7321
3.7321
undefined
y1
y2
.96593
.86603
.70711
.5
.25882
undefined
3.8637
2
1.4142
1.1547
1.0353
1
x؇
0
15
30
45
60
75
90
y2
y1
0
x
1
0
27. Concept Check What value of A between 0РЉ
and 90РЉ will produce the output shown on the
graphing calculator screen?
9 28.
A student was asked to give the exact value of sin 45РЉ. Using a calculator, he gave
the answer .7071067812. The teacher did not give him credit. What was the
teacher’s reason for this?
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504 CHAPTER 5 Trigonometric Functions
29.
Н©
НЄ
Н™2 Н™2
,
; 45РЉ
2
2
30. y а·‡ Н™3x
31. y а·‡
Н™3
x
3
32. 30РЉ 33. 60РЉ
34. a а·‡ 12; b а·‡ 12Н™3;
d а·‡ 12Н™3; c а·‡ 12Н™6
9Н™3
9
35. x а·‡
;yа·‡ ;
2
2
3Н™3
; w а·‡ 3Н™3
zа·‡
2
7Н™3
14Н™3
36. m а·‡
;aа·‡
;
3
3
14Н™3
14Н™6
;qа·‡
nа·‡
3
3
37. p а·‡ 15; r а·‡ 15Н™2;
q а·‡ 5Н™6; t а·‡ 10Н™6
s2
s2Н™3
38. A а·‡
39. A а·‡
4
2
40. C 41. F 42. A 43. B
44. D 45. B
29. With a graphing calculator, find the coordinates of the point of intersection of
y а·‡ x and y а·‡ Н™1 ПЄ x 2. These coordinates are the cosine and sine of what angle
between 0РЉ and 90РЉ?
Concept Check
Work each problem.
30. Find the equation of the line passing through the origin and making a 60РЉ angle with
the x-axis.
31. Find the equation of the line passing through the origin and making a 30РЉ angle with
the x-axis.
32. What angle does the line y а·‡
Н™3
3 x
make with the positive x-axis?
33. What angle does the line y а·‡ Н™3x make with the positive x-axis?
Find the exact value of each part labeled with a variable in each figure.
34.
35.
45В°
24
9
y
c
b
30В°
60В°
z
x
60В°
a
w
d
36.
37.
q
45В° 30В°
45В°
a
7
15
t
r
q
n
60В°
90В°
m
45В°
p
Find a formula for the area of each figure in terms of s.
38.
39.
60В°
45В°
s
s
60В°
s
60В°
45В°
s
s
Match each angle in Column I with its reference angle in Column II. Choices may be
used once, more than once, or not at all. See Example 3.
I
II
A. 45РЉ
B. 60РЉ
43. ПЄ60РЉ
C. 82РЉ
D. 30РЉ
45. 480РЉ
E. 38РЉ
F. 32РЉ
40. 98РЉ
41. 212РЉ
42. ПЄ135РЉ
44. 750РЉ
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5.3 Evaluating Trigonometric Functions 505
Н™3
; Н™3
3
Н™2 Н™2
51.
;
; Н™2; Н™2
2
2
Н™3 Н™3 2Н™3
52.
;
;
2
3
3
1
Н™3
53. ПЄ ; ПЄ
; ПЄ2
2
3
54. ПЄ1; ПЄ1
1
2Н™3
55.
; ПЄН™3; ПЄ
2
3
2Н™3
Н™3
56. ПЄ
;ПЄ
2
3
Н™3
57. Н™3;
3
In Exercises 58 – 63, we give, in
order, sine, cosine, tangent,
cotangent, secant, and cosecant.
Н™3 1
Н™3
58. ПЄ
; ; ПЄН™3; ПЄ
;
2 2
3
2Н™3
2; ПЄ
3
Н™2 Н™2
59. ПЄ
;
; ПЄ1, ПЄ1;
2
2
Н™2; ПЄН™2
Н™2 Н™2
60.
;
; 1; 1; Н™2; Н™2
2
2
2Н™3
Н™3 1
Н™3
61.
; ; Н™3;
; 2;
2 2
3
3
1
Н™3 Н™3
62. ПЄ ; ПЄ
;
; Н™3;
2
2
3
2Н™3
; ПЄ2
ПЄ
3
1 Н™3 Н™3
2Н™3
63.
;
;
; Н™3;
;2
2 2
3
3
Н™2
Н™3
64. ПЄ
65. ПЄ
2
2
Н™3
66. Н™3 67.
2
1 П© Н™3
68. false;
1 69. true
2
1
70. true 71. false;
Н™3
2
50.
9 Give a short explanation in Exercises 46 – 49.
46. In Example 4, why was 2 a good choice for r? Could any other positive number have
been used?
47. Explain how the reference angle is used to find values of the trigonometric functions
for an angle in quadrant III.
48. Explain why two coterminal angles have the same values for their trigonometric
functions.
49. If two angles have the same values for each of the six trigonometric functions, must
the angles be coterminal? Explain your reasoning.
Complete the table with exact trigonometric function values. Do not use a calculator.
See Examples 2, 4, and 5.
вђЄ
sin вђЄ
cos вђЄ
50.
30РЉ
1
2
Н™3
2
51.
45РЉ
52.
60РЉ
53.
120РЉ
Н™3
2
54.
135РЉ
Н™2
2
55.
150РЉ
56.
210РЉ
ПЄ
57.
240РЉ
ПЄ
tan вђЄ
cot вђЄ
sec вђЄ
2Н™3
3
1
1
2
Н™3
ПЄ
Н™2
2
ПЄ
Н™3
2
2
2Н™3
3
ПЄН™2
ПЄ
ПЄ
Н™2
Н™3
3
Н™3
3
Н™3
2
2
1
ПЄН™3
1
2
csc вђЄ
2
ПЄ2
Н™3
1
2
ПЄ2
ПЄ
2Н™3
3
Find exact values of the six trigonometric functions for each angle. Rationalize denominators when applicable. See Examples 2, 4, and 5.
58. 300РЉ
59. 315РЉ
60. 405РЉ
61. ПЄ300РЉ
62. ПЄ510РЉ
63. 750РЉ
Find the exact value of each expression. See Example 5.
64. sin 1305РЉ
65. cosН‘ПЄ510РЉН’
66. tanН‘ПЄ1020РЉН’
Tell whether each statement is true or false. If false, tell why.
68. sin 30РЉ П© sin 60РЉ а·‡ sinН‘30РЉ П© 60РЉН’
69. sinН‘30РЉ П© 60РЉН’ а·‡ sin 30РЉ Рё cos 60РЉ П© sin 60РЉ Рё cos 30РЉ
70. cos 60РЉ а·‡ 2 cos2 30РЉ ПЄ 1
71. cos 60РЉ а·‡ 2 cos 30РЉ
67. sin 1500РЉ
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506 CHAPTER 5 Trigonometric Functions
Н™3
0 73. true
2
74. .5657728 75. .6252427
76. 1.1342773 77. 1.0273488
78. 1.7768146 79. 15.055723
80. .4771588 81. 1.4887142
82. ПЄ5.7297416 83. .6743024
84. 1.9074147 85. .9999905
86. .9668234 87. .4327386
88. .6494076 89. .2308682
90. 57.997172РЉ 91. 55.845496РЉ
92. 30.502748РЉ 93. 38.491580РЉ
94. 30РЉ; 150РЉ 95. 30РЉ; 330РЉ
96. 120РЉ; 300РЉ 97. 135РЉ; 225РЉ
98. 120РЉ; 300РЉ 99. 45РЉ; 315РЉ
100. A: 68.94 mph; B: 65.78 mph
72. false;
72. sin 120РЉ а·‡ sin 150РЉ ПЄ sin 30РЉ
73. sin 120РЉ а·‡ sin 180РЉ Рё cos 60РЉ ПЄ sin 60РЉ Рё cos 180РЉ
Use a calculator to find a decimal approximation for each value. Give as many digits as
your calculator displays. In Exercises 86 –89, simplify the expression before using the
calculator. See Example 6.
74. tan 29РЉ 30Р€
75. sin 38РЉ 42Р€
76. cot 41РЉ 24Р€
77. sec 13РЉ 15Р€
78. csc 145РЉ 45Р€
79. cot 183РЉ 48Р€
80. cos 421РЉ 30Р€
81. sec 312РЉ 12Р€
82. tanН‘ПЄ80РЉ 6Р€Н’
83. sinН‘ПЄ317РЉ 36Р€Н’
84. cotН‘ПЄ512РЉ 20Р€Н’
85. cosН‘ПЄ15Р€Н’
1
86.
sec 14.8РЉ
1
87.
cot 23.4РЉ
88.
89.
sin 33РЉ
cos 33РЉ
cos 77РЉ
sin 77РЉ
Find a values of вђЄ in Н“0РЉ, 90РЉН” that satisfies each statement. Leave answers in decimal
degrees. See Example 7.
90. sin вђЄ а·‡ .84802194
91. tan вђЄ а·‡ 1.4739716
92. sec вђЄ а·‡ 1.1606249
93. cot вђЄ а·‡ 1.2575516
Find all values of вђЄ, if вђЄ is in the interval Н“0РЉ, 360РЉН’ and has the given function value. See
Example 8.
94. sin вђЄ а·‡
1
2
95. cos вђЄ а·‡
97. sec вђЄ а·‡ ПЄН™2
Н™3
2
98. cot вђЄ а·‡ ПЄ
Н™3
3
96. tan вђЄ а·‡ ПЄН™3
99. cos вђЄ а·‡
Н™2
2
Work each problem.
100. Measuring Speed by Radar Any offset between a stationary radar gun and a
moving target creates a “cosine effect” that reduces the radar mileage reading by
the cosine of the angle between the gun and the vehicle. That is, the radar speed
reading is the product of the actual reading and the cosine of the angle. Find the
radar readings for Auto A and Auto B shown in the figure. (Source: “Working
Knowledge,” Fischetti, M., Scientific American, March 2001.)
Radar gun
Auto A
10В° angle
Actual speed: 70 mph
POLIC
Auto B
9 101.
Measuring Speed by Radar In Exercise 100,
we saw that the mileage reported by a radar gun
is reduced by the cosine of angle вђЄ, shown in
the figure. In the figure, r represents reduced
speed and a represents the actual speed. Use the
figure to show why this “cosine effect” occurs.
E
20В° angle
Actual speed: 70 mph
Radar gun
r
вђЄ
Auto
POLI
CE
a
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5.3 Evaluating Trigonometric Functions 507
102. (a) 2 П« 10 8 m per sec
(b) 2 П« 10 8 m per sec
103. (a) 19РЉ (b) 50РЉ
104. 48.7РЉ 105. 7.9РЉ
106. (a) approximately 155 ft
(b) approximately 194 ft
(Modeling) Speed of Light When a light
ray travels from one medium, such as air,
to another medium, such as water or glass,
the speed of the light changes, and the direction in which the ray is traveling
changes. (This is why a fish under water is
in a different position than it appears to
be.) These changes are given by Snell’s
law
Medium 1
If this medium is
less dense, light
travels at a faster
speed, c1.
Оё1
Medium 2
Оё2
If this medium is
more dense, light
travels at a slower
speed, c2.
c 1 sin вђЄ1
,
а·‡
c 2 sin вђЄ2
where c 1 is the speed of light in the first medium, c 2 is the speed of light in the second
medium, and вђЄ1 and вђЄ2 are the angles shown in the figure. (Source: The Physics
Classroom, www.glenbrook.k12.il.us) In Exercises 102 and 103, assume that c 1 а·‡
3 П« 10 8 m per sec.
102. Find the speed of light in the second medium for each of the following.
(a) вђЄ1 а·‡ 46РЉ, вђЄ2 а·‡ 31РЉ
(b) вђЄ1 а·‡ 39РЉ, вђЄ2 а·‡ 28РЉ
103. Find вђЄ2 for each of the following values of вђЄ1 and c 2. Round to the nearest degree.
(a) вђЄ1 а·‡ 40РЉ, c 2 а·‡ 1.5 П« 10 8 m per sec (b) вђЄ1 а·‡ 62РЉ, c 2 а·‡ 2.6 П« 10 8 m per sec
(Modeling) Fish’s View of the World The figure shows a fish’s view of the world above the
surface of the water. (Source: Walker, J., “The
Amateur Scientist,” Scientific American, March
1984.) Suppose that a light ray comes from
the horizon, enters the water, and strikes the
fish’s eye.
Ray
from
horizon
104. Assume that this ray gives a value of 90РЉ for
angle ␪1 in the formula for Snell’s law. (In a
practical situation, this angle would probably
be a little less than 90РЉ.) The speed of light
in water is about 2.254 П« 10 8 m per sec.
Find angle вђЄ2 .
Apparent
horizon
Ray from zenith
Window’s
center
вђЄ2
105. Suppose an object is located at a true angle of 29.6РЉ above the horizon. Find the
apparent angle above the horizon to a fish.
106. (Modeling) Braking Distance If aerodynamic resistance is ignored, the braking
distance D (in feet) for an automobile to change its velocity from V1 to V2 (feet per
second) can be modeled using the equation
Dа·‡
1.05Н‘V 12 ПЄ V 22Н’
.
64.4Н‘K 1 П© K 2 П© sin вђЄН’
K 1 is a constant determined by the efficiency of the brakes and tires, K 2 is a constant determined by the rolling resistance of the automobile, and вђЄ is the grade of
the highway. (Source: Mannering, F. and W. Kilareski, Principles of Highway
Engineering and Traffic Analysis, 2nd Edition, John Wiley & Sons, 1998.)
(a) Compute the number of feet required to slow a car from 55 mph to 30 mph
while traveling uphill with a grade of вђЄ а·‡ 3.5РЉ. Let K 1 а·‡ .4 and K 2 а·‡ .02.
(Hint: Change miles per hour to feet per second.)
(b) Repeat part (a) with вђЄ а·‡ ПЄ2РЉ.
9 (c) How is braking distance affected by grade вђЄ ? Does this agree with your
driving experience?
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508 CHAPTER 5 Trigonometric Functions
107. approximately 78 mph
108. ПЄ100.5 lb 109. 65.96 lb
110. 2771 lb 111. ПЄ2.87РЉ
112. 2200-lb car on 2РЉ
uphill grade
113. (a) 703 ft (b) 1701 ft
(c) R would decrease.
114. 55 mph
107. (Modeling) Car’s Speed at Collision Refer to Exercise 106. An automobile is
traveling at 90 mph on a highway with a downhill grade of вђЄ а·‡ ПЄ3.5РЉ. The driver
sees a stalled truck in the road 200 ft away and immediately applies the brakes.
Assuming that a collision cannot be avoided, how fast (in miles per hour) is the
car traveling when it hits the truck? (Use the same values for K 1 and K 2 as in
Exercise 106.)
(Modeling) Grade Resistance See Example 9 to work Exercises 108 –112.
108. What is the grade resistance of a 2400-lb car traveling on a ПЄ2.4РЉ downhill grade?
109. What is the grade resistance of a 2100-lb car traveling on a 1.8РЉ uphill grade?
110. A car traveling on a ПЄ3РЉ downhill grade has a grade resistance of ПЄ145 lb. What
is the weight of the car?
111. A 2600-lb car traveling downhill has a grade resistance of ПЄ130 lb. What is the
angle of the grade?
112. Which has the greater grade resistance: a 2200-lb car on a 2РЉ uphill grade or a
2000-lb car on a 2.2РЉ uphill grade?
113. (Modeling) Design of Highway Curves When highway curves are designed, the
outside of the curve is often slightly elevated or inclined above the inside of the
curve. See the figure. This inclination is called superelevation. For safety reasons,
it is important that both the curve’s radius and superelevation are correct for a
given speed limit.
вђЄ
If an automobile is traveling at velocity V (in feet per second), the safe radius R
for a curve with superelevation вђЄ is modeled by the formula
Rа·‡
V2
,
gН‘ f П© tan вђЄН’
where f and g are constants. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, 2nd Edition, John Wiley &
Sons, 1998.)
(a) A roadway is being designed for automobiles traveling at 45 mph. If вђЄ а·‡ 3РЉ,
g а·‡ 32.2, and f а·‡ .14, calculate R.
(b) What should the radius of the curve be if the speed in part (a) is increased to
70 mph?
(c) How would increasing the angle вђЄ affect the results? Verify your answer by
repeating parts (a) and (b) with вђЄ а·‡ 4РЉ.
114. (Modeling) Speed Limit on a Curve Refer to Exercise 113 and use the same
values for f and g. A highway curve has radius R а·‡ 1150 ft and a superelevation
of вђЄ а·‡ 2.1РЉ. What should the speed limit (in miles per hour) be for this curve?
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5.4 Solving Right Triangles
509
5.4 Solving Right Triangles
Significant Digits
Applications
d
в– Solving Triangles
15 ft
в– Angles of Elevation or Depression
в– Bearing
в– Further
Significant Digits Suppose we quickly measure a room as 15 ft by 18 ft. See
Figure 38. To calculate the length of a diagonal of the room, we can use the
Pythagorean theorem.
d 2 а·‡ 152 П© 182
(Section 1.5)
d а·‡ 549
2
18 ft
Figure 38
TEACHING TIP Make the analogy
that this discussion is similar to
the idea that a chain is only as
strong as its weakest link.
d а·‡ Н™549 П· 23.430749
Should this answer be given as the length of the diagonal of the room? Of course
not. The number 23.430749 contains six decimal places, while the original data
of 15 ft and 18 ft are only accurate to the nearest foot. Since the results of
a problem can be no more accurate than the least accurate number in any calculation, we really should say that the diagonal of the 15-by-18-ft room is
about 23 ft.
If a wall measured to the nearest foot is 18 ft long, this actually means that
the wall has length between 17.5 ft and 18.5 ft. If the wall is measured more accurately as 18.3 ft long, then its length is really between 18.25 ft and 18.35 ft. A
measurement of 18.00 ft would indicate that the length of the wall is between
17.995 ft and 18.005 ft. The measurement 18 ft is said to have two significant
digits of accuracy; 18.0 has three significant digits, and 18.00 has four.
A significant digit is a digit obtained by actual measurement. A number
that represents the result of counting, or a number that results from theoretical
work and is not the result of a measurement, is an exact number. There are
50 states in the United States, so in that statement, 50 is an exact number.
Most values obtained for trigonometric functions are approximations, and
virtually all measurements are approximations. When performing calculations
involving approximate numbers, start by determining the number that has the
least number of significant digits. Round your final answer to the same number
of significant digits as this number. Remember that your answer is no more
accurate than the least accurate number in your calculation.
Use the following table to determine the significant digits in angle measure.
Significant Digits for Angles
Number of
Significant Digits
Angle Measure to Nearest:
2
Degree
3
Ten minutes, or nearest tenth of a degree
4
Minute, or nearest hundredth of a degree
5
Tenth of a minute, or nearest thousandth of a degree
For example, an angle measuring 52РЉ 30Р€ has three significant digits (assuming
that 30Р€ is measured to the nearest ten minutes).
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510 CHAPTER 5 Trigonometric Functions
B
Solving Triangles To solve a triangle means to find the measures of all the
angles and sides of the triangle. As shown in Figure 39, we use a to represent the
length of the side opposite angle A, b for the length of the side opposite angle B,
and so on. In a right triangle, the letter c is reserved for the hypotenuse.
c
a
C
A
b
EXAMPLE 1 Solving a Right Triangle Given an Angle and a Side
Solve right triangle ABC, if A а·‡ 34РЉ 30Р€ and c а·‡ 12.7 in. See Figure 40.
When solving triangles, a labeled
sketch is an important aid.
Solution
Figure 39
B
To solve the triangle, find the measures of the remaining sides and
angles. To find the value of a, use a trigonometric function involving the known
values of angle A and side c. Since the sine of angle A is given by the quotient of
the side opposite A and the hypotenuse, use sin A.
c = 12.7 in.
a
sin A а·‡
C
sin 34РЉ 30Р€ а·‡
34° 30′
A
b
Figure 40
a
c
sin A а·‡
a
12.7
A а·‡ 34РЉ 30Р€, c а·‡ 12.7
side opposite
hypotenuse
(Section 5.3)
a а·‡ 12.7 sin 34РЉ 30Р€
Multiply by 12.7; rewrite.
a а·‡ 12.7 sin 34.5РЉ
Convert to decimal degrees.
a а·‡ 12.7Н‘.56640624Н’ Use a calculator.
a П· 7.19 in.
Looking Ahead to Calculus
The derivatives of the parametric
equations x а·‡ f Н‘tН’ and y а·‡ gН‘tН’ often
represent the rate of change of physical
quantities, such as velocities. When x
and y are related by an equation, the
derivatives are called related rates because a change in one causes a related
change in the other. Determining these
rates in calculus often requires solving
a right triangle. Some problems that
ask for the maximum or minimum
value of a quantity also involve solving a right triangle.
Three significant digits
To find the value of b, we could use the Pythagorean theorem. It is better, however, to use the information given in the problem rather than a result just calculated. If a mistake is made in finding a, then b also would be incorrect. Also,
rounding more than once may cause the result to be less accurate. Use cos A.
cos A а·‡
cos 34РЉ 30Р€ а·‡
b
c
cos A а·‡
side adjacent
hypotenuse
(Section 5.3)
b
12.7
b а·‡ 12.7 cos 34РЉ 30Р€
b П· 10.5 in.
Once b is found, the Pythagorean theorem can be used as a check. All that remains
to solve triangle ABC is to find the measure of angle B. Since A П© B а·‡ 90РЉ,
B а·‡ 90РЉ ПЄ A
TEACHING TIP When solving triangles, encourage students to
refer to a sketch of the problem to
see if their answer is reasonable.
B а·‡ 89РЉ 60Р€ ПЄ 34РЉ 30Р€
A а·‡ 34РЉ 30Р€
B а·‡ 55РЉ 30Р€.
Now try Exercise 17.
NOTE
In Example 1, we could have found the measure of angle B first, and
then used the trigonometric function values of B to find the unknown sides. The
process of solving a right triangle can usually be done in several ways, each producing the correct answer. To maintain accuracy, always use given information
as much as possible, and avoid rounding off in intermediate steps.
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5.4 Solving Right Triangles
511
EXAMPLE 2 Solving a Right Triangle Given Two Sides
Solve right triangle ABC if a а·‡ 29.43 cm and c а·‡ 53.58 cm.
Solution
We draw a sketch showing the given information, as in Figure 41.
One way to begin is to find angle A by using the sine function.
B
c = 53.58 cm
a = 29.43 cm
sin A а·‡
C
side opposite 29.43
а·‡
П· .5492721165
hypotenuse
53.58
A
b
Using the sinПЄ1 function on a calculator, we find that A П· 33.32РЉ. The measure
of B is approximately
Figure 41
90РЉ ПЄ 33.32РЉ а·‡ 56.68РЉ.
We now find b from the Pythagorean theorem, using a 2 П© b2 а·‡ c2, or
b а·‡ c2 ПЄ a 2. Since c а·‡ 53.58 and a а·‡ 29.43,
2
b2 а·‡ c 2 ПЄ a2
Pythagorean theorem (Section 1.5)
b2 а·‡ 53.582 ПЄ 29.432
b П· 44.77 cm.
Now try Exercise 21.
Y
Angle of
elevation
X
Horizontal
(a)
X
Angles of Elevation or Depression Many applications of right triangles
involve angles of elevation or depression. The angle of elevation from point X
to point Y (above X) is the acute angle formed by ray XY and a horizontal ray
with endpoint at X. See Figure 42(a). The angle of depression from point X to
point Y (below X) is the acute angle formed by ray XY and a horizontal ray with
endpoint X. See Figure 42(b).
Horizontal
Angle of
depression
CAUTION
Be careful when interpreting the angle of depression. Both the
angle of elevation and the angle of depression are measured between the line of
sight and a horizontal line.
Y
(b)
Figure 42
TEACHING TIP Point out that when
point X is below point Y, then the
angle of elevation from X to Y is
the same as the angle of depression from Y to X, since they are the
alternate interior angles resulting
from the parallel horizontal lines
being cut by a transversal (the line
of sight).
To solve applied trigonometry problems, follow the same procedure as solving a triangle.
Solving an Applied Trigonometry Problem
Step 1 Draw a sketch, and label it with the given information. Label the
quantity to be found with a variable.
Step 2 Use the sketch to write an equation relating the given quantities to
the variable.
Step 3 Solve the equation, and check that your answer makes sense.
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512 CHAPTER 5 Trigonometric Functions
EXAMPLE 3 Finding the Angle of Elevation When Lengths Are Known
The length of the shadow of a building 34.09 m tall is 37.62 m. Find the angle of
elevation of the sun.
Solution
As shown in Figure 43, the angle of elevation of the sun is angle B.
Since the side opposite B and the side adjacent to B are known, use the tangent
ratio to find B.
34.09 m
37.62 m
tan B а·‡
B
34.09
,
37.62
so
B а·‡ tanПЄ1
34.09
П· 42.18РЉ. (Section 5.3)
37.62
The angle of elevation of the sun is 42.18В°.
Figure 43
Now try Exercise 29.
TEACHING TIP As an extension of
the type of problem in Example 3,
mention that professional surveyors often determine distances
by measuring a zenith angle. A
zenith angle is the angle made
with an imaginary line perpendicular to Earth and the line of
sight to the object being measured. This provides a more accurate measurement.
Bearing Other applications of right triangles involve bearing, an important
idea in navigation. There are two methods for expressing bearing. When a single
angle is given, such as 164В°, it is understood that the bearing is measured in a
clockwise direction from due north. Several sample bearings using this first
method are shown in Figure 44.
N
N
N
N
32В°
164В°
229В°
304В°
Figure 44
EXAMPLE 4 Solving a Problem Involving Bearing (First Method)
Radar stations A and B are on an east-west line, 3.7 km apart. Station A detects a
plane at C, on a bearing of 61В°. Station B simultaneously detects the same plane,
on a bearing of 331В°. Find the distance from A to C.
Solution
Draw a sketch showing the given information, as in Figure 45. Since
a line drawn due north is perpendicular to an east-west line, right angles are
formed at A and B, so angles CAB and CBA can be found as shown in Figure 45.
Angle C is a right angle because angles CAB and CBA are complementary. Find
distance b by using the cosine function.
N
N
C
61В°
b
61В°
29В°
A
3.7 km
B
331В°
Figure 45
cos 29РЉ а·‡
b
3.7
3.7 cos 29РЉ а·‡ b
b П· 3.2 km Use a calculator; round to the nearest tenth.
Now try Exercise 45.
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5.4 Solving Right Triangles
TEACHING TIP Emphasize the need
to draw fairly accurate sketches
describing each application problem. Encourage students to use
their sketches to check that their
answers are reasonable.
513
CAUTION
The importance of a correctly labeled sketch when solving applications like that in Example 4 cannot be overemphasized. Some of the necessary
information is often not directly stated in the problem and can be determined
only from the sketch.
The second method for expressing bearing starts with a north-south line and
uses an acute angle to show the direction, either east or west, from this line.
Figure 46 shows several sample bearings using this system. Either N or S always
comes first, followed by an acute angle, and then E or W.
N
N
42В°
S
S
N 42В° E
52В°
40В°
31В°
S 31В° E
S 40В° W
N 52В° W
Figure 46
EXAMPLE 5 Solving a Problem Involving Bearing (Second Method)
The bearing from A to C is S 52В° E. The bearing from A to B is N 84В° E. The
bearing from B to C is S 38В° W. A plane flying at 250 mph takes 2.4 hr to go
from A to B. Find the distance from A to C.
Solution
N
N
D
96В°
84В°
A
c = 600 mi
46В°
38В°
44В°
52В°
b
S
C
Figure 47
S
B
Make a sketch. First draw the two bearings from point A. Choose a
point B on the bearing N 84В° E from A, and draw the bearing to C. Point C will
be located where the bearing lines from A and B intersect, as shown in
Figure 47.
Since the bearing from A to B is N 84В° E, angle ABD is 180РЉ ПЄ 84РЉ а·‡ 96РЉ.
Thus, angle ABC is 46В°. Also, angle BAC is 180РЉ ПЄ Н‘84РЉ П© 52РЉН’ а·‡ 44РЉ. Angle C
is 180РЉ ПЄ Н‘44РЉ П© 46РЉН’ а·‡ 90РЉ. From the statement of the problem, a plane flying
at 250 mph takes 2.4 hr to go from A to B. The distance from A to B is the product of rate and time, or
c а·‡ rate П« time а·‡ 250Н‘2.4Н’ а·‡ 600 mi.
(Section 1.2)
To find b, the distance from A to C, use the sine. (The cosine could also
be used.)
sin 46РЉ а·‡
b
c
sin 46РЉ а·‡
b
600
600 sin 46РЉ а·‡ b
b П· 430 mi
Now try Exercise 47.
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514 CHAPTER 5 Trigonometric Functions
Further Applications
EXAMPLE 6 Using Trigonometry to Measure a Distance
P
вђЄ
d
Figure 48
b/2
Q
b/2
A method that surveyors use to determine a small distance d between two points
P and Q is called the subtense bar method. The subtense bar with length b is
centered at Q and situated perpendicular to the line of sight between P and Q.
See Figure 48. Angle вђЄ is measured, and then the distance d can be determined.
(a) Find d when вђЄ а·‡ 1РЉ 23Р€ 12Р‰ and b а·‡ 2.0000 m.
(b) Angle вђЄ usually cannot be measured more accurately than to the nearest 1Р‰.
How much change would there be in the value of d if вђЄ were measured 1Р‰
larger?
Solution
(a) From Figure 48, we see that
cot
вђЄ
d
а·‡ b
2
2
dа·‡
вђЄ
b
cot . Multiply; rewrite.
2
2
вђЄ
Let b а·‡ 2. To evaluate 2 , we change вђЄ to decimal degrees.
1РЉ 23Р€ 12Р‰ а·‡ 1.386667РЉ (Section 5.1)
Then
dа·‡
1.386667РЉ
2
cot
П· 82.6341 m.
2
2
(b) Since вђЄ is 1Р‰ larger, use вђЄ а·‡ 1РЉ 23Р€ 13Р‰ П· 1.386944РЉ.
dа·‡
1.386944РЉ
2
cot
П· 82.6176 m
2
2
The difference is 82.6341 ПЄ 82.6176 П· .0170 m.
Now try Exercise 59.
EXAMPLE 7 Solving a Problem Involving Angles of Elevation
Francisco needs to know the height of a tree. From a given point on the ground,
he finds that the angle of elevation to the top of the tree is 36.7В°. He then moves
back 50 ft. From the second point, the angle of elevation to the top of the tree is
22.2В°. See Figure 49. Find the height of the tree.
B
h
22.2В°
D
A
36.7В°
x
50 ft
Figure 49
C
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5.4 Solving Right Triangles
Algebraic Solution
Graphing Calculator Solution*
Figure 49 shows two unknowns: x, the distance from the center of the
trunk of the tree to the point where the first observation was made, and
h, the height of the tree. Since nothing is given about the length of the
hypotenuse of either triangle ABC or triangle BCD, use a ratio that
does not involve the hypotenuse — namely the tangent. See Figure 50
in the Graphing Calculator Solution.
In Figure 50, we superimposed Figure 49 on coordinate axes with the origin at D. By definition, the tangent of
the angle between the x-axis and
the graph of a line with equation
y а·‡ mx П© b is the slope of the line, m.
For line DB, m а·‡ tan 22.2РЉ. Since b
equals 0, the equation of line DB is
y1 а·‡ Н‘tan 22.2РЉН’x. The equation of line
AB is y2 а·‡ Н‘tan 36.7РЉН’x П© b. Since
b 0 here, we use the point AН‘50, 0Н’
and the point-slope form to find the
equation.
In triangle ABC,
tan 36.7РЉ а·‡
h
x
In triangle BCD,
tan 22.2РЉ а·‡
h
or h а·‡ Н‘50 П© xН’ tan 22.2РЉ.
50 П© x
or h а·‡ x tan 36.7РЉ.
Each expression equals h, so the expressions must be equal.
x tan 36.7РЉ а·‡ Н‘50 П© xН’ tan 22.2РЉ
Solve for x.
x tan 36.7РЉ а·‡ 50 tan 22.2РЉ П© x tan 22.2РЉ
Distributive property (Section R.1)
x tan 36.7РЉ ПЄ x tan 22.2РЉ а·‡ 50 tan 22.2РЉ
Get x-terms on one side. (Section 1.1)
xН‘tan 36.7РЉ ПЄ tan 22.2РЉН’ а·‡ 50 tan 22.2РЉ
Factor out x. (Section R.4)
y2 ПЄ y1 а·‡ mН‘x ПЄ x1Н’
(Section 2.4)
y2 ПЄ 0 а·‡ mН‘x ПЄ 50Н’
x1 а·‡ 50, y1 а·‡ 0
y2 а·‡ tan 36.7РЉН‘x ПЄ 50Н’
Lines y1 and y2 are graphed in Figure 51. The y-coordinate of the point of
intersection of the graphs gives the
length of BC, or h. Thus, h П· 45.
y
B
50 tan 22.2РЉ
xа·‡
tan 36.7РЉ ПЄ tan 22.2РЉ
h
Divide by the coefficient of x.
We saw above that h а·‡ x tan 36.7РЉ. Substituting for x,
Н©
НЄ
50 tan 22.2РЉ
tan 36.7РЉ.
hа·‡
tan 36.7РЉ ПЄ tan 22.2РЉ
22.2В°
D
A
Figure 50
75
tan 36.7РЉ а·‡ .74537703
and
x
C
50 ft
From a calculator,
so
36.7В°
x
and
tan 22.2РЉ а·‡ .40809244
tan 36.7РЉ ПЄ tan 22.2РЉ а·‡ .74537703 ПЄ .40809244 а·‡ .33728459
hа·‡
Н©
НЄ
50Н‘.40809244Н’
.74537703 П· 45.
.33728459
The height of the tree is approximately 45 ft.
–10
150
–20
Figure 51
Now try Exercise 53.
NOTE
In practice, we usually do not write down intermediate calculator
approximation steps. We did in Example 7 so you could follow the steps more
easily.
*Source: Adapted with permission from “Letter to the Editor,” by Robert Ruzich (Mathematics Teacher,
Volume 88, Number 1). Copyright б®Љ 1995 by the National Council of Teachers of Mathematics.
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516 CHAPTER 5 Trigonometric Functions
5.4 Exercises
1. 20,385.5 to 20,386.5
2. 28,999.5 to 29,000.5
3. 8958.5 to 8959.5 7. .05
8. .5 9. B а·‡ 53В° 40Р€;
a а·‡ 571 m; b а·‡ 777 m
10. B а·‡ 58В° 20Р€; c а·‡ 68.4 km;
b а·‡ 58.2 km 11. M а·‡ 38.8В°;
n а·‡ 154 m; p а·‡ 198 m
12. Y а·‡ 42.2В°; x а·‡ 66.4 cm;
y а·‡ 60.2 cm
Concept Check Refer to the discussion of accuracy and significant digits in this section to work Exercises 1 –8.
1. Leading NFL Receiver At the end of the 2001 National Football League season,
Oakland Raider Jerry Rice was the leading career receiver with 20,386 yd. State the
range represented by this number. (Source: The World Almanac and Book of Facts,
2003.)
2. Height of Mt. Everest When Mt. Everest was first surveyed, the surveyors
obtained a height of 29,000 ft to the nearest foot. State the range represented by this
number. (The surveyors thought no one would believe a measurement of 29,000 ft,
so they reported it as 29,002.) (Source: Dunham, W., The Mathematical Universe,
John Wiley & Sons, 1994.)
3. Longest Vehicular Tunnel The E. Johnson Memorial Tunnel in Colorado, which
measures 8959 ft, is one of the longest land vehicular tunnels in the United States.
What is the range of this number? (Source: The World Almanac and Book of Facts,
2003.)
9
9
9
4. Top WNBA Scorer Women’s National Basketball Association player Tamika
Catchings of the Indiana Fever received the 2002 award for most points scored, 594.
Is it appropriate to consider this number as between 593.5 and 594.5? Why or why
not? (Source: The World Almanac and Book of Facts, 2003.)
5. Circumference of a Circle The formula for the circumference of a circle is
C а·‡ 2вђІr. Suppose you use the вђІ key on your calculator to find the circumference
of a circle with radius 54.98 cm, getting 345.44953. Since 2 has only one significant
digit, the answer should be given as 3 П« 10 2, or 300 cm. Is this conclusion correct?
If not, explain how the answer should be given.
6. Explain the difference between a measurement of 23.0 ft and a measurement of
23.00 ft.
7. If h is the actual height of a building and the height is measured as 58.6 ft, then
.
Н‰ h ПЄ 58.6 Н‰ Х…
8. If w is the actual weight of a car and the weight is measured as 15.00 П« 10 2 lb, then
.
Н‰ w ПЄ 1500 Н‰ Х…
Solve each right triangle. See Example 1.
9.
10.
B
c
964 m
a
A
36° 20′
C
11.
B
P
Y
p
51.2В°
124 m
89.6 cm
N
C
b
12.
M
n
31° 40′
A
b
35.9 km
X
47.8В°
y
x
Z
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517
5.4 Solving Right Triangles
14. The other acute angle requires
the least work to find.
17. B а·‡ 62.00В°; a а·‡ 8.17 ft;
b а·‡ 15.4 ft 18. A а·‡ 44.00В°;
a а·‡ 20.6 m; b а·‡ 21.4 m
19. A а·‡ 17.00В°; a а·‡ 39.1 in.;
c а·‡ 134 in. 20. B а·‡ 29.00В°;
a а·‡ 70.7 cm; c а·‡ 80.9 cm
21. c а·‡ 85.9 yd; A а·‡ 62В° 50Р€;
B а·‡ 27В° 10Р€ 22. c а·‡ 1080 m;
A а·‡ 63В° 00Р€; B а·‡ 27В° 00Р€
23. The angle of elevation from X
to Y is 90В° whenever Y is directly
above X. 24. no
27. It should be shown as an
angle measured from due north.
28. It should be shown measured
from north (or south) in the east
(or west) direction. 29. 9.35 m
30. 33.4 m
9 13.
Can a right triangle be solved if we are given measures of its two acute angles and
no side lengths? Explain.
14. Concept Check If we are given an acute angle and a side in a right triangle, what
unknown part of the triangle requires the least work to find?
9 15.
9 16.
Explain why you can always solve a right triangle if you know the measures of one
side and one acute angle.
Explain why you can always solve a right triangle if you know the lengths of two
sides.
Solve each right triangle. In each case, C а·‡ 90В°. If angle information is given in degrees
and minutes, give answers in the same way. If given in decimal degrees, do likewise in
answers. When two sides are given, give angles in degrees and minutes. See Examples 1
and 2.
17. A а·‡ 28.00В°, c а·‡ 17.4 ft
18. B а·‡ 46.00В°, c а·‡ 29.7 m
19. B а·‡ 73.00В°, b а·‡ 128 in.
20. A а·‡ 61В° 00Р€, b а·‡ 39.2 cm
21. a а·‡ 76.4 yd, b а·‡ 39.3 yd
22. a а·‡ 958 m, b а·‡ 489 m
23. Concept Check
When is an angle of elevation equal to 90В°?
24. Concept Check
Can an angle of elevation be more than 90В°?
9 25.
A
Explain why the angle of depression DAB has
the same measure as the angle of elevation
ABC in the figure.
D
C
B
AD is parallel to BC.
9 26.
Why is angle CAB not an angle of depression in the figure for Exercise 25?
27. Concept Check When bearing is given as a single angle measure, how is the angle
represented in a sketch?
28. Concept Check When bearing is given as N (or S), the angle measure, then E (or
W), how is the angle represented in a sketch?
Solve each problem involving triangles.
29. Height of a Ladder on a Wall A 13.5-m fire truck
ladder is leaning against a wall. Find the distance d the
ladder goes up the wall (above the top of the fire truck)
if the ladder makes an angle of 43В° 50Р€ with the horizontal.
13.5 m
d
43° 50′
30. Distance Across a Lake To find the
distance RS across a lake, a surveyor
lays off RT а·‡ 53.1 m, with angle
T а·‡ 32В° 10Р€ and angle S а·‡ 57В° 50Р€.
Find length RS.
S
Lake
R
T
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518 CHAPTER 5 Trigonometric Functions
31. 128 ft 32. 1,730,000 mi
33. 26.92 in. 34. 134.7 cm
35. 22В°
31. Height of a Building From a window 30 ft above the street, the angle of elevation
to the top of the building across the street is 50.0В° and the angle of depression to the
base of this building is 20.0В°. Find the height of the building across the street.
50.0В°
20.0В°
30.0 ft
32. Diameter of the Sun To determine the diameter of the sun, an astronomer might
sight with a transit (a device used by surveyors for measuring angles) first to one
edge of the sun and then to the other, finding that the included angle equals 1В° 4Р€.
Assuming that the distance from Earth to the sun is 92,919,800 mi, calculate the
diameter of the sun.
Sun
Earth
Not to
scale
33. Side Lengths of a Triangle The length of the base of an isosceles triangle is
42.36 in. Each base angle is 38.12В°. Find the length of each of the two equal sides
of the triangle. (Hint: Divide the triangle into two right triangles.)
34. Altitude of a Triangle Find the altitude of an isosceles triangle having base
184.2 cm if the angle opposite the base is 68В° 44Р€.
Solve each problem involving an angle of elevation or depression. See Example 3.
35. Angle of Elevation of Pyramid of the Sun The Pyramid of the Sun in the ancient
Mexican city of Teotihuacan was the largest and most important structure in the city.
The base is a square with sides 700 ft long, and the height of the pyramid is 200 ft.
Find the angle of elevation of the edge indicated in the figure to two significant digits. (Hint: The base of the triangle in the figure is half the diagonal of the square base
of the pyramid.)
200 ft
вђЄ
700 ft
700 ft
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5.4 Solving Right Triangles
36. 583 ft 37. 28.0 m
38. 469 m 39. 13.3 ft
40. 42,600 ft 41. 146 m
519
36. Cloud Ceiling The U.S. Weather Bureau defines a cloud ceiling as the altitude of
the lowest clouds that cover more than half the sky. To determine a cloud ceiling, a
powerful searchlight projects a circle of light vertically on the bottom of the cloud.
An observer sights the circle of light in the crosshairs of a tube called a clinometer.
A pendant hanging vertically from the tube and resting on a protractor gives the
angle of elevation. Find the cloud ceiling if the searchlight is located 1000 ft from
the observer and the angle of elevation is 30.0В° as measured with a clinometer at eyeheight 6 ft. (Assume three significant digits.)
Cloud
Searchlight
30.0В°
1000 ft
Observer
6 ft
37. Height of a Tower The shadow of a vertical tower is 40.6 m long when the angle
of elevation of the sun is 34.6В°. Find the height of the tower.
38. Distance from the Ground to the Top of a Building The angle of depression from
the top of a building to a point on the ground is 32В° 30Р€. How far is the point on the
ground from the top of the building if the building is 252 m high?
39. Length of a Shadow Suppose that the angle of elevation of the sun is 23.4В°. Find
the length of the shadow cast by Diane Carr, who is 5.75 ft tall.
23.4В°
5.75 ft
40. Airplane Distance An airplane is flying 10,500 ft above the level ground. The
angle of depression from the plane to the base of a tree is 13В° 50Р€. How far horizontally must the plane fly to be directly over the tree?
10,500 ft
41. Height of a Building The angle of elevation
from the top of a small building to the top of a
nearby taller building is 46В° 40Р€, while the angle
of depression to the bottom is 14В° 10Р€. If the
smaller building is 28.0 m high, find the height
of the taller building.
x
46° 40′
28.0 m
14° 10′
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520 CHAPTER 5 Trigonometric Functions
42.
(b)
45.
47.
37В° 35Р€ 43. (a) 29,008 ft
shorter 44. 34.0 mi
220 mi 46. 150 km
47 nautical mi
42. Angle of Depression of a Light A company safety committee has recommended
that a floodlight be mounted in a parking
lot so as to illuminate the employee exit.
Find the angle of depression of the light.
39.82 ft
Employee
exit
51.74 ft
43. Height of Mt. Everest The highest
i
4m
mountain peak in the world is Mt.
13
0
.
27
Everest, located in the Himalayas. The
вђЄ
height of this enormous mountain was
determined in 1856 by surveyors using
14,545 ft
trigonometry long before it was first
climbed in 1953. This difficult measurement had to be done from a great distance.
At an altitude of 14,545 ft on a different mountain, the straight line distance to the
peak of Mt. Everest is 27.0134 mi and its angle of elevation is вђЄ а·‡ 5.82В°. (Source:
Dunham, W., The Mathematical Universe, John Wiley & Sons, 1994.)
(a) Approximate the height (in feet) of Mt. Everest.
(b) In the actual measurement, Mt. Everest was over 100 mi away and the curvature
of Earth had to be taken into account. Would the curvature of Earth make the
peak appear taller or shorter than it actually is?
44. Error in Measurement A degree may seem like a very small unit, but an error of
one degree in measuring an angle may be very significant. For example, suppose a
laser beam directed toward the visible center of the moon misses its assigned target
by 30 sec. How far is it (in miles) from its assigned target? Take the distance from
the surface of Earth to that of the moon to be 234,000 mi. (Source: A Sourcebook of
Applications of School Mathematics by Donald Bushaw et al. Copyright В© 1980 by
The Mathematical Association of America.)
Work each problem. Assume the course of a plane or ship is on the indicated bearing.
See Examples 4 and 5.
45. Distance Flown by a Plane A plane flies 1.3 hr at 110 mph on a bearing of 40В°. It
then turns and flies 1.5 hr at the same speed on a bearing of 130В°. How far is the
plane from its starting point?
N
130В°
N
40В°
x
46. Distance Traveled by a Ship A ship travels 50 km on a bearing of 27В°, then travels on a bearing of 117В° for 140 km. Find the distance traveled from the starting
point to the ending point.
47. Distance Between Two Ships Two ships leave a port at the same time. The first
ship sails on a bearing of 40В° at 18 knots (nautical miles per hour) and the second at
a bearing of 130В° at 26 knots. How far apart are they after 1.5 hr?
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5.4 Solving Right Triangles
48.
50.
52.
54.
120 mi 49. 130 mi
148 mi 51. 2.01 mi
114 ft 53. 147 m
5.18 m
521
48. Distance Between Two Cities The bearing from Winston-Salem, North Carolina,
to Danville, Virginia, is N 42В° E. The bearing from Danville to Goldsboro, North
Carolina, is S 48В° E. A car driven by Mark Ferrari, traveling at 60 mph, takes 1 hr
to go from Winston-Salem to Danville and 1.8 hr to go from Danville to Goldsboro.
Find the distance from Winston-Salem to Goldsboro.
49. Distance Between Two Cities The bearing from Atlanta to Macon is S 27В° E, and
the bearing from Macon to Augusta is N 63В° E. An automobile traveling at 60 mph
needs 1 14 hr to go from Atlanta to Macon and 1 34 hr to go from Macon to Augusta.
Find the distance from Atlanta to Augusta.
50. Distance Between Two Ships A cruise ship leaves port and sails on a bearing of
N 28В° 10Р€ E. Another ship leaves the same port at the same time and sails on a bearing of S 61В° 50Р€ E. If the first ship sails at 24.0 mph and the second sails at 28.0 mph,
find the distance between the two ships after 4 hr.
51. Distance Between Transmitters Radio direction
finders are set up at two points A and B, which
are 2.50 mi apart on an east-west line. From A,
it is found that the bearing of a signal from a
radio transmitter is N 36В° 20Р€ E, while from B
the bearing of the same signal is N 53В° 40Р€ W.
Find the distance of the transmitter from B.
N
Transmitter
N
36° 20′
53° 40′
2.50 mi
A
B
In Exercises 52 –55, use the method of Example 7. Drawing a sketch for the problems
where one is not given may be helpful.
52. Height of a Pyramid The angle of elevation from a point on the ground to the top
of a pyramid is 35В° 30Р€. The angle of elevation from a point 135 ft farther back to
the top of the pyramid is 21В° 10Р€. Find the height of the pyramid.
x
35° 30′
21° 10′
135 ft
53. Distance Between a Whale and a Lighthouse Debbie Glockner-Ferrari, a whale
researcher, is watching a whale approach directly toward her as she observes from
the top of a lighthouse. When she first begins watching the whale, the angle of
depression of the whale is 15В° 50Р€. Just as the whale turns away from the lighthouse,
the angle of depression is 35В° 40Р€. If the height of the lighthouse is 68.7 m, find the
distance traveled by the whale as it approaches the lighthouse.
15° 50′
35° 40′
68.7 m
x
54. Height of an Antenna A scanner antenna is on top of the center of a house. The
angle of elevation from a point 28.0 m from the center of the house to the top of the
antenna is 27В° 10Р€, and the angle of elevation to the bottom of the antenna is 18В° 10Р€.
Find the height of the antenna.
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522 CHAPTER 5 Trigonometric Functions
55. 2.47 km 56. 1.95 mi
57. 10.8 ft 58. A а·‡ 35.987В° or
35° 59Ј 10ЈЈ; B ෇ 54.013° or
54° 00Ј 50ЈЈ 59. (a) 323 ft
вђЄ
(b) R 1 ПЄ cos
2
Н©
НЄ
55. Height of Mt. Whitney The angle of elevation from Lone Pine to the top of
Mt. Whitney is 10В° 50Р€. Van Dong Le, traveling 7.00 km from Lone Pine along a
straight, level road toward Mt. Whitney, finds the angle of elevation to be 22В° 40Р€.
Find the height of the top of Mt. Whitney above the level of the road.
Solve each problem.
56. Height of a Plane Above Earth Find the
minimum height h above the surface of Earth
so that a pilot at point A in the figure can
see an object on the horizon at C, 125 mi
away. Assume that the radius of Earth is
4.00 П« 10 3 mi.
C
A
125 mi
пЈј
пЈґ
пЈЅ
пЈґ
h
пЈѕ
B
Not to
scale
57. Distance of a Plant from a Fence In one area, the lowest angle of elevation of the
sun in winter is 23В° 20Р€. Find the minimum distance x that a plant needing full sun
can be placed from a fence 4.65 ft high.
4.65 ft
23° 20′
x
Plant
58. Distance Through a Tunnel A tunnel is to be dug from A to B. Both A and B are
visible from C. If AC is 1.4923 mi and BC is 1.0837 mi, and if C is 90В°, find the measures of angles A and B.
Tunnel
A
B
1.4923 mi
1.0837 mi
C
59. (Modeling) Highway Curves A basic highway curve
connecting two straight sections of road is often circular. In the figure, the points P and S mark the beginning
and end of the curve. Let Q be the point of intersection
where the two straight sections of highway leading into
the curve would meet if extended. The radius of the
curve is R, and the central angle вђЄ denotes how many
degrees the curve turns. (Source: Mannering, F. and
W. Kilareski, Principles of Highway Engineering and
Traffic Analysis, 2nd Edition, John Wiley & Sons, 1998.)
P
Q
d
S
N
R
M
вђЄ
вђЄ 2
2
C
R
(a) If R а·‡ 965 ft and вђЄ а·‡ 37В°, find the distance d
between P and Q.
(b) Find an expression in terms of R and вђЄ for the distance between points M and N.
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5.4 Solving Right Triangles
60. 84.7 m 61. (a) 23.4 ft
(b) 48.3 ft (c) The faster the
speed, the more land needs to be
cleared inside the curve.
60. Length of a Side of a Piece of Land
figure. Find x.
523
A piece of land has the shape shown in the
198.4 m
x
30° 50′
52° 20′
61. (Modeling) Stopping Distance on a Curve Refer to Exercise 59. When an automobile travels along a circular curve, objects like trees and buildings situated on the
inside of the curve can obstruct a driver’s vision. These obstructions prevent the driver from seeing sufficiently far down the highway to ensure a safe stopping distance.
In the figure, the minimum distance d that should be cleared on the inside of the highway is modeled by the equation
Н©
d а·‡ R 1 ПЄ cos
НЄ
вђЄ
.
2
(Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and
Traffic Analysis, 2nd Edition, John Wiley & Sons, 1998.)
d
R
вђЄ
R
Not to scale
(a) It can be shown that if вђЄ is measured in degrees, then вђЄ П· 57.3S
R , where S is safe
stopping distance for the given speed limit. Compute d for a 55 mph speed limit
if S а·‡ 336 ft and R а·‡ 600 ft.
(b) Compute d for a 65 mph speed limit if S а·‡ 485 ft and R а·‡ 600 ft.
(c) How does the speed limit affect the amount of land that should be cleared on the
inside of the curve?
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524 CHAPTER 5 Trigonometric Functions
Chapter 5 Summary
KEY TERMS
5.1 line
degree
acute angle
right angle
obtuse angle
straight angle
complementary angles
supplementary angles
minute
second
line segment (or
segment)
ray
angle
initial side
terminal side
vertex of an angle
positive angle
negative angle
angle in standard
position
quadrantal angle
coterminal angles
5.2 sine
cosine
tangent
cotangent
secant
cosecant
5.3 side opposite
side adjacent
reference angle
5.4 significant digit
exact number
angle of elevation
angle of depression
bearing
NEW SYMBOLS
вђЄ Greek letter theta
Ш‡ degree
Ш…
Ш†
minute
second
QUICK REVIEW
CONCEPTS
EXAMPLES
5.1 Angles
Types of Angles
If вђЄ а·‡ 32В°, then angle вђЄ is an acute angle.
If вђЄ а·‡ 90В°, then angle вђЄ is a right angle.
If вђЄ а·‡ 148В°, then angle вђЄ is an obtuse angle.
вђЄ
вђЄ
If вђЄ а·‡ 180В°, then angle вђЄ is a straight angle.
Acute angle
0В° < вђЄ < 90В°
Right angle
вђЄ = 90В°
вђЄ
вђЄ
Straight angle
вђЄ = 180В°
Obtuse angle
90В° < вђЄ < 180В°
5.2 Trigonometric Functions
Definitions of the Trigonometric Functions
Let Н‘x, yН’ be a point other than the origin on the terminal side
of an angle вђЄ in standard position. Let r а·‡ Н™x 2 П© y 2, the
distance from the origin to Н‘x, yН’. Then
If a point Н‘ПЄ2, 3Н’ is on the terminal side of angle вђЄ in
standard position, then x а·‡ ПЄ2, y а·‡ 3, r а·‡ Н™4 П© 9 а·‡
Н™13, and
sin ␪ ‫؍‬
y
r
y
͑x ‫؍‬
Нћ 0Н’
x
sin вђЄ а·‡
3Н™13
,
13
cos вђЄ а·‡ ПЄ
2Н™13
,
13
tan вђЄ а·‡ ПЄ
3
2
csc ␪ ‫؍‬
r
r
x
Нћ 0Н’.
͑y ‫؍‬
͞ 0͒ sec ␪ ‫͑ ؍‬x ‫؍‬
͞ 0͒ cot ␪ ‫ ͑ ؍‬y ‫؍‬
y
x
y
csc вђЄ а·‡
Н™13
,
3
sec вђЄ а·‡ ПЄ
Н™13
,
2
cot вђЄ а·‡ ПЄ
2
.
3
cos ␪ ‫؍‬
x
r
tan ␪ ‫؍‬
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CHAPTER 5
CONCEPTS
Summary 525
EXAMPLES
Reciprocal Identities
sin ␪ ‫؍‬
1
csc вђЄ
cos ␪ ‫؍‬
1
sec вђЄ
tan ␪ ‫؍‬
1
cot вђЄ
csc ␪ ‫؍‬
1
sin вђЄ
sec ␪ ‫؍‬
1
cos вђЄ
cot ␪ ‫؍‬
1
tan вђЄ
The preceding examples satisfy these identities.
Pythagorean Identities
sin2 ␪ ؉ cos2 ␪ ‫ ؍‬1
Using the preceding trigonometric values,
tan2 ␪ ؉ 1 ‫ ؍‬sec2 ␪
Н© НЄ Н© НЄ
Н© НЄ
Н© НЄ
3Н™13
13
1 ؉ cot2 ␪ ‫ ؍‬csc2 ␪
2
П© ПЄ
ПЄ
2Н™13
13
3
2
2
а·‡
2
П©1а·‡
1П© ПЄ
2
3
9
4
П©
а·‡ 1,
13 13
Н© НЄ
Н© НЄ
9
4
13
Н™13 2
,
П© а·‡
а·‡ ПЄ
4
4
4
2
2
а·‡1П©
4
13
Н™13 2
.
а·‡
а·‡
9
9
3
Quotient Identities
sin вђЄ
‫ ؍‬tan ␪
cos вђЄ
From the preceding trigonometric values,
cos вђЄ
‫ ؍‬cot ␪
sin вђЄ
Н©
НЄ
3Н™13
sin вђЄ
3Н™13
13
3
а·‡ 13 а·‡
ПЄ
а·‡ ПЄ а·‡ tan вђЄ.
cos вђЄ ПЄ 2Н™13
13
2
2Н™13
13
5.3 Evaluating Trigonometric Functions
Right-Triangle-Based Definitions of the Trigonometric
Functions
For any acute angle A in standard position,
sin A ‫؍‬
y
side opposite
hypotenuse
r
‫؍‬
csc A ‫؍ ؍‬
r
hypotenuse
y
side opposite
cos A ‫؍‬
x
side adjacent
hypotenuse
r
‫؍‬
sec A ‫؍ ؍‬
r
hypotenuse
x
side adjacent
tan A ‫؍‬
y
side opposite
x
side adjacent
.
‫؍‬
cot A ‫؍ ؍‬
x
side adjacent
y
side opposite
Side
opposite 7
A B
C
Side adjacent to A
24
A
25
Hypotenuse
sin A а·‡
7
25
cos A а·‡
24
25
tan A а·‡
7
24
csc A а·‡
25
7
sec A а·‡
25
24
cot A а·‡
24
7
Function Values of Special Angles
вђЄ
30В°
sin вђЄ
1
2
cos вђЄ
Н™3
2
tan вђЄ
Н™3
3
cot вђЄ
Н™3
sec вђЄ
2Н™3
3
csc вђЄ
60В°
1
2
Н™2
2
Н™2
2
1
1
Н™2
Н™2
60В°
Н™3
2
1
2
Н™3
Н™3
3
2
2Н™3
3
1
в€љ2
30В°
в€љ3
45В°
45В°
2
45В°
1
(continued)
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526 CHAPTER 5 Trigonometric Functions
CONCEPTS
EXAMPLES
Reference Angle вђЄ Р€ for вђЄ in (0В°, 360В°Н’
вђЄ in Quadrant
I
II
III
IV
вђЄ Р€ is
вђЄ
180В° ПЄ вђЄ
вђЄ ПЄ 180В°
360В° ПЄ вђЄ
Finding Trigonometric Function Values for Any Angle
Step 1
Step 2
Step 3
Step 4
Add or subtract 360В° as many times as needed to get
an angle greater than 0В° but less than 360В°.
Find the reference angle вђЄР€.
Find the trigonometric function values for вђЄР€.
Determine the correct signs for the values found in
Step 3.
Quadrant I: For вђЄ а·‡ 25В°, вђЄР€ а·‡ 25В°
Quadrant II: For вђЄ а·‡ 152В°, вђЄР€ а·‡ 28В°
Quadrant III: For вђЄ а·‡ 200В°, вђЄР€ а·‡ 20В°
Quadrant IV: For вђЄ а·‡ 320В°, вђЄР€ а·‡ 40В°
Find sin 1050В°.
1050В° ПЄ 2Н‘360В°Н’ а·‡ 330В°
Thus, вђЄР€ а·‡ 30В°.
sin 1050В° а·‡ ПЄsin 30В° а·‡ ПЄ
1
2
5.4 Solving Right Triangles
Solving an Applied Trigonometry Problem
Find the angle of elevation of the sun if a 48.6-ft flagpole
casts a shadow 63.1 ft long.
Step 1
Step 1
Draw a sketch, and label it with the given information. Label the quantity to be found with a variable.
See the sketch. We must find вђЄ.
Sun
48.6 Flagpole
ft
вђЄ
Shadow
63.1 ft
Step 2
Step 3
Use the sketch to write an equation relating the
given quantities to the variable.
Solve the equation, and check that your answer
makes sense.
Figures 44 and 46 on pages 512 and 513 illustrate the two
methods for expressing bearing.
Step 2
Step 3
tan вђЄ а·‡
48.6
П· .770206
63.1
вђЄ а·‡ tanПЄ1 .770206 П· 37.6В°
The angle of elevation is 37.6В°.
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CHAPTER 5
Review Exercises 527
Chapter 5 Review Exercises
1. complement: 55В°; supplement:
145° 2. 35° 15Ј 0ЈЈ
3. 59.5916В° 4. (a) 186В°
(b) 200В° 5. 270В° П© n Рё 360В°
In Exercises 6–11, we give, in
order, sine, cosine, tangent,
cotangent, secant, and cosecant.
Н™2
Н™2
6. ПЄ
;ПЄ
; 1; 1; ПЄН™2 ;
2
2
Н™3 1
ПЄН™2 7. ПЄ
; ; ПЄН™3 ;
2 2
2Н™3
Н™3
8. 0; ПЄ1;
ПЄ
; 2; ПЄ
3
3
0; undefined; ПЄ1; undefined
5
4 3
4
3 5
9. ПЄ ; ; ПЄ ; ПЄ ; ; ПЄ
5 5
3
4 3
4
9
2Н™85 9Н™85
2
10. ПЄ
;
;ПЄ ;ПЄ ;
85
85
9
2
Н™85 Н™85
Н™2 Н™2
11.
;ПЄ
;ПЄ
;
9
2
2
2
ПЄ1; ПЄ1; ПЄН™2 ; Н™2 12. tangent
y
and secant 13.
(–1, 5)
5
–1 0
y = –5x, x ≤ 0
вђЄ
x
5Н™26
14. sin вђЄ а·‡
;
26
Н™26
cos вђЄ а·‡ ПЄ
26
15. (a) impossible (b) possible
(c) possible
In Exercises 16 and 17, we give,
in order, sine, cosine, tangent,
cotangent, secant, and cosecant.
Н™66
Н™3 Н™22
16.
;ПЄ
;ПЄ
;
5
5
22
5Н™22 5Н™3
Н™66
ПЄ
;ПЄ
;
3
22
3
5 Н™39 5Н™39
Н™39
17. ПЄ
;ПЄ ;
;
;
8
8
5
39
8Н™39
8
18. IV; negative
ПЄ ;ПЄ
5
39
In Exercises 20, 21, and 25 – 27,
we give, in order, sine, cosine,
tangent, cotangent, secant, and
cosecant.
60 11 60 11 61 61
20.
; ; ; ; ;
61 61 11 60 11 60
8 15 8 15 17 17
21.
, , , , ,
17 17 15 8 15 8
1. Give the complement and the supplement of an angle of 35В°.
2. Convert 32.25В° to degrees, minutes, and seconds.
3. Convert 59° 35Ј 30ЈЈ to decimal degrees.
4. Find the angle of smallest possible positive measure coterminal with each angle.
(a) ПЄ174В°
(b) 560В°
5. Let n represent any integer, and write an expression for all angles coterminal with an
angle of 270В°.
Find the six trigonometric function values for each angle. If a value is undefined, say so.
y
6.
y
7.
8. 180В°
Оё
Оё
0
x
0
x
(1, –√3)
(–3, –3)
Find the values of the six trigonometric functions for an angle in standard position having each point on its terminal side.
9. Н‘3, ПЄ4Н’
11. Н‘ ПЄ2Н™2, 2Н™2 Н’
10. Н‘9, ПЄ2Н’
12. Concept Check If the terminal side of a quadrantal angle lies along the y-axis,
which of its trigonometric functions are undefined?
In Exercises 13 and 14, consider an angle вђЄ in standard position whose terminal side has
the equation y а·‡ ПЄ5x, with x Х… 0.
13. Sketch вђЄ and use an arrow to show the rotation if 0В° Х… вђЄ ПЅ 360В°.
14. Find the exact values of sin вђЄ and cos вђЄ.
15. Decide whether each statement is possible or impossible.
(a) sec вђЄ а·‡ ПЄ
2
3
(b) tan вђЄ а·‡ 1.4
(c) csc вђЄ а·‡ 5
Find all six trigonometric function values for each angle. Rationalize denominators
when applicable.
16. sin вђЄ а·‡
Н™3
and cos вђЄ ПЅ 0
5
17. cos вђЄ а·‡ ПЄ
5
, with вђЄ in quadrant III
8
18. Concept Check If, for some particular angle вђЄ, sin вђЄ ПЅ 0 and cos вђЄ Пѕ 0, in what
quadrant must вђЄ lie? What is the sign of tan вђЄ ?
9 19. Explain how you would find the cotangent of an angle вђЄ whose tangent is 1.6778490
using a calculator. Then find cot вђЄ.
Find the values of the six trigonometric functions for each angle A.
20.
11
A
60
61
21.
8
17
15
A
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528 CHAPTER 5 Trigonometric Functions
1 Н™3 Н™3
,
,
,
2 2
3
2Н™3
Н™2 Н™2
, 2; Row 2:
,
,
Н™3 ,
3
2
2
Н™3 1
1, 1, Н™2, Н™2; Row 3:
, ,
2 2
2Н™3
Н™3
, 2,
24. (a) 80В°
Н™3,
3
3
Н™3 1
; ; ПЄН™3 ;
(b) 5В° 25. ПЄ
2 2
2Н™3
Н™3
ПЄ
; 2; ПЄ
3
3
Н™2 Н™2
;ПЄ
; ПЄ1; ПЄ1;
26.
2
2
1 Н™3
ПЄН™2 ; Н™2 27. ПЄ ;
;
2 2
2Н™3
Н™3
; ПЄН™3 ;
; ПЄ2
ПЄ
3
3
28. .95371695 29. ПЄ1.3563417
30. 5.6712818 31. .20834446
32. 150В°; 330В° 33. D
34. 55.673870В° 35. 41.635092В°
36. 42.971835В°, 137.02817В°
38. ПЄ.29237170; ПЄ.9530476; III
39. B а·‡ 31В° 30Р€; a а·‡ 638;
b а·‡ 391 40. B а·‡ 50.28В°;
a а·‡ 32.38 m; c а·‡ 50.66 m
23. Row 1:
9 22. Explain why, in the figure, the cosine of angle A is
A
equal to the sine of angle B.
c
b
a
C
B
23. Complete the following table of exact function values.
вђЄ
sin вђЄ
cos вђЄ
tan вђЄ
cot вђЄ
sec вђЄ
csc вђЄ
30В°
45В°
60В°
24. Give the reference angle for each angle measure.
(b) ПЄ365В°
(a) 100В°
Find exact values of the six trigonometric functions for each angle. Do not use a calculator. Rationalize denominators when applicable.
26. ПЄ225В°
25. 300В°
27. ПЄ390В°
Use a calculator to find each value.
28. sin 72В° 30Р€
29. sec 222В° 30Р€
30. tanН‘ПЄ100В°Н’
31. tan 11.7689В°
32. Find all values of вђЄ, if вђЄ is in the interval Н“0В°, 360В°Н’ and tan вђЄ а·‡ ПЄ Н™3
3 .
33. Concept Check Which one of the following cannot be exactly determined using
the methods of this chapter?
A. cos 135В°
B. cotН‘ПЄ45В°Н’
C. sin 300В°
D. tan 140В°
Use a calculator to find each value of вђЄ, where вђЄ is in the interval Н“0В°, 90В°Н’. Give answers
in decimal degrees.
34. sin вђЄ а·‡ .82584121
35. cot вђЄ а·‡ 1.1249386
36. Find two angles in the interval Н“0, 360В°Н’ that satisfy sin вђЄ а·‡ .68163876.
to use a calculator to find the value of cot 25В°. However, instead of
9 37. A student wants
1
entering tan 25, he enters tanПЄ1 25. Assuming the calculator is in degree mode, will
this produce the correct answer? Explain.
38. For вђЄ а·‡ 1997В°, use a calculator to find cos вђЄ and sin вђЄ. Use your results to decide in
which quadrant the angle lies.
Solve each right triangle.
40. A а·‡ 39.72В°, b а·‡ 38.97 m
39. B
a
C
c = 748
58° 30′
b
A
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CHAPTER 5
41.
43.
45.
47.
(b)
73.7 ft 42. 34 ft
20.4 m 44. 433 ft
19,600 ft 46. 1200 m
110 km 48. (a) 716 mi
1104 mi
Review Exercises 529
Solve each problem.
41. Height of a Tower The angle of elevation from a point
93.2 ft from the base of a tower to the top of the tower is
38В° 20Р€. Find the height of the tower.
38° 20′
93.2 ft
42. Length of a Shadow The length of a building’s shadow is 48 ft when the angle of
elevation of the sun is 35.3В°. Find the height of the building.
43. Height of a Tower The angle of depression of a television tower to a point on the ground 36.0 m from the
bottom of the tower is 29.5В°. Find the height of the
tower.
29.5В°
36.0 m
44. Find h as indicated in the figure.
h
29.5В°
392 ft
49.2В°
45. Height of Mt. Kilimanjaro From a point A the angle of elevation of Mt.
Kilimanjaro in Africa is 13.7В°, and from a point B directly behind A, the angle of
elevation is 10.4В°. If the distance between A and B is 5 mi, approximate the height
of Mt. Kilimanjaro to the nearest hundred feet.
10.4В°
B
5 mi
13.7В°
A
46. Distance Between Two Points The bearing of B from C is 254В°. The bearing of A
from C is 344В°. The bearing of A from B is 32В°. The distance from A to C is 780 m.
Find the distance from A to B.
47. Distance a Ship Sails The bearing from point A to point B is S 55В° E and from
point B to point C is N 35В° E. If a ship sails from A to B, a distance of 80 km, and
then from B to C, a distance of 74 km, how far is it from A to C?
48. (Modeling) Height of a Satellite Artificial satellites that orbit Earth often use VHF
signals to communicate with the ground. VHF signals travel in straight lines. The
height h of the satellite above Earth and the time T that the satellite can communicate with a fixed location on the ground are related by the model
hа·‡R
Н©
НЄ
1
ПЄ1 ,
cos 180T
P
where R а·‡ 3955 mi is the radius of Earth and P is the period for the satellite to orbit
Earth. (Source: Schlosser, W., T. Schmidt-Kaler, and E. Milone, Challenges of
Astronomy, Springer-Verlag, 1991.)
(a) Find h when T а·‡ 25 min and P а·‡ 140 min. (Evaluate the cosine function in
degree mode.)
(b) What is the value of h if T is increased to 30?
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530 CHAPTER 5 Trigonometric Functions
Chapter 5 Test
1. 203В° 2. (a) П© (b) ПЄ
5Н™29
;
3. III 4. sin вђЄ а·‡ ПЄ
29
2Н™29
5
cos вђЄ а·‡
; tan вђЄ а·‡ ПЄ
29
2
3
5. 120В° 6. sin вђЄ а·‡ ПЄ ;
5
3
4
tan вђЄ а·‡ ПЄ ; cot вђЄ а·‡ ПЄ ;
4
3
5
5
sec вђЄ а·‡ ; csc вђЄ а·‡ ПЄ
4
3
7. x а·‡ 4; y а·‡ 4Н™3 ; z а·‡ 4Н™2 ;
w а·‡ 8 8. ПЄН™3 9. (a) false
(b) false 10. (a) .97939940
(b) ПЄ1.9056082 (c) 1.9362132
11. 221.8В°; 318.2В°
12. B а·‡ 31В° 30Р€; a а·‡ 638;
b а·‡ 391 13. 15.5 ft
14. 92 km 15. 448 m
1. Find the angle of smallest positive measure coterminal with ПЄ157В°.
2. Suppose вђЄ is in the interval Н‘90В°, 180В°Н’. Find the sign of each value.
вђЄ
(a) cos
(b) cotН‘вђЄ П© 180В°Н’
2
3. If cos вђЄ ПЅ 0 and cot вђЄ Пѕ 0, in what quadrant does вђЄ lie?
4. If Н‘2, ПЄ5Н’ is on the terminal side of an angle вђЄ in standard position, find sin вђЄ, cos вђЄ,
and tan вђЄ.
5. What angle does the line y а·‡ ПЄН™3 x make with the positive x-axis?
6. If cos вђЄ а·‡ 45 and вђЄ is in quadrant IV, find the values of the other trigonometric functions of вђЄ.
7. Find the exact values of each part of the triangle labeled with a letter.
z
w
4
45В°
x
30В°
y
8. Find the exact value of cotН‘ПЄ750В°Н’.
9. Decide whether each statement is true or false.
(a) cos 40В° а·‡ 2 cos 20В°
(b) sin 10В° П© sin 10В° а·‡ sin 20В°
10. Use a calculator to approximate each value.
(a) sin 78В° 21Р€
(b) tan 117.689В°
(c) sec 58.9041В°
11. Find the two values of вђЄ to the nearest tenth in Н“0В°, 360В°Н’, if sin вђЄ а·‡ ПЄ 23 .
12. Solve the triangle.
A
c = 748
B
a
58° 30′ b
C
13. Height of a Flagpole To measure the height of a flagpole, Amado Carillo found
that the angle of elevation from a point 24.7 ft from the base to the top is 32В° 10Р€.
What is the height of the flagpole?
14. Distance of a Ship from a Pier A ship leaves a pier on a bearing of S 62В° E and
travels for 75 km. It then turns and continues on a bearing of N 28В° E for 53 km.
How far is the ship from the pier?
15. Find h as indicated in the figure.
h
41.2В° 52.5В°
168 m
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CHAPTER 5
Quantitative Reasoning
531
Chapter 5 Quantitative Reasoning
Can trigonometry be used to win an Olympic medal?
A shot-putter trying to improve performance may wonder: Is there an optimal
angle to aim for, or is the velocity (speed) at which the ball is thrown more
important? The figure shows the path of a steel ball thrown by a shot-putter. The
distance D depends on initial velocity v, height h, and angle вђЄ.
вђЄ
h
D
One model developed for this situation gives D as
Dа·‡
v 2 sin вђЄ cos вђЄ П© v cos вђЄ Н™Н‘v sin вђЄН’2 П© 64h
.
32
Typical ranges for the variables are v: 33 – 46 ft per sec; h: 6 – 8 ft; and ␪ :
40°–45°. (Source: Kreighbaum, E. and K. Barthels, Biomechanics, Allyn &
Bacon, 1996.)
1. To see how angle вђЄ affects distance D, let v а·‡ 44 ft per sec and h а·‡ 7 ft.
Calculate D for вђЄ а·‡ 40В°, 42В°, and 45В°. How does distance D change as вђЄ
increases?
2. To see how velocity v affects distance D, let h а·‡ 7 and вђЄ а·‡ 42В°. Calculate D
for v а·‡ 43, 44, and 45 ft per sec. How does distance D change as v increases?
3. Which affects distance D more, v or вђЄ ? What should the shot-putter do to
improve performance?
1. 67.00 ft; 67.14 ft; 66.84 ft; D increases and then decreases. 2. 64.40 ft; 67.14 ft; 69.93 ft; D increases. 3. v; The shot-putter
should concentrate on achieving as large a value of v as possible.