Download Beecher - Algebra and Trigonometry 3e HQ

Summer 2012
L.E.A.D. Ambassador Team 3
(Advanced Algebra and Trigonometry Course Content)
“Algebra and Trigonometry” (Third Edition)
Beecher, Penna, Bittinger
Addision Wesley (February 2, 2007)
BBEPMC05_0321279115.QXP
432
Chapter 5
1/10/05
1:00 PM
Page 432
• The Trigonometric Functions
2.1
5.1
Trigonometric
Polynomial
Functions
Functionsand
of
Acute
Modeling
Angles
Determine the six trigonometric ratios for a given acute angle of a
right triangle.
Determine the trigonometric function values of 30, 45, and 60.
Using a calculator, find function values for any acute angle, and given a
function value of an acute angle, find the angle.
Given the function values of an acute angle, find the function values of
its complement.
The Trigonometric Ratios
We begin our study of trigonometry by considering right triangles and
acute angles measured in degrees. An acute angle is an angle with measure greater than 0 and less than 90. Greek letters such as (alpha),
(beta), (gamma), (theta), and (phi) are often used to denote an
angle. Consider a right triangle with one of its acute angles labeled .
The side opposite the right angle is called the hypotenuse. The other
sides of the triangle are referenced by their position relative to the acute
angle . One side is opposite and one is adjacent to .
Hypotenuse
Side opposite u
u
Side adjacent to u
The lengths of the sides of the triangle are used to define the six
trigonometric ratios:
sine (sin),
cosine (cos),
tangent (tan),
Hypotenuse
Opposite u
u
Adjacent to u
Hypotenuse
u
Adjacent to u
Figure 2
The sine of is the length of the side opposite divided by the length
of the hypotenuse (see Fig. 1):
sin Figure 1
Opposite u
cosecant (csc),
secant (sec),
cotangent (cot).
length of side opposite .
length of hypotenuse
The ratio depends on the measure of angle and thus is a function of .
The notation sin actually means sin , where sin, or sine, is the name
of the function.
The cosine of is the length of the side adjacent to divided by the
length of the hypotenuse (see Fig. 2):
cos length of side adjacent to .
length of hypotenuse
The six trigonometric ratios, or trigonometric functions, are defined as
follows.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 433
Section 5.1
•
Trigonometric Functions of Acute Angles
433
Trigonometric Function Values of an Acute Angle Let be an acute angle of a right triangle. Then the six trigonometric
functions of are as follows:
side opposite ,
hypotenuse
side adjacent to cos ,
hypotenuse
side opposite tan ,
side adjacent to sin Hypotenuse
Opposite u
u
Adjacent to u
hypotenuse
,
side opposite hypotenuse
sec ,
side adjacent to side adjacent to cot .
side opposite csc EXAMPLE 1 In the right triangle shown at left, find the six trigonometric function values of (a) and (b) .
a
13
12
u
5
Solution
We use the definitions.
opp 12
hyp 13
csc ,
,
a) sin hyp 13
opp 12
adj
hyp 13
5
sec cos ,
,
hyp 13
adj
5
opp 12
adj
5
cot tan ,
adj
5
opp 12
opp
5
,
hyp 13
adj
12
cos ,
hyp 13
opp
5
tan ,
adj
12
b) sin hyp 13
,
opp
5
hyp 13
sec ,
adj
12
adj
12
cot opp
5
The references to
opposite, adjacent, and
hypotenuse are relative
to .
csc The references to
opposite, adjacent, and
hypotenuse are relative
to .
12
In Example 1(a), we note that the value of sin , 13 , is the reciprocal
of
the value of csc . Likewise, we see the same reciprocal relationship
between the values of cos and sec and between the values of tan and
cot . For any angle, the cosecant, secant, and cotangent values are the
reciprocals of the sine, cosine, and tangent function values, respectively.
13
12 ,
Reciprocal Functions
1
1
csc sec ,
,
cos sin cot 1
tan Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
434
Chapter 5
1/10/05
1:00 PM
Page 434
• The Trigonometric Functions
If we know the values of the sine, cosine, and tangent functions of an
angle, we can use these reciprocal relationships to find the values of the
cosecant, secant, and cotangent functions of that angle.
Study Tip
Success on the next exam can
be planned. Include study time
(even if only 30 minutes a day) in
your daily schedule and commit to
making this time a priority. Choose
a time when you are most alert
and a setting in which you can
concentrate. You will be surprised
how much more you can learn and
retain if study time is included
each day rather than in one long
session before the exam.
4
3
4
EXAMPLE 2 Given that sin 5 , cos 5 , and tan 3 , find csc ,
sec , and cot .
Solution
Using the reciprocal relationships, we have
csc 1
1
5
,
sin 4
4
5
cot 1
1
3
.
tan 4
4
3
sec 1
1
5
,
cos 3
3
5
and
Triangles are said to be similar if their corresponding angles have
the same measure. In similar triangles, the lengths of corresponding sides
are in the same ratio. The right triangles shown below are similar. Note
that the corresponding angles are equal and the length of each side of the
second triangle is four times the length of the corresponding side of the
first triangle.
a
20
12
3
a
5
b
4
b
16
Let’s observe the sine, cosine, and tangent values of in each triangle.
Can we expect corresponding function values to be the same?
FIRST TRIANGLE
SECOND TRIANGLE
3
5
4
cos 5
3
tan 4
sin sin 12
20
16
cos 20
12
tan 16
3
5
4
5
3
4
For the two triangles, the corresponding values of sin , cos , and
tan are the same. The lengths of the sides are proportional — thus the
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 435
Section 5.1
•
435
Trigonometric Functions of Acute Angles
ratios are the same. This must be the case because in order for the sine,
cosine, and tangent to be functions, there must be only one output (the
ratio) for each input (the angle ).
The trigonometric function values of depend only on the measure
of the angle, not on the size of the triangle.
The Six Functions Related
We can find the other five trigonometric function values of an acute angle
when one of the function-value ratios is known.
EXAMPLE 3 If sin 67 and is an acute angle, find the other five
trigonometric function values of .
Solution
6
7
pythagorean theorem
review section 1.1.
We know from the definition of the sine function that the ratio
opp
.
hyp
is
Using this information, let’s consider a right triangle in which the
hypotenuse has length 7 and the side opposite has length 6. To find the
length of the side adjacent to , we recall the Pythagorean theorem:
7
6
b
a2 b 2 c 2
a 2 62 72
a2 36 49
a2 49 36 13
a 13.
a
We now use the lengths of the three sides to find the other five ratios:
6
,
7
13
cos ,
7
6
tan ,
13
sin 7
b
13
6
csc sec or
613
,
13
7
,
6
7
,
13
13
cot .
6
or
713
,
13
Function Values of 30, 45, and 60
In Examples 1 and 3, we found the trigonometric function values of an
acute angle of a right triangle when the lengths of the three sides were
known. In most situations, we are asked to find the function values when
the measure of the acute angle is given. For certain special angles such as
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
436
Chapter 5
1/10/05
1:00 PM
Page 436
• The Trigonometric Functions
30, 45, and 60, which are frequently seen in applications, we can use
geometry to determine the function values.
A right triangle with a 45 angle actually has two 45 angles. Thus the
triangle is isosceles, and the legs are the same length. Let’s consider such
a triangle whose legs have length 1. Then we can find the length of its
hypotenuse, c, using the Pythagorean theorem as follows:
12 12 c 2,
or c 2 2,
or c 2.
Such a triangle is shown below. From this diagram, we can easily determine the trigonometric function values of 45.
1
opp
2
0.7071,
hyp 2
2
adj
1
2
cos 45 0.7071,
hyp 2
2
opp
1
1
tan 45 adj
1
sin 45 45
2
1
45
1
It is sufficient to find only the function values of the sine, cosine, and
tangent, since the others are their reciprocals.
It is also possible to determine the function values of 30 and 60. A
right triangle with 30 and 60 acute angles is half of an equilateral triangle, as shown in the following figure. Thus if we choose an equilateral
triangle whose sides have length 2 and take half of it, we obtain a right
triangle that has a hypotenuse of length 2 and a leg of length 1. The other
leg has length a, which can be found as follows:
2
a
1
30
2
60
1
a2 12 22
a2 1 4
a2 3
a 3.
We can now determine the function values of 30 and 60:
1
0.5,
2
3
cos 30 0.8660,
2
1
3
tan 30 0.5774,
3
3
sin 30 3
30
2
60
1
3
0.8660,
2
1
cos 60 0.5,
2
3
3 1.7321.
tan 60 1
sin 60 Since we will often use the function values of 30, 45, and 60, either the
triangles that yield them or the values themselves should be memorized.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 437
Section 5.1
•
Trigonometric Functions of Acute Angles
30
45
60
sin
12
22
32
cos
32
22
12
tan
33
1
3
437
Let’s now use what we have learned about trigonometric functions of
special angles to solve problems. We will consider such applications in
greater detail in Section 5.2.
EXAMPLE 4 Height of a Hot-air Balloon. As a hot-air balloon began to
rise, the ground crew drove 1.2 mi to an observation station. The initial
observation from the station estimated the angle between the ground and
the line of sight to the balloon to be 30. Approximately how high was the
balloon at that point? (We are assuming that the wind velocity was low
and that the balloon rose vertically for the first few minutes.)
Solution We begin with a drawing of the situation. We know the measure
of an acute angle and the length of its adjacent side.
h
30°
1.2 mi
Since we want to determine the length of the opposite side, we can use
the tangent ratio, or the cotangent ratio. Here we use the tangent ratio:
opp
h
adj
1.2
1.2 tan 30 h
3
3
1.2
h
Substituting; tan 30 3
3
0.7 h.
tan 30 The balloon is approximately 0.7 mi, or 3696 ft, high.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
438
Chapter 5
1/10/05
1:00 PM
Page 438
• The Trigonometric Functions
Function Values of Any Acute Angle
Historically, the measure of an angle has been expressed in degrees,
minutes, and seconds. One minute, denoted 1, is such that 60 1,
1
or 1 60 1. One second, denoted 1
, is such that 60
1, or
1
1
60 1. Then 61 degrees, 27 minutes, 4 seconds could be written as
61274
. This DMS form was common before the widespread use of
scientific calculators. Now the preferred notation is to express fractional
parts of degrees in decimal degree form. Although the DMS
notation
is still widely used in navigation, we will most often use the decimal form
in this text.
Most calculators can convert DMS
notation to decimal degree
notation and vice versa. Procedures among calculators vary.
Normal Sci Eng
Float 0123456789
Radian Degree
Func Par Pol Seq
Connected Dot
Sequential Simul
Real abi reˆθ i
Full Horiz G –T
GCM
EXAMPLE 5
Convert 54230
to decimal degree notation.
Solution We can use a graphing calculator set in DEGREE mode to convert
between DMS
form and decimal degree form. (See window at left.)
To convert DMS
form to decimal degree form, we enter 54230
using the ANGLE menu for the degree and minute symbols and ALPHA for the symbol representing seconds. Pressing ENTER gives us
54230
5.71,
rounded to the nearest hundredth of a degree.
5°42'30"
5.708333333
Without a calculator, we can convert as follows:
54230
5 42 30
30 5 42 60
42.5 60
1 EXAMPLE 6
1
42.5 ; 42.5 60
60
42.5 0.71
60
5.71.
GCM
1
30 ; 30 60
60
30 0.5
60
5 42.5
5 1 Convert 72.18 to DMS
notation.
Solution To convert decimal degree form to DMS
form, we enter 72.18
and access the DMS feature in the ANGLE menu. The result is
72.18 721048
.
72.18
DMS
72°10'48"
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 439
Section 5.1
•
Trigonometric Functions of Acute Angles
439
Without a calculator, we can convert as follows:
72.18 72 0.18 1
72 0.18 60
72 10.8
72 10 0.8 1
72 10 0.8 60
72 10 48
721048
.
1 60
1 60
So far we have measured angles using degrees. Another useful unit
for angle measure is the radian, which we will study in Section 5.4. Calculators work with either degrees or radians. Be sure to use whichever mode is
appropriate. In this section, we use the degree mode.
Keep in mind the difference between an exact answer and an approximation. For example,
sin 60 3
.
2
This is exact!
But using a calculator, you get an answer like
sin 60 0.8660254038.
This is an approximation!
Calculators generally provide values only of the sine, cosine, and
tangent functions. You can find values of the cosecant, secant, and cotangent by taking reciprocals of the sine, cosine, and tangent functions,
respectively.
GCM
EXAMPLE 7 Find the trigonometric function value, rounded to four
decimal places, of each of the following.
a) tan 29.7
b) sec 48
c) sin 841039
Solution
a) We check to be sure that the calculator is in DEGREE mode. The function
value is
tan 29.7 0.5703899297
Rounded to four decimal places
0.5704.
b) The secant function value can be found by taking the reciprocal of the
cosine function value:
sec 48 1
1.49447655 1.4945.
cos 48
c) We enter sin 841039
. The result is
sin 841039
0.9948409474 0.9948.
We can use the TABLE feature on a graphing calculator to find an angle
for which we know a trigonometric function value.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
440
Chapter 5
1/10/05
1:00 PM
Page 440
• The Trigonometric Functions
GCM
EXAMPLE 8 Find the acute angle, to the nearest tenth of a degree,
whose sine value is approximately 0.20113.
Solution With a graphing calculator set in DEGREE mode, we first enter
the equation y sin x . With a minimum value of 0 and a step-value of 0.1,
we scroll through the table of values looking for the y-value closest to
0.20113.
X
11.1
11.2
11.3
11.4
11.5
11.6
11.7
X 11.6
Y1
.19252
.19423
.19595
.19766
.19937
.20108
.20279
sin 11.6 0.20108
We find that 11.6 is the angle whose sine value is about 0.20113.
The quickest way to find the angle with a calculator is to use an inverse
function key. (We first studied inverse functions in Section 4.1 and will consider inverse trigonometric functions in Section 6.4.) First check to be sure
that your calculator is in DEGREE mode. Usually two keys must be pressed in
sequence. For this example, if we press
2ND
SIN
.20113
ENTER
,
we find that the acute angle whose sine is 0.20113 is approximately
11.60304613, or 11.6.
EXAMPLE 9 Ladder Safety. A paint crew has purchased new 30-ft extension ladders. The manufacturer states that the safest placement on a
wall is to extend the ladder to 25 ft and to position the base 6.5 ft from
the wall. (Source : R. D. Werner Co., Inc.) What angle does the ladder
make with the ground in this position?
25 ft
6.5 ft
Solution We make a drawing and then use the most convenient trigonometric function. Because we know the length of the side adjacent to and
the length of the hypotenuse, we choose the cosine function.
From the definition of the cosine function, we have
cos adj
6.5 ft
0.26.
hyp
25 ft
Using a calculator, we find the acute angle whose cosine is 0.26:
74.92993786.
Pressing 2ND
COS 0.26 ENTER
Thus when the ladder is in its safest position, it makes an angle of about
75 with the ground.
Cofunctions and Complements
We recall that two angles are complementary whenever the sum of their
measures is 90. Each is the complement of the other. In a right triangle,
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 441
Section 5.1
•
Trigonometric Functions of Acute Angles
441
the acute angles are complementary, since the sum of all three angle
measures is 180 and the right angle accounts for 90 of this total. Thus
if one acute angle of a right triangle is , the other is 90 .
The six trigonometric function values of each of the acute angles in
the triangle below are listed at the right. Note that 53 and 37 are complementary angles since 53 37 90.
53
37
sin 37 0.6018
cos 37 0.7986
tan 37 0.7536
csc 37 1.6616
sec 37 1.2521
cot 37 1.3270
sin 53 0.7986
cos 53 0.6018
tan 53 1.3270
csc 53 1.2521
sec 53 1.6616
cot 53 0.7536
Try this with the acute, complementary angles 20.3 and 69.7 as well.
What pattern do you observe? Look for this same pattern in Example 1
earlier in this section.
Note that the sine of an angle is also the cosine of the angle’s complement. Similarly, the tangent of an angle is the cotangent of the angle’s
complement, and the secant of an angle is the cosecant of the angle’s
complement. These pairs of functions are called cofunctions. A list of
cofunction identities follows.
Cofunction Identities
90 u
sin cos 90 ,
tan cot 90 ,
sec csc 90 ,
u
cos sin 90 ,
cot tan 90 ,
csc sec 90 EXAMPLE 10 Given that sin 18 0.3090 , cos 18 0.9511 , and
tan 18 0.3249, find the six trigonometric function values of 72.
Solution
Using reciprocal relationships, we know that
1
3.2361,
sin 18
1
sec 18 1.0515,
cos 18
1
cot 18 3.0777.
tan 18
csc 18 and
Since 72 and 18 are complementary, we have
sin 72 cos 18 0.9511,
tan 72 cot 18 3.0777,
sec 72 csc 18 3.2361,
cos 72 sin 18 0.3090,
cot 72 tan 18 0.3249,
csc 72 sec 18 1.0515.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
442
Chapter 5
5.1
1/10/05
1:00 PM
Page 442
• The Trigonometric Functions
Exercise Set
In Exercises 1–6, find the six trigonometric function
values of the specified angle.
1.
22
1
, cos , and
3
3
tan 22, find csc , sec , and cot .
8. Given that sin 2.
0.4
0.3
17
8
3.
Given a function value of an acute angle, find the other
five trigonometric function values.
24
9. sin 25 10. cos 0.7 b
15
0.5
11. tan 2
f
4.
e
6
6
3
u
u
33
5. 7
φ
4
15. cos 12. cot 13
13. csc 1.5
a
14. sec 17 5
5
16. sin 10
11
Find the exact function value.
3
17. cos 45 2
18. tan 30
3
2
2
19. sec 60 2
20. sin 45
2
3
21. cot 60
22. csc 45 2
3
23. sin 30 12
24. cos 60 12
23
25. tan 45 1
26. sec 30
3
3
27. csc 30 2
28. cot 60
3
29. Distance Across a River. Find the distance a across
the river. 62.4 m
B
6. Grill
c
30°
A
a
36 m
9
C
8.2
7. Given that sin tan 2
5
, cos , and
3
3
5
, find csc , sec , and cot .
2
30. Distance Between Bases. A baseball diamond is
actually a square 90 ft on a side. If a line is drawn
from third base to first base, then a right triangle
Answers to Exercises 1–16 can be found on p. IA-31.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 443
Section 5.1
QPH is formed, where QPH is 45. Using a
trigonometric function, find the distance from
third base to first base. 127.3 ft
•
443
Trigonometric Functions of Acute Angles
67. csc 89.5 1.0000
68. sec 35.28 1.2250
69. cot 30256
1.7032
70. sin 59.2 0.8590
Find the acute angle , to the nearest tenth of a degree,
for the given function value.
71. sin 0.5125 30.8
72. tan 2.032 63.8
P
Q
(third)
45
(first)
h
90 ft
90 ft
GCM
H
Convert to decimal degree notation. Round to two
decimal places.
31. 943 9.72
32. 5215 52.25
73. tan 0.2226 12.5
74. cos 0.3842 67.4
75. sin 0.9022 64.4
76. tan 3.056 71.9
77. cos 0.6879 46.5
78. sin 0.4005
23.6
79. cot 2.127 25.2
1
Hint: tan .
cot 80. csc 1.147 60.7
81. sec 1.279 38.6
82. cot 1.351 36.5
33. 3550
35.01
34. 6453 64.88
35. 32 3.03
36. 194723
19.79
37. 493846
49.65
38. 761134
76.19
GCM
39. 155
0.25
40. 682
68.00
GCM
41. 553
5.01
42. 4410
0.74
Find the exact acute angle for the given function
value.
2
3
83. sin 45
84. cot 60
2
3
85. cos Convert to degrees, minutes, and seconds. Round to the
nearest second.
43. 17.6 1736
44. 20.14 20824
45. 83.025 83130
46. 67.84 675024
47. 11.75 1145
48. 29.8 2948
49. 47.8268 474936
50. 0.253 01511
51. 0.9 54
52. 30.2505 30152
53. 39.45 3927
54. 2.4 224
Find the function value. Round to four decimal places.
55. cos 51 0.6293
56. cot 17 3.2709
57. tan 413 0.0737
58. sin 26.1 0.4399
59. sec 38.43 1.2765
60. cos 741040
61. cos 40.35 0.7621
62. csc 45.2 1.4093
63. sin 69 0.9336
64. tan 6348 2.0323
65. tan 85.4 12.4288
66. cos 4 0.9976
1
2
87. tan 1
89. csc 60
45
23
3
86. sin 1
2
88. cos 3
30
2
30
90. tan 3
60
91. cot 3 30
60
92. sec 2 45
Use the cofunction and reciprocal identities to complete
each of the following.
1
93. cos 20 sin; sec
70 _____ 20
26 1
cos; csc
_____ 64
95. tan 52 cot
1
38; cot
_____ 52
96. sec 13 csc
1
77; cos
_____ 13
94. sin 64 0.2727
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
444
Chapter 5
1/10/05
1:00 PM
Page 444
• The Trigonometric Functions
97. Given that
sin 65 0.9063,
tan 65 2.1445,
sec 65 2.3662,
cos 65 0.4226,
cot 65 0.4663,
csc 65 1.1034,
find the six function values of 25. 109. 5x 625 [4.5] 4
110. log 3x 1 log x 1 2 [4.5] 101
97
111. log7 x 3 [4.5] 343
98. Given that
sin 8 0.1392,
tan 8 0.1405,
sec 8 1.0098,
Solve.
108. e t 10,000 [4.5] 9.21
cos 8 0.9903,
cot 8 7.1154,
csc 8 7.1853,
find the six function values of 82. Synthesis
112. Given that cos 0.9651, find csc 90° .
1.0362
113. Given that sec 1.5304, find sin 90 .
0.6534
99. Given that sin 71105
0.9465,
cos 71105
0.3228, and tan 71105
2.9321,
find the six function values of 184955
. 100. Given that sin 38.7 0.6252, cos 38.7 0.7804,
and tan 38.7 0.8012, find the six function values
of 51.3. 114. Find the six trigonometric function values of .
q
1
q
a
115. Show that the area of this right triangle is
101. Given that sin 82 p, cos 82 q, and
tan 82 r , find the six function values of 8 in
terms of p, q, and r. 1
2 bc
sin A. A
c
Collaborative Discussion and Writing
102. Explain the difference between reciprocal functions
and cofunctions.
103. Explain why it is not necessary to memorize the
function values for both 30 and 60.
B
b
a
C
116. Show that the area of this triangle is
1
2 ab
sin .
Skill Maintenance
a
Make a hand-drawn graph of the function. Then check
your work using a graphing calculator.
104. fx 2x 105. fx e x/2 106. gx log2 x
107. hx ln x
u
b
Let h the height of the triangle. Then Area where sin 1
bh,
2
1
h
, or h a sin , so Area ab sin .
a
2
Answers to Exercises 97–101, 104–107, 114, and 115 can be found on p. IA-31.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 445
Section 5.2
5.2
Applications of
Right Triangles
•
Applications of Right Triangles
445
Solve right triangles.
Solve applied problems involving right triangles and trigonometric
functions.
Solving Right Triangles
Now that we can find function values for any acute angle, it is possible to
solve right triangles. To solve a triangle means to find the lengths of all
sides and the measures of all angles.
EXAMPLE 1 In ABC (shown at left), find a, b, and B, where a and b
represent lengths of sides and B represents the measure of B. Here we
use standard lettering for naming the sides and angles of a right triangle:
Side a is opposite angle A, side b is opposite angle B, where a and b are the
legs, and side c, the hypotenuse, is opposite angle C, the right angle.
B
106.2
A
Solution
In ABC , we know three of the measures:
A 61.7,
B ?,
C 90,
a
a ?,
b ?,
c 106.2.
Since the sum of the angle measures of any triangle is 180 and C 90,
the sum of A and B is 90. Thus,
61.7
b
C
B 90 A 90 61.7 28.3.
We are given an acute angle and the hypotenuse. This suggests that
we can use the sine and cosine ratios to find a and b, respectively:
sin 61.7 opp
a
hyp 106.2
and
cos 61.7 adj
b
.
hyp 106.2
Solving for a and b, we get
a 106.2 sin 61.7 and b 106.2 cos 61.7
a 93.5
b 50.3.
Thus,
A 61.7,
B 28.3,
C 90,
a 93.5,
b 50.3,
c 106.2.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
446
Chapter 5
1/10/05
Page 446
• The Trigonometric Functions
d
E
1:00 PM
F
EXAMPLE 2
Solution
13
In DEF (shown at left), find D and F. Then find d.
In DEF , we know three of the measures:
D ?,
E 90,
F ?,
23
D
d ?,
e 23,
f 13.
We know the side adjacent to D and the hypotenuse. This suggests the
use of the cosine ratio:
cos D 13
adj
.
hyp 23
We now find the angle whose cosine is
of a degree,
D 55.58.
Pressing 2ND
13
23 .
To the nearest hundredth
COS 1323 ENTER
Since the sum of D and F is 90, we can find F by subtracting:
F 90 D 90 55.58 34.42.
We could use the Pythagorean theorem to find d, but we will use a
trigonometric function here. We could use cos F, sin D, or the tangent or
cotangent ratios for either D or F. Let’s use tan D:
tan D opp
d
,
adj
13
or
tan 55.58 d
.
13
Then
d 13 tan 55.58 19.
The six measures are
D 55.58,
E 90,
F 34.42,
d 19,
e 23,
f 13.
Applications
Right triangles can be used to model and solve many applied problems in
the real world.
North Rim
6.2 mi
b
South Rim
c
50°
EXAMPLE 3 Hiking at the Grand Canyon. A backpacker hiking east
along the North Rim of the Grand Canyon notices an unusual rock formation directly across the canyon. She decides to continue watching the
landmark while hiking along the rim. In 2 hr, she has gone 6.2 mi due
east and the landmark is still visible but at approximately a 50 angle to
the North Rim. (See the figure at left.)
a) How many miles is she from the rock formation?
b) How far is it across the canyon from her starting point?
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 447
Section 5.2
•
Applications of Right Triangles
447
Solution
a) We know the side adjacent to the 50 angle and want to find the hypotenuse. We can use the cosine function:
6.2 mi
c
6.2 mi
9.6 mi.
c
cos 50
cos 50 After hiking 6.2 mi, she is approximately 9.6 mi from the rock formation.
b) We know the side adjacent to the 50 angle and want to find the opposite
side. We can use the tangent function:
b
6.2 mi
b 6.2 mi tan 50 7.4 mi.
tan 50 Thus it is approximately 7.4 mi across the canyon from her starting
point.
Many applications with right triangles involve an angle of elevation
or an angle of depression. The angle between the horizontal and a line of
sight above the horizontal is called an angle of elevation. The angle between the horizontal and a line of sight below the horizontal is called an
angle of depression. For example, suppose that you are looking straight
ahead and then you move your eyes up to look at an approaching airplane. The angle that your eyes pass through is an angle of elevation. If
the pilot of the plane is looking forward and then looks down, the pilot’s
eyes pass through an angle of depression.
Horizontal
Angle of
depression
Angle of
elevation
Horizontal
A
Angle of
depression
475 ft
850 ft
B
Angle of
elevation
C
EXAMPLE 4 Aerial Photography. An aerial photographer who photographs farm properties for a real estate company has determined from
experience that the best photo is taken at a height of approximately 475 ft
and a distance of 850 ft from the farmhouse. What is the angle of depression from the plane to the house?
Solution When parallel lines are cut by a transversal, alternate interior
angles are equal. Thus the angle of depression from the plane to the
house, , is equal to the angle of elevation from the house to the plane,
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
448
Chapter 5
1/10/05
1:00 PM
Page 448
• The Trigonometric Functions
so we can use the right triangle shown in the figure. Since we know the
side opposite B and the hypotenuse, we can find by using the sine
function. We first find sin :
sin sin B 475 ft
0.5588.
850 ft
Using a calculator in DEGREE mode, we find the acute angle whose sine is approximately 0.5588:
34. Pressing
2nd
SIN
0.5588 ENTER
Thus the angle of depression is approximately 34.
EXAMPLE 5 Cloud Height. To measure cloud height at night, a vertical
beam of light is directed on a spot on the cloud. From a point 135 ft away
from the light source, the angle of elevation to the spot is found to be
67.35. Find the height of the cloud.
Solution
From the figure, we have
h
h
135 ft
h 135 ft tan 67.35 324 ft.
tan 67.35 67.35°
135 ft
The height of the cloud is about 324 ft.
Some applications of trigonometry involve the concept of direction,
or bearing. In this text we present two ways of giving direction, the first
below and the second in Exercise Set 5.3.
Bearing: First-Type
One method of giving direction, or bearing, involves reference to a
north – south line using an acute angle. For example, N55W means
55 west of north and S67E means 67 east of south.
N
N
N
N55W
N
N60E
W
E
W
E
W
E
W
S67E
S
S
S
S35W S
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
E
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 449
Section 5.2
•
Applications of Right Triangles
449
EXAMPLE 6 Distance to a Forest Fire. A forest ranger at point A sights
a fire directly south. A second ranger at point B, 7.5 mi east, sights the
same fire at a bearing of S2723W. How far from A is the fire?
N
A
W
7.5 mi
B
E
6237
d
2723
Fire
S
F
S2723W
Solution
We first find the complement of 2723:
B 90 2723
6237
62.62.
Angle B is opposite side d in the right triangle.
From the figure shown above, we see that the desired distance d is part of
a right triangle. We have
d
tan 62.62
7.5 mi
d 7.5 mi tan 62.62 14.5 mi.
The forest ranger at point A is about 14.5 mi from the fire.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
450
Chapter 5
1/10/05
1:00 PM
Page 450
• The Trigonometric Functions
EXAMPLE 7 Comiskey Park. In the new Comiskey Park, the home
of the Chicago White Sox baseball team, the first row of seats in the
upper deck is farther away from home plate than the last row of seats
in the old Comiskey Park. Although there is no obstructed view in the
new park, some of the fans still complain about the present distance
from home plate to the upper deck of seats. (Source : Chicago Tribune,
September 19, 1993) From a seat in the last row of the upper deck
directly behind the batter, the angle of depression to home plate is
29.9, and the angle of depression to the pitcher’s mound is 24.2.
Find (a) the viewing distance to home plate and (b) the viewing distance to the pitcher’s mound.
24.2
29.9
d2
Pitcher
Batter
u2
h
d1
u1
x
60.5 ft
Study Tip
Tutoring is available to students
using this text. The AddisonWesley Math Tutor Center, staffed
by mathematics instructors, can be
reached by telephone, fax, or
e-mail. When you are having
difficulty with an exercise, this
live tutoring can be a valuable
resource. These instructors have a
copy of your text and are familiar
with the content objectives in
this course.
Solution From geometry we know that 1 29.9 and 2 24.2. The
standard distance from home plate to the pitcher’s mound is 60.5 ft. In
the drawing, we let d1 be the viewing distance to home plate, d2 the viewing distance to the pitcher’s mound, h the elevation of the last row, and x
the horizontal distance from the batter to a point directly below the seat
in the last row of the upper deck.
We begin by determining the distance x. We use the tangent function
with 1 29.9 and 2 24.2:
tan 29.9 h
x
and
tan 24.2 h
x 60.5
or
h x tan 29.9 and
h x 60.5 tan 24.2.
Then substituting x tan 29.9 for h in the second equation, we obtain
x tan 29.9 x 60.5 tan 24.2.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 451
Section 5.2
•
Applications of Right Triangles
451
Solving for x, we get
x tan 29.9 x tan 24.2 60.5 tan 24.2
x tan 29.9 x tan 24.2 x tan 24.2 60.5 tan 24.2 x tan 24.2
xtan 29.9 tan 24.2 60.5 tan 24.2
60.5 tan 24.2
x
tan 29.9 tan 24.2
x 216.5.
We can then find d1 and d2 using the cosine function:
cos 29.9 216.5
d1
and cos 24.2 216.5 60.5
d2
or
216.5
and
cos 29.9
d1 249.7
d1 277
cos 24.2
d2 303.7.
d2 The distance to home plate is about 250 ft,* and the distance to the
pitcher’s mound is about 304 ft.
*In the old Comiskey Park, the distance to home plate was only 150 ft.
,
,
T 63.3,
s 0.38, t 0.34
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:00 PM
Page 451
Section 5.2
5.2
•
Applications of Right Triangles
Exercise Set
In Exercises 1–6, solve the right triangle.
2.
1.
F
5.
6.
P
P 4738,
n 34.4,
p 25.4
A
6
d
30
E
f
10
b
F 60, d 3, f 5.2
M
45
C
a
126
c
B
s
67.3
B
A 22.7, a 52.7,
c 136.6
f
N
p
G
H 6126,
f 36.2,
h 31.8
H
4222
R
a
A
17.3
23.2
4.
C
2834
In Exercises 7–16, solve the right triangle. (Standard
lettering has been used.)
B
A 45, a 7.1, b 7.1
3.
h
F
n
D
451
c
26.7
a
T
t
0.17
S
T 63.3,
s 0.38, t 0.34
A
b
C
7. A 8743, a 9.73 B 217, b 0.39, c 9.74
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
452
Chapter 5
1/10/05
1:00 PM
Page 452
• The Trigonometric Functions
8. a 12.5, b 18.3 A 34.3, B 55.7, c 22.2
9. b 100, c 450 A 77.2, B 12.8, a 439
10. B 56.5, c 0.0447
A 33.5, a 0.0247, b 0.0373
11. A 47.58, c 48.3
B 42.42, a 35.7, b 32.6
20. Height of a Tree. A supervisor must train a new
team of loggers to estimate the heights of trees.
As an example, she walks off 40 ft from the base of
a tree and estimates the angle of elevation to the
tree’s peak to be 70. Approximately how tall is
the tree? About 110 ft
12. B 20.6, a 7.5
A 69.4, b 2.8, c 8.0
13. A 35, b 40
B 55, a 28.0, c 48.8
14. B 69.3, b 93.4
A 20.7, a 35.3, c 99.8
15. b 1.86, c 4.02
A 62.4, B 27.6, a 3.56
16. a 10.2, c 20.4
A 30, B 60, b 17.7
17. Safety Line to Raft. Each spring Bryan uses his
vacation time to ready his lake property for the
summer. He wants to run a new safety line from
point B on the shore to the corner of the anchored
diving raft. The current safety line, which runs
perpendicular to the shore line to point A, is 40 ft
long. He estimates the angle from B to the corner of
the raft to be 50. Approximately how much rope
does he need for the new safety line if he allows 5 ft
of rope at each end to fasten the rope?
About 62.2 ft
70°
40 ft
21. Sand Dunes National Park. While visiting the
Sand Dunes National Park in Colorado, Cole
approximated the angle of elevation to the top of a
sand dune to be 20. After walking 800 ft closer, he
guessed that the angle of elevation had increased by
15. Approximately how tall is the dune he was
observing? About 606 ft
Diving
raft
40 ft
B
50°
Shoreline
A
18. Enclosing an Area. Alicia is enclosing a triangular
area in a corner of her fenced rectangular backyard
for her Labrador retriever. In order for a certain
tree to be included in this pen, one side needs to be
14.5 ft and make a 53 angle with the new side.
How long is the new side? About 24.1 ft
20°
800 ft
22. Tee Shirt Design. A new tee shirt design is to have
a regular octagon inscribed in a circle, as shown
in the figure. Each side of the octagon is to be
3.5 in. long. Find the radius of the circumscribed
circle. About 4.6 in.
3.5 in.
19. Easel Display. A marketing group is designing an
easel to display posters advertising their newest
products. They want the easel to be 6 ft tall and the
back of it to fit flush against a wall. For optimal eye
contact, the best angle between the front and back
legs of the easel is 23. How far from the wall should
the front legs be placed in order to obtain this angle?
About 2.5 ft
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
r
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 453
Section 5.2
23. Inscribed Pentagon. A regular pentagon is inscribed
in a circle of radius 15.8 cm. Find the perimeter of
the pentagon. About 92.9 cm
24. Height of a Weather Balloon. A weather balloon is
directly west of two observing stations that are
10 mi apart. The angles of elevation of the balloon
from the two stations are 17.6 and 78.2. How high
is the balloon? About 3.4 mi
•
Applications of Right Triangles
453
29. Distance from a Lighthouse. From the top of a
lighthouse 55 ft above sea level, the angle of
depression to a small boat is 11.3. How far from
the foot of the lighthouse is the boat? About 275 ft
25. Height of a Kite. For a science fair project, a
group of students tested different materials used
to construct kites. Their instructor provided an
instrument that accurately measures the angle of
elevation. In one of the tests, the angle of elevation
was 63.4 with 670 ft of string out. Assuming the
string was taut, how high was the kite? About 599 ft
30. Lightning Detection. In extremely large forests, it
is not cost-effective to position forest rangers in
towers or to use small aircraft to continually watch
for fires. Since lightning is a frequent cause of fire,
lightning detectors are now commonly used instead.
These devices not only give a bearing on the
location but also measure the intensity of the
lightning. A detector at point Q is situated 15 mi
west of a central fire station at point R. The bearing
from Q to where lightning hits due south of R is
S37.6E. How far is the hit from point R?
26. Height of a Building. A window washer on a ladder
looks at a nearby building 100 ft away, noting that
the angle of elevation of the top of the building is
18.7 and the angle of depression of the bottom of
the building is 6.5. How tall is the nearby building?
31. Lobster Boat. A lobster boat is situated due west of
a lighthouse. A barge is 12 km south of the lobster
boat. From the barge, the bearing to the lighthouse
is N6320E. How far is the lobster boat from the
lighthouse? About 24 km
About 19.5 mi
About 45 ft
18.7°
100 ft
6.5°
12 km
North
63° 20'
27. Distance Between Towns. From a hot-air balloon
2 km high, the angles of depression to two towns in
line with the balloon are 81.2 and 13.5. How far
apart are the towns? About 8 km
28. Angle of Elevation. What is the angle of elevation of
the sun when a 35-ft mast casts a 20-ft shadow?
About 60.3
32. Length of an Antenna. A vertical antenna is
mounted atop a 50-ft pole. From a point on level
ground 75 ft from the base of the pole, the antenna
subtends an angle of 10.5. Find the length of the
antenna. About 23 ft
35 ft
10.5°
50 ft
20 ft
75 ft
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
454
Chapter 5
1/10/05
1:01 PM
Page 454
• The Trigonometric Functions
Collaborative Discussion and Writing
33. In this section, the trigonometric functions have
been defined as functions of acute angles. Thus the
set of angles whose measures are greater than 0 and
less than 90 is the domain for each function. What
appear to be the ranges for the sine, the cosine, and
the tangent functions given this domain?
41. Construction of Picnic Pavilions. A construction
company is mass-producing picnic pavilions for
national parks, as shown in the figure. The rafter
ends are to be sawed in such a way that they will be
vertical when in place. The front is 8 ft high, the
back is 6 12 ft high, and the distance between the
front and back is 8 ft. At what angle should the
rafters be cut? Cut so that 79.38
34. Explain in your own words five ways in which
length c can be determined in this triangle. Which
way seems the most efficient?
u
u
c
14
8 ft
6
6 q ft
Skill Maintenance
Find the distance between the points.
35. 8, 2 and 6, 4 [1.1] 102, or about 14.142
36. 9, 3 and 0, 0 [1.1] 310, or about 9.487
37. Convert to an exponential equation:
log 0.001 3. [4.3] 10 3 0.001
38. Convert to a logarithmic equation: e 4 t .
[4.3] ln t 4
Synthesis
39. Find h, to the nearest tenth.
3.3
8 ft
42. Diameter of a Pipe. A V-gauge is used to find the
diameter of a pipe. The advantage of such a device is
that it is rugged, it is accurate, and it has no moving
parts to break down. In the figure, the measure of
angle AVB is 54. A pipe is placed in the V-shaped
slot and the distance VP is used to estimate the
diameter. The line VP is calibrated by listing as its
units the corresponding diameters. This, in effect,
establishes a function between VP and d.
A
Q
h
d
V
36
P
B
7
40. Find a, to the nearest tenth.
5.9
5
72
a
a) Suppose that the diameter of a pipe is 2 cm. What
is the distance VP? About 1.96 cm
b) Suppose that the distance VP is 3.93 cm. What is
the diameter of the pipe? About 4.00 cm
c) Find a formula for d in terms of VP. d 1.02 VP
d) Find a formula for VP in terms of d. VP 0.98d
43. Sound of an Airplane. It is a common experience to
hear the sound of a low-flying airplane and look at
the wrong place in the sky to see the plane. Suppose
that a plane is traveling directly at you at a speed of
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 455
Section 5.2
200 mph and an altitude of 3000 ft, and you hear the
sound at what seems to be an angle of inclination
of 20. At what angle should you actually look in
order to see the plane? Consider the speed of sound
to be 1100 ftsec. 27
Perceived location
of plane
P
3000 ft
Actual location
of plane when heard
•
Applications of Right Triangles
mountain and a line drawn from the top of the
mountain to the horizon, as shown in the figure.
The height of Mt. Shasta in California is 14,162 ft.
From the top of Mt. Shasta, one can see the horizon
on the Pacific Ocean. The angle formed between a
line to the horizon and the vertical is found to be
8753. Use this information to estimate the radius
of the earth, in miles. R 3928 mi
A
455
R 14,162
u 8753
20°
V
R
44. Measuring the Radius of the Earth. One way to
measure the radius of the earth is to climb to the
top of a mountain whose height above sea level is
known and measure the angle between a vertical
line to the center of the earth from the top of the
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 455
Section 5.3
5.3
Trigonometric
Functions of
Any Angle
•
Trigonometric Functions of Any Angle
Find angles that are coterminal with a given angle and find the
complement and the supplement of a given angle.
Determine the six trigonometric function values for any angle in
standard position when the coordinates of a point on the terminal
side are given.
Find the function values for any angle whose terminal side lies on
an axis.
Find the function values for an angle whose terminal side makes
an angle of 30, 45, or 60 with the x-axis.
Use a calculator to find function values and angles.
Angles, Rotations, and Degree Measure
An angle is a familiar figure in the world around us.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
455
BBEPMC05_0321279115.QXP
456
Chapter 5
1/10/05
1:01 PM
Page 456
• The Trigonometric Functions
An angle is the union of two rays with a common endpoint called
the vertex. In trigonometry, we often think of an angle as a rotation. To
do so, think of locating a ray along the positive x-axis with its endpoint
at the origin. This ray is called the initial side of the angle. Though we
leave that ray fixed, think of making a copy of it and rotating it. A rotation counterclockwise is a positive rotation, and a rotation clockwise
is a negative rotation. The ray at the end of the rotation is called the
terminal side of the angle. The angle formed is said to be in standard
position.
y Terminal side
y
Vertex
y
A positive
rotation
(or angle)
x
x
x
Initial side
A negative
rotation
(or angle)
The measure of an angle or rotation may be given in degrees. The
Babylonians developed the idea of dividing the circumference of a circle
into 360 equal parts, or degrees. If we let the measure of one of these
parts be 1, then one complete positive revolution or rotation has a measure of 360. One half of a revolution has a measure of 180, one fourth of
a revolution has a measure of 90, and so on. We can also speak of an
angle of measure 60, 135, 330, or 420. The terminal sides of these
angles lie in quadrants I, II, IV, and I, respectively. The negative rotations 30, 110, and 225 represent angles with terminal sides in
quadrants IV, III, and II, respectively.
y
y
y
90
180
x
x
360
y
y
270
x
270
330
30 x
x
y
60
x
90
y
135
225
y
420
610
x
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x
110
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 457
Section 5.3
•
Trigonometric Functions of Any Angle
457
If two or more angles have the same terminal side, the angles are said
to be coterminal. To find angles coterminal with a given angle, we add or
subtract multiples of 360. For example, 420, shown above, has the same
terminal side as 60, since 420 360 60. Thus we say that angles of
measure 60 and 420 are coterminal. The negative rotation that measures
300 is also coterminal with 60 because 60 360 300. The set of
all angles coterminal with 60 can be expressed as 60 n 360, where
n is an integer. Other examples of coterminal angles shown above are 90
and 270, 90 and 270, 135 and 225, 30 and 330, and 110
and 610.
EXAMPLE 1 Find two positive and two negative angles that are coterminal with (a) 51 and (b) 7.
Solution
a) We add and subtract multiples of 360. Many answers are possible.
y
y
51
51
x
x
411
1131
51 360 411
51 3(360) 1131
y
309
y
51
669
51
x
51 360 309
x
51 2(360) 669
Thus angles of measure 411, 1131, 309, and 669 are coterminal
with 51.
b) We have the following:
7 360 353,
7 360 367,
7 2360 713,
7 10360 3607.
Thus angles of measure 353, 713, 367, and 3607 are coterminal
with 7.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
458
Chapter 5
1/10/05
1:01 PM
Page 458
• The Trigonometric Functions
Angles can be classified by their measures, as seen in the following
figure.
161
180
Obtuse: 90 u 180
Straight : u 180
90
24
Right: u 90
Acute: 0 u 90
Recall that two acute angles are complementary if their sum is 90.
For example, angles that measure 10 and 80 are complementary because
10 80 90. Two positive angles are supplementary if their sum is
180. For example, angles that measure 45 and 135 are supplementary
because 45 135 180.
10
80
Complementary angles
EXAMPLE 2
Solution
45
135
Supplementary angles
Find the complement and the supplement of 71.46.
We have
90 71.46 18.54,
180 71.46 108.54.
Thus the complement of 71.46 is 18.54 and the supplement is 108.54.
y
Trigonometric Functions of Angles or Rotations
x 2 y 2 r2
r y
u
x
P(x, y)
x
Many applied problems in trigonometry involve the use of angles that are
not acute. Thus we need to extend the domains of the trigonometric
functions defined in Section 5.1 to angles, or rotations, of any size. To do
this, we first consider a right triangle with one vertex at the origin of a
coordinate system and one vertex on the positive x-axis. (See the figure
at left.) The other vertex is at P, a point on the circle whose center is at
the origin and whose radius r is the length of the hypotenuse of the triangle. This triangle is a reference triangle for angle , which is in standard position. Note that y is the length of the side opposite and x is the
length of the side adjacent to .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 459
Section 5.3
•
Trigonometric Functions of Any Angle
459
Recalling the definitions in Section 5.1, we note that three of the
trigonometric functions of angle are defined as follows:
sin opp
y
,
hyp
r
cos adj
x
,
hyp
r
tan opp
y
.
adj
x
Since x and y are the coordinates of the point P and the length of the
radius is the length of the hypotenuse, we can also define these functions
as follows:
y-coordinate
,
radius
x-coordinate
cos ,
radius
y-coordinate
.
tan x-coordinate
sin We will use these definitions for functions of angles of any measure.
The following figures show angles whose terminal sides lie in quadrants II, III, and IV.
y
P(x, y)
y
r
x
y
u
y
u
x
x
y
P(x, y)
x
x
r
u
r
x
y
P(x, y)
A reference triangle can be drawn for angles in any quadrant, as
shown. Note that the angle is in standard position; that is, it is always
measured from the positive half of the x-axis. The point Px, y is a
point, other than the vertex, on the terminal side of the angle. Each of its
two coordinates may be positive, negative, or zero, depending on the location of the terminal side. The length of the radius, which is also the
length of the hypotenuse of the reference triangle, is always considered
positive. Note that x 2 y 2 r 2, or r x 2 y 2. Regardless of the
location of P, we have the following definitions.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
460
Chapter 5
1/10/05
x
Page 460
• The Trigonometric Functions
y
P(x, y)
r
y
1:01 PM
u
x
Trigonometric Functions of Any Angle Suppose that Px, y is any point other than the vertex on the
terminal side of any angle in standard position, and r is the
radius, or distance from the origin to Px, y. Then the
trigonometric functions are defined as follows:
y-coordinate
radius
x-coordinate
cos radius
y-coordinate
tan x-coordinate
y
,
r
x
,
r
y
,
x
sin radius
y-coordinate
radius
sec x-coordinate
x-coordinate
cot y-coordinate
csc r
,
y
r
,
x
x
.
y
Values of the trigonometric functions can be positive, negative, or
zero, depending on where the terminal side of the angle lies. The length
of the radius is always positive. Thus the signs of the function values
depend only on the coordinates of the point P on the terminal side of
the angle. In the first quadrant, all function values are positive because
both coordinates are positive. In the second quadrant, first coordinates
are negative and second coordinates are positive; thus only the sine and
the cosecant values are positive. Similarly, we can determine the signs of
the function values in the third and fourth quadrants. Because of the
reciprocal relationships, we need to learn only the signs for the sine, cosine,
and tangent functions.
Positive: sin
Negative: cos, tan
y
II (, )
(, )
III
Positive: tan
Negative: sin, cos
Positive: All
Negative: None
I
(, )
x
(, )
IV
Positive: cos
Negative: sin, tan
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 461
Section 5.3
b)
y
Trigonometric Functions of Any Angle
461
Find the six trigonometric function values for each angle
EXAMPLE 3
shown.
a)
•
c)
y
y
(1, 3)
u
3
4
3
(4, 3)
r
u
1
x
r
u
r
1
x
1
x
(1, 1)
Solution
a) We first determine r, the distance from the origin 0, 0 to the point
4, 3. The distance between 0, 0 and any point x, y on the terminal side of the angle is
r x 02 y 02
x 2 y 2.
Substituting 4 for x and 3 for y, we find
r 42 32
16 9 25 5.
Using the definitions of the trigonometric functions, we can now find
the function values of . We substitute 4 for x, 3 for y, and 5 for r :
y
3
3
,
r
5
5
x
4
4
cos ,
r
5
5
3
3
y
tan ,
x
4
4
sin r
5
5
,
y
3
3
r
5
5
sec ,
x
4
4
4
4
x
.
cot y
3
3
csc As expected, the tangent and the cotangent values are positive and the
other four are negative. This is true for all angles in quadrant III.
b) We first determine r, the distance from the origin to the point 1, 1:
r 12 12 1 1 2.
Substituting 1 for x, 1 for y, and 2 for r, we find
y
1
2
,
r
2
2
x
1
2
cos ,
r
2
2
y
1
tan 1,
x
1
sin r
2
2,
y
1
r
2
sec 2,
x
1
x
1
1.
cot y
1
csc Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
462
Chapter 5
1/10/05
1:01 PM
Page 462
• The Trigonometric Functions
c) We determine r, the distance from the origin to the point 1, 3 :
r 12 3 2 1 3 4 2.
Substituting 1 for x, 3 for y, and 2 for r, we find the trigonometric
function values of are
3
,
2
1
1
cos ,
2
2
3
tan 3,
1
sin y
(8, 4)
(4, 2)
108642
u
4 6 8 10 x
csc 2
23
,
3
3
2
sec 2,
1
1
3
cot .
3
3
Any point other than the origin on the terminal side of an angle
in standard position can be used to determine the trigonometric function values of that angle. The function values are the same regardless
of which point is used. To illustrate this, let’s consider an angle in
1
standard position whose terminal side lies on the line y 2 x . We
can determine two second-quadrant solutions of the equation, find the
length r for each point, and then compare the sine, cosine, and tangent
function values using each point.
1
If x 4, then y 2 4 2.
y qx
1
If x 8, then y 2 8 4.
For 4, 2, r 42 22 20 25.
For 8, 4, r 82 42 80 45.
Using 4, 2 and r 25, we find that
2
1
5
,
25 5
5
2
1
tan .
4
2
sin and
cos 4
2
25
,
25 5
5
cos 8
2
25
,
45 5
5
Using 8, 4 and r 45, we find that
4
1
5
,
45 5
5
4
1
tan .
8
2
sin and
We see that the function values are the same using either point. Any
point other than the origin on the terminal side of an angle can be used
to determine the trigonometric function values.
The trigonometric function values of depend only on the angle,
not on the choice of the point on the terminal side that is used to
compute them.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 463
Section 5.3
•
Trigonometric Functions of Any Angle
463
The Six Functions Related
When we know one of the function values of an angle, we can find the
other five if we know the quadrant in which the terminal side lies. The
procedure is to sketch a reference triangle in the appropriate quadrant,
use the Pythagorean theorem as needed to find the lengths of its sides,
and then find the ratios of the sides.
2
EXAMPLE 4 Given that tan 3 and is in the second quadrant,
find the other function values.
Solution
We first sketch a second-quadrant angle. Since
tan Expressing 2
2
as
since is in
3 3
quadrant II
we make the legs lengths 2 and 3. The hypotenuse must then have length
22 32, or 13. Now we read off the appropriate ratios:
y
(3, 2)
2
y
2
2
,
x
3
3
13
sin u
3
2
,
or
13
3
cos ,
13
2
tan ,
3
x
13
,
2
13
sec ,
3
213
,
13
313
or ,
13
csc cot 3
.
2
Terminal Side on an Axis
An angle whose terminal side falls on one of the axes is a quadrantal
angle. One of the coordinates of any point on that side is 0. The definitions of the trigonometric functions still apply, but in some cases, function values will not be defined because a denominator will be 0.
EXAMPLE 5 Find the sine, cosine, and tangent values for 90, 180,
270, and 360.
Solution We first make a drawing of each angle in standard position
and label a point on the terminal side. Since the function values are
the same for all points on the terminal side, we choose 0, 1, 1, 0,
0, 1, and 1, 0 for convenience. Note that r 1 for each choice.
y
y
y
y
(0, 1)
90
180
x
(1, 0)
(1, 0)
270
x
x
x
360
(0, 1)
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
464
Chapter 5
1/10/05
1:01 PM
Page 464
• The Trigonometric Functions
Then by the definitions we get
1
1,
1
0
cos 90 0,
1
1
tan 90 , Not defined
0
sin 90 0
0,
1
1
cos 180 1,
1
0
tan 180 0,
1
sin 180 1
1,
1
0
cos 270 0,
1
1
, Not defined
tan 270 0
0
0,
1
1
cos 360 1,
1
0
tan 360 0.
1
sin 360 sin 270 In Example 5, all the values can be found using a calculator, but you
will find that it is convenient to be able to compute them mentally. It is
also helpful to note that coterminal angles have the same function values.
For example, 0 and 360 are coterminal; thus, sin 0 0, cos 0 1, and
tan 0 0.
EXAMPLE 6
Find each of the following.
a) sin 90
b) csc 540
Solution
a) We note that 90 is coterminal with 270. Thus,
sin 90 sin 270 1
1.
1
b) Since 540 180 360, 540 and 180 are coterminal. Thus,
csc 540 csc 180 1
1
,
sin 180
0
which is not defined.
Trigonometric values can always be checked using a calculator. When
the value is undefined, the calculator will display an ERROR message.
ERR: DIVIDE BY 0
1: Quit
2: Goto
Reference Angles: 30, 45, and 60
We can also mentally determine trigonometric function values whenever
the terminal side makes a 30, 45, or 60 angle with the x-axis. Consider,
for example, an angle of 150. The terminal side makes a 30 angle with
the x-axis, since 180 150 30.
y
150
P(3, 1)
Reference
2 150
angle
1
30
180
O
N 3
P(3, 1)
2
30
3
1
N
x
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 465
Section 5.3
•
Trigonometric Functions of Any Angle
465
As the figure shows, ONP is congruent to ONP; therefore, the
ratios of the sides of the two triangles are the same. Thus the trigonometric function values are the same except perhaps for the sign. We
could determine the function values directly from ONP , but this is not
necessary. If we remember that in quadrant II, the sine is positive and
the cosine and the tangent are negative, we can simply use the function
values of 30 that we already know and prefix the appropriate sign. Thus,
sin 150 sin 30 1
,
2
3
,
2
1
tan 150 tan 30 ,
3
cos 150 cos 30 and
or 3
.
3
Triangle ONP is the reference triangle and the acute angle NOP is
called a reference angle.
Reference Angle
The reference angle for an angle is the acute angle formed by the
terminal side of the angle and the x-axis.
EXAMPLE 7 Find the sine, cosine, and tangent function values for each
of the following.
b) 780
a) 225
Solution
a) We draw a figure showing the terminal side of a 225 angle. The reference angle is 225 180, or 45.
y
y
2
45 1
Reference
angle
45
225
x
45
1
x
225
Recall from Section 5.1 that sin 45 22, cos 45 22,
and tan 45 1. Also note that in the third quadrant, the sine and the
cosine are negative and the tangent is positive. Thus we have
sin 225 2
,
2
cos 225 2
,
2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
and tan 225 1.
BBEPMC05_0321279115.QXP
466
Chapter 5
1/10/05
1:01 PM
Page 466
• The Trigonometric Functions
b) We draw a figure showing the terminal side of a 780 angle. Since
780 2360 60, we know that 780 and 60 are coterminal.
y
y
2
60
60
3
1
x
x
60
Reference
angle
780
780
The reference angle for 60 is the acute angle formed by the
terminal side of the angle and the x-axis. Thus the reference angle
for 60 is 60. We know that since 780 is a fourth-quadrant angle,
the cosine is positive and the sine and the tangent are negative. Recalling that sin 60 32, cos 60 12, and tan 60 3, we have
3
,
2
tan 780 3.
sin 780 and
cos 780 1
,
2
Function Values for Any Angle
When the terminal side of an angle falls on one of the axes or makes a
30, 45, or 60 angle with the x-axis, we can find exact function values
without the use of a calculator. But this group is only a small subset of all
angles. Using a calculator, we can approximate the trigonometric function values of any angle. In fact, we can approximate or find exact function values of all angles without using a reference angle.
EXAMPLE 8 Find each of the following function values using a calculator and round the answer to four decimal places, where appropriate.
a)
c)
e)
g)
cos(112)
1/cos(500)
tan(83.4)
cos 112
tan 83.4
cos 2400
cot 135
Solution
.3746065934
1.305407289
8.64274761
b) sec 500
d) csc 351.75
f ) sin 175409
Using a calculator set in DEGREE mode, we find the values.
a) cos 112 0.3746
1
1.3054
b) sec 500 cos 500
c) tan 83.4 8.6427
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 467
Section 5.3
•
Trigonometric Functions of Any Angle
467
1
6.9690
sin 351.75
e) cos 2400 0.5
f ) sin 175409
0.0755
1
1
g) cot 135 tan 135
d) csc 351.75 1/sin(351.75)
6.968999424
cos(2400)
.5
sin(175409
)
.0755153443
In many applications, we have a trigonometric function value and
want to find the measure of a corresponding angle. When only acute
angles are considered, there is only one angle for each trigonometric
function value. This is not the case when we extend the domain of the
trigonometric functions to the set of all angles. For a given function
value, there is an infinite number of angles that have that function value.
There can be two such angles for each value in the range from 0 to 360.
To determine a unique answer in the interval 0, 360, the quadrant in
which the terminal side lies must be specified.
The calculator gives the reference angle as an output for each function value that is entered as an input. Knowing the reference angle
and the quadrant in which the terminal side lies, we can find the specified angle.
EXAMPLE 9
find .
Given the function value and the quadrant restriction,
a) sin 0.2812, 90 180
b) cot 0.1611, 270 360
Solution
a) We first sketch the angle in the second quadrant. We use the calculator to find the acute angle (reference angle) whose sine is 0.2812. The
reference angle is approximately 16.33. We find the angle by subtracting 16.33 from 180:
y
u
16.33
x
Thus, 163.67.
b) We begin by sketching the angle in the fourth quadrant. Because the
tangent and cotangent values are reciprocals, we know that
y
u
180 16.33 163.67.
x
80.85
tan 1
6.2073.
0.1611
We use the calculator to find the acute angle (reference angle) whose
tangent is 6.2073, ignoring the fact that tan is negative. The reference
angle is approximately 80.85. We find angle by subtracting 80.85
from 360:
360 80.85 279.15.
Thus, 279.15.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
468
Chapter 5
5.3
1/10/05
1:01 PM
Page 468
• The Trigonometric Functions
Exercise Set
The terminal side of angle in standard position lies on
the given line in the given quadrant. Find sin , cos ,
and tan .
29. 2x 3y 0; quadrant IV For angles of the following measures, state in which
quadrant the terminal side lies. It helps to sketch the
angle in standard position.
1. 187 III
2. 14.3 IV
3. 24515 III
4. 120 III
30. 4x y 0; quadrant II 5. 800 I
6. 1075 IV
31. 5x 4y 0; quadrant I
7. 460.5 III
8. 315 IV
32. y 0.8x ; quadrant III
9. 912 II
10. 131560
I
12. 345.14 I
11. 537 II
Find two positive angles and two negative angles that
are coterminal with the given angle. Answers may vary.
13. 74 14. 81 A function value and a quadrant are given. Find the
other five function values. Give exact answers.
1
33. sin , quadrant III 3
34. tan 5, quadrant I
15. 115.3 16. 27510 35. cot 2, quadrant IV 17. 180 18. 310 36. cos Find the complement and the supplement.
19. 17.11 72.89, 162.89 20. 4738 4222, 13222
21. 12314
22. 9.038 80.962, 170.962°
23. 45.2 44.8, 134.8
24. 67.31 22.69, 112.69
Find the six trigonometric function values for the angle
shown.
25.
26. y
y
(12, 5)
r
b
x
u
x
r
(7, 3)
27.
28.
y
y
x
(23, 4)
r
3
, quadrant IV 5
38. sin 5
, quadrant III
13
Find the reference angle and the exact function value if
it exists.
45; 2
3
39. cos 150 30°; 40. sec 225
2
2
2
Not defined
41. tan 135 45; 1
42. sin 45 45°; 43. sin 7560 0
44. tan 270
2
45. cos 495 45°; 2
47. csc 210 30; 2
48. sin 300 60; 49. cot 570 30; 3
50. cos 120
55. cos 180 1
56. csc 90 1
57. tan 240 60; 3
58. cot 180
46. tan 675 45; 1
3
2
60; 21
3
51. tan 330 30; 52. cot 855 45; 1
3
53. sec 90 Not defined 54. sin 90 1
(9, 1)
f
37. cos 4
, quadrant II 5
r
a
x
Answers to Exercises 13–18, 21, and 25–38 can be found on pp. IA-31 and IA-32.
Not defined
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 469
Section 5.3
59. sin 495 45°;
2
2
60. sin 1050 30; 12
61. csc 225 45; 2
62. sin 450 1
63. cos 0 1
64. tan 480 60; 3
65. cot 90 0
66. sec 315 45; 2
67. cos 90 0
68. sin 135 45; 69. cos 270 0
•
Trigonometric Functions of Any Angle
In aerial navigation, directions are
given in degrees clockwise from north. Thus, east is 90 ,
south is 180 , and west is 270. Several aerial directions
or bearings are given below.
Aerial Navigation.
0
N
2
2
70. tan 0 0
74. 620 75. 215 76. 290 77. 272 78. 91 Use a calculator in Exercises 79 – 82, but do not use the
trigonometric function keys.
79. Given that
sin 41 0.6561,
cos 41 0.7547,
tan 41 0.8693,
find the trigonometric function values for 319. 0
N
50
W
270
Find the signs of the six trigonometric function values
for the given angles.
71. 319 72. 57 73. 194 469
E
90
S
180
0
N
0
N
W
270
W 350
270
E
90
S
180
S
180
83. An airplane flies 150 km from an airport in a
direction of 120. How far east of the airport
is the plane then? How far south?
East: about 130 km; south: 75 km
0°
N
82. Given that
sin 35 0.5736,
cos 35 0.8192,
tan 35 0.7002,
find the trigonometric function values for 215. 120°
W
270°
E
90°
150 km
S
180°
81. Given that
find the trigonometric function values for 115. E
90
223
find the trigonometric function values for 333. sin 65 0.9063,
cos 65 0.4226,
tan 65 2.1445,
E
90
S
180
80. Given that
sin 27 0.4540,
cos 27 0.8910,
tan 27 0.5095,
115
W
270
84. An airplane leaves an airport and travels for 100 mi
in a direction of 300. How far north of the airport
is the plane then? How far west?
North: 50 mi; west: about 87 mi
0°
N
100 mi
W
270°
E
90°
300°
S
180°
Answers to Exercises 71– 82 can be found on p. IA-32.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
470
Chapter 5
1/10/05
1:01 PM
Page 470
• The Trigonometric Functions
85. An airplane travels at 150 kmh for 2 hr in a
direction of 138 from Omaha. At the end of this
time, how far south of Omaha is the plane?
Determine the domain and the range of the function.
x4
111. fx x2
86. An airplane travels at 120 kmh for 2 hr in a
direction of 319 from Chicago. At the end of this
time, how far north of Chicago is the plane?
112. gx About 223 km
About 181 km
Find the function value. Round to four decimal places.
87. tan 310.8 1.1585
88. cos 205.5 0.9026
89. cot 146.15 1.4910
90. sin 16.4 0.2823
91. sin 11842 0.8771
92. cos 27345 0.0654
x2 9
2x 7x 15
2
Find the zeros of the function.
[2.3] 2, 3
113. fx 12 x [2.1] 12 114. gx x 2 x 6
Find the x-intercepts of the graph of the function.
115. fx 12 x
116. gx x 2 x 6
[2.1] 12, 0
[2.3] 2, 0, 3, 0
93. cos 295.8 0.4352
94. tan 1086.2 0.1086
Synthesis
95. cos 5417 0.9563
96. sec 24055 2.0573
97. csc 520 2.9238
98. sin 3824 0.6947
117. Valve Cap on a Bicycle. The valve cap on a bicycle
wheel is 12.5 in. from the center of the wheel.
From the position shown, the wheel starts to roll.
After the wheel has turned 390, how far above the
ground is the valve cap? Assume that the outer
radius of the tire is 13.375 in. 19.625 in.
Given the function value and the quadrant restriction,
find .
INTERVAL
270, 360
275.4
100. tan 0.2460
180, 270
193.8
101. cos 0.9388
180, 270
200.1
102. sec 1.0485
90, 180
162.5
103. tan 3.0545
270, 360
288.1
104. sin 0.4313
180, 270
205.6
105. csc 1.0480
0, 90
72.6
106. cos 0.0990
90, 180
95.7
FUNCTION VALUE
99. sin 0.9956
13.375 in.
12.5 in.
118. Seats of a Ferris Wheel. The seats of a ferris wheel
are 35 ft from the center of the wheel. When you
board the wheel, you are 5 ft above the ground.
After you have rotated through an angle of 765,
how far above the ground are you? About 15.3 ft
Collaborative Discussion and Writing
107. Why do the function values of depend only on
the angle and not on the choice of a point on the
terminal side?
108. Why is the domain of the tangent function
different from the domains of the sine and the
cosine functions?
35 ft
Skill Maintenance
Graph the function. Sketch and label any vertical
asymptotes.
1
109. fx 2
110. gx x 3 2x 1
x 25
5 ft
Answers to Exercises 109 – 112 can be found on p. IA-32.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 471
Section 5. 4
•
471
Radians, Arc Length, and Angular Speed
Find points on the unit circle determined by real numbers.
Convert between radian and degree measure; find coterminal,
complementary, and supplementary angles.
Find the length of an arc of a circle; find the measure of a central angle
of a circle.
Convert between linear speed and angular speed.
5.4
Radians, Arc
Length, and
Angular Speed
Another useful unit of angle measure is called a radian. To introduce
radian measure, we use a circle centered at the origin with a radius of
length 1. Such a circle is called a unit circle. Its equation is x 2 y 2 1.
y
(x, y)
1
circles
review section 1.1.
x
x2 y2 1
Distances on the Unit Circle
The circumference of a circle of radius r is 2 r . Thus for the unit circle,
where r 1, the circumference is 2. If a point starts at A and travels
around the circle (Fig. 1), it will travel a distance of 2. If it travels
1
halfway around the circle (Fig. 2), it will travel a distance of 2 2, or .
y
y
y
y
p
C
A
B
A
x
x
u
D
d
B
A
x
A
x
2p
Figure 1
Figure 2
Figure 3
Figure 4
If a point C travels 18 of the way around the circle (Fig. 3), it will
1
1
travel a distance of 8 2, or 4. Note that C is 4 of the way from A to B.
1
If a point D travels 6 of the way around the circle (Fig. 4), it will travel
1
1
a distance of 6 2, or 3. Note that D is 3 of the way from A to B.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
472
Chapter 5
1/10/05
1:01 PM
Page 472
• The Trigonometric Functions
EXAMPLE 1 How far will a point travel if it goes (a) 41 , (b) 121 , (c) 83 , and
(d) 65 of the way around the unit circle?
Solution
1
1
1
a) 4 of the total distance around the circle is 4 2, which is 2 ,
or 2.
1
1
b) The distance will be 12 2, which is 6 , or 6.
3
3
c) The distance will be 8 2, which is 4 , or 34.
5
5
d) The distance will be 6 2, which is 3 , or 53. Think of 53 as
2
3 .
These distances are illustrated in the following figures.
y
y
y
y
f
q
p
A
x
x
x
x
A point may travel completely around the circle and then continue.
1
For example, if it goes around once and then continues 4 of the way
1
around, it will have traveled a distance of 2 4 2, or 52 (Fig. 5).
Every real number determines a point on the unit circle. For the positive
number 10, for example, we start at A and travel counterclockwise a
distance of 10. The point at which we stop is the point “determined”
by the number 10. Note that 2 6.28 and that 10 1.62 . Thus
3
the point for 10 travels around the unit circle about 1 5 times (Fig. 6).
y
y
r
10
A
x
Figure 5
x
Figure 6
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 473
Section 5. 4
•
473
Radians, Arc Length, and Angular Speed
For a negative number, we move clockwise around the circle. Points
for 4 and 32 are shown in the figure below. The number 0 determines the point A.
y
y
w
A
A
x
x
d
EXAMPLE 2 On the unit circle, mark the point determined by each of
the following real numbers.
a)
9
4
b) 7
6
Solution
1
a) Think of 94 as 2 4 . (See the figure on the left below.) Since
94 0, the point moves counterclockwise. The point goes com1
pletely around once and then continues 4 of the way from A to B.
y
y
k
B
A
x
F
A
B
D
A
x
b) The number 76 is negative, so the point moves clockwise. From
6
A to B, the distance is , or 6 , so we need to go beyond B another
distance of 6, clockwise. (See the figure on the right above.)
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
474
Chapter 5
1/10/05
1:01 PM
Page 474
• The Trigonometric Functions
Radian Measure
Degree measure is a common unit of angle measure in many everyday
applications. But in many scientific fields and in mathematics (calculus,
in particular), there is another commonly used unit of measure called
the radian.
Consider the unit circle. Recall that this circle has radius 1. Suppose
we measure, moving counterclockwise, an arc of length 1, and mark a
point T on the circle.
y
T
Arc length is 1
u 1 radian
r1
x
1
If we draw a ray from the origin through T, we have formed an angle.
The measure of that angle is 1 radian. The word radian comes from the
word radius. Thus measuring 1 “radius” along the circumference of the
circle determines an angle whose measure is 1 radian. One radian is
about 57.3. Angles that measure 2 radians, 3 radians, and 6 radians are
shown below.
1 radian 57.3
y
y
u 2 radians
1
y
u 3 radians
1
u 6 radians
1
1
1
1
1
1
1
1
x
1
x
x
1
1
1
When we make a complete (counterclockwise) revolution, the terminal side coincides with the initial side on the positive x-axis. We then
have an angle whose measure is 2 radians, or about 6.28 radians, which
is the circumference of the circle:
2 r 2 1 2.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 475
Section 5. 4
•
Radians, Arc Length, and Angular Speed
475
Thus a rotation of 360 (1 revolution) has a measure of 2 radians. A half
revolution is a rotation of 180, or radians. A quarter revolution is a rotation of 90, or 2 radians, and so on.
y
u 90
q radians
1.57 radians
1
y
u 180
p radians
3.14 radians
1
x
y
u 270
w radians
4.71 radians
1
x
y
u 360
2p radians
6.28 radians
x
1
To convert between degrees and radians, we first note that
360 2 radians.
It follows that
180 radians.
To make conversions, we multiply by 1, noting that:
Converting Between Degree and Radian Measure
radians
180
1.
180
radians
radians
To convert from degree to radian measure, multiply by
.
180
180
To convert from radian to degree measure, multiply by
.
radians
GCM
EXAMPLE 3
Convert each of the following to radians.
a) 120
b) 297.25
Solution
a) 120 120 radians
180
Multiplying by 1
120
radians
180
2
radians, or about 2.09 radians
3
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x
BBEPMC05_0321279115.QXP
476
Chapter 5
1/10/05
1:01 PM
Page 476
• The Trigonometric Functions
b) 297.25 297.25 radians
180
297.25
radians
180
297.25
radians
180
5.19 radians
120°
We also can use a calculator set in RADIAN mode to convert the angle
measures. We enter the angle measure followed by ° (degrees) from the
ANGLE menu.
2.094395102
297.25°
5.187991202
GCM
Convert each of the following to degrees.
EXAMPLE 4
a)
3
radians
4
b) 8.5 radians
Solution
3
3
180
radians Multiplying by 1
a)
radians 4
4
radians
3
3
180 180 135
4
4
180
b) 8.5 radians 8.5 radians radians
8.5180
487.01
(3/4) r
135
8.5 r
487.0141259
With a calculator set in DEGREE mode, we can enter the angle measure
followed by r(radians) from the ANGLE menu.
The radian – degree equivalents of the most commonly used angle
measures are illustrated in the following figures.
y
y
135
f
180 p
h
225
90 60
45
qu
30
d
A 0
w
2p 360 x
j
270
315
270
315
w
j
h
2p 360
180 p
0
x
A
d
f
30
q u
45
135
90 60
225
When a rotation is given in radians, the word “radians” is optional
and is most often omitted. Thus if no unit is given for a rotation, the
rotation is understood to be in radians.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 477
Section 5. 4
•
Radians, Arc Length, and Angular Speed
477
We can also find coterminal, complementary, and supplementary
angles in radian measure just as we did for degree measure in Section 5.3.
EXAMPLE 5 Find a positive angle and a negative angle that are coterminal with 23. Many answers are possible.
Solution To find angles coterminal with a given angle, we add or subtract multiples of 2 :
2
2 6 8
2 ,
3
3
3
3
16
2
2 18
32 .
3
3
3
3
y
y
8p
3
i
i
16p
3
x
x
Thus, 83 and 163 are two of the many angles coterminal with
23.
EXAMPLE 6
Solution
Find the complement and the supplement of 6.
Since 90 equals 2 radians, the complement of 6 is
3
2
, or
2
6
6
6
6
.
3
Since 180 equals radians, the supplement of 6 is
6
5
.
6
6
6
6
Thus the complement of 6 is 3 and the supplement is 56.
Arc Length and Central Angles
y
s
s1
x
1
r
Radian measure can be determined using a circle other than a unit
circle. In the figure at left, a unit circle (with radius 1) is shown along
with another circle (with radius r, r 1). The angle shown is a central
angle of both circles.
From geometry, we know that the arcs that the angle subtends have
their lengths in the same ratio as the radii of the circles. The radii of the
circles are r and 1. The corresponding arc lengths are s and s1. Thus we
have the proportion
r
s
,
s1
1
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
478
Chapter 5
1/10/05
1:01 PM
Page 478
• The Trigonometric Functions
which also can be written as
s1
s
.
1
r
Now s1 is the radian measure of the rotation in question. It is common to
use a Greek letter, such as , for the measure of an angle or rotation and
the letter s for arc length. Adopting this convention, we rewrite the proportion above as
s
.
r
In any circle, the measure (in radians) of a central angle, the arc length
the angle subtends, and the length of the radius are related in this fashion.
Or, in general, the following is true.
Radian Measure
The radian measure of a rotation is the ratio of the distance s
traveled by a point at a radius r from the center of rotation, to the
length of the radius r:
s
.
r
y
r
u
s
x
When using the formula sr , must be in radians and s and
r must be expressed in the same unit.
EXAMPLE 7 Find the measure of a rotation in radians when a point
2 m from the center of rotation travels 4 m.
Solution
We have
s
r
4m
2.
2m
The unit is understood to be radians.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 479
Section 5. 4
y
Radians, Arc Length, and Angular Speed
479
EXAMPLE 8 Find the length of an arc of a circle of radius 5 cm associated with an angle of 3 radians.
s
r = 5 cm
•
Solution
θ =
3
x
We have
s
,
r
or s r .
Thus s 5 cm 3, or about 5.24 cm.
Linear Speed and Angular Speed
Linear speed is defined to be distance traveled per unit of time. If we
use v for linear speed, s for distance, and t for time, then
v
s
.
t
Similarly, angular speed is defined to be amount of rotation per unit
of time. For example, we might speak of the angular speed of a bicycle
wheel as 150 revolutions per minute or the angular speed of the earth
as 2 radians per day. The Greek letter (omega) is generally used for
angular speed. Thus for a rotation and time t, angular speed is defined as
.
t
As an example of how these definitions can be applied, let’s consider
the refurbished carousel at the Children’s Museum in Indianapolis, Indiana. It consists of three circular rows of animals. All animals, regardless of the row, travel at the same angular speed. But the animals in the
outer row travel at a greater linear speed than those in the inner rows.
What is the relationship between the linear speed v and the angular
speed ?
To develop the relationship we seek, recall that, for rotations measured in radians, sr . This is equivalent to
s r .
We divide by time, t, to obtain
s
r
Dividing by t
t
t
s
r
t
t
v
Now st is linear speed v and t is angular speed . Thus we have the
relationship we seek,
v r.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
480
Chapter 5
1/10/05
1:01 PM
Page 480
• The Trigonometric Functions
Linear Speed in Terms of Angular Speed
The linear speed v of a point a distance r from the center of rotation
is given by
v r,
where is the angular speed in radians per unit of time.
For the formula v r, the units of distance for v and r must be the
same, must be in radians per unit of time, and the units of time for
v and must be the same.
EXAMPLE 9 Linear Speed of an Earth Satellite. An earth satellite in
circular orbit 1200 km high makes one complete revolution every
90 min. What is its linear speed? Use 6400 km for the length of a radius
of the earth.
Solution
6400 km
To use the formula v r, we need to know r and :
r 6400 km 1200 km
1200 km
7600 km,
2
.
t
90 min 45 min
Radius of earth plus height
of satellite
We have, as usual, omitted
the word radians.
Now, using v r, we have
v 7600 km 7600 km
km
531
.
45 min
45
min
min
Thus the linear speed of the satellite is approximately 531 kmmin.
EXAMPLE 10 Angular Speed of a Capstan. An anchor is hoisted at a
rate of 2 ftsec as the chain is wound around a capstan with a 1.8-yd
diameter. What is the angular speed of the capstan?
1.8 yd
Capstan
Chain
Anchor
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 481
Section 5. 4
•
Radians, Arc Length, and Angular Speed
481
Solution We will use the formula v r in the form vr , taking
care to use the proper units. Since v is given in feet per second, we need
r in feet:
1.8
3 ft
d
yd 2.7 ft.
2
2
1 yd
r
Then will be in radians per second:
2 ftsec 2 ft
1
v
0.741sec.
r
2.7 ft
sec 2.7 ft
Thus the angular speed is approximately 0.741 radiansec.
The formulas t and v r can be used in combination to
find distances and angles in various situations involving rotational
motion.
EXAMPLE 11 Angle of Revolution. A 2004 Tundra V8 is traveling at
a speed of 65 mph. Its tires have an outside diameter of 30.56 in. Find
the angle through which a tire turns in 10 sec.
30.56 in.
Solution Recall that t , or t . Thus we can find if we know
and t. To find , we use the formula v r. The linear speed v of a
point on the outside of the tire is the speed of the Tundra, 65 mph. For
convenience, we first convert 65 mph to feet per second:
mi
1 hr
1 min 5280 ft
hr 60 min 60 sec 1 mi
ft
95.333
.
sec
v 65
The radius of the tire is half the diameter. Now r d2 30.56 in.2 15.28 in. We will convert to feet, since v is in feet per second:
r 15.28 in. 1 ft
12 in.
15.28
ft 1.27 ft.
12
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
482
Chapter 5
1/10/05
1:01 PM
Page 482
• The Trigonometric Functions
Using v r, we have
95.333
ft
1.27 ft ,
sec
so
Study Tip
The Student’s Solutions Manual
is an excellent resource if you
need additional help with an
exercise in the exercise sets. It
contains worked-out solutions to
the odd-numbered exercises in
the exercise sets.
95.333 ftsec 75.07
.
1.27 ft
sec
Then in 10 sec,
t 75.07
10 sec 751.
sec
Thus the angle, in radians, through which a tire turns in 10 sec is 751.
Answers to Exercises 1–4 can be found on p. IA-32.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
482
Chapter 5
5.4
1/10/05
1:01 PM
Page 482
• The Trigonometric Functions
Exercise Set
For each of Exercises 1– 4, sketch a unit circle and mark
the points determined by the given real numbers.
3
3
1. a)
b)
c)
4
2
4
17
11
d) e)
f)
4
4
2. a)
2
9
d)
4
6
10
d)
6
3. a)
4. a) 2
5
d) 2
5
b)
4
13
e)
4
2
3
14
e)
6
b)
3
b) 4
17
e) 6
Find two real numbers between 2 and 2 that
determine each of the points on the unit circle.
y
5.
M
1
x
c) 2
23
f)
4
7
6
23
f)
4
Q
P
N
y
6.
c)
5
c) 6
9
f) 4
N
M
x
Q
2 4
, ;
3
3
3
N: , ;
2
2
5 3
P: , ;
4
4
11
Q:
,
6
6
M:
M: , ;
7
N: , ;
4
4
4 2
P: , ;
3
3
7 5
Q: , 6
6
P
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Answers to Exercises 1– 4 can be found on p. IA-32.
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 483
Section 5. 4
For Exercises 7 and 8, sketch a unit circle and mark the
approximate location of the point determined by the
given real number.
7. a) 2.4 b) 7.5
c) 32
d) 320
8. a) 0.25 c) 47
b) 1.8
d) 500
Find a positive angle and a negative angle that are
coterminal with the given angle. Answers may vary.
5 11 9 7
9.
,
10.
,
4
3
4
4
3
3
7
11.
6
13. 19 5
,
6
6
2
3
4 8
,
3
3
12. 14. 3 , 3
4
5 11
,
4
4
Find the complement and the supplement.
5
15.
16.
3
12
17.
3
8
18.
4
19.
12
20.
6
•
41. 345 6.02
42. 75 1.31
43. 95 1.66
44. 24.8 0.43
483
Convert to degree measure. Round the answer to two
decimal places.
3
7
45. 135
46.
210
4
6
3
47. 8 1440
48. 49. 1 57.30
50. 17.6 1008.41
51. 2.347 134.47
52. 25 1432.39
53.
5
225
4
57.
60
54. 6 1080
55. 90 5156.62
56. 37.12 2126.82
2
51.43
7
58.
9
20
59. Certain positive angles are marked here in degrees.
Find the corresponding radian measures. y
90 60
45
30
135
0
360 x
180
Convert to radian measure. Leave the answer in terms
of .
21. 75 512
22. 30 6
23. 200 109
Radians, Arc Length, and Angular Speed
315
225
270
24. 135 34
25. 214.6 214.6180 26. 37.71 37.71180
27. 180 28. 90 2
29. 12.5 572
30. 6.3 7200
31. 340 179
32. 60 3
60. Certain negative angles are marked here in degrees.
Find the corresponding radian measures. Convert to radian measure. Round the answer to two
decimal places.
33. 240 4.19
34. 15 0.26
35. 60 1.05
y
270
225
315
360
0
x
180 p
36. 145 2.53
37. 117.8 2.06
38. 231.2 4.04
39. 1.354 0.02
40. 584 10.19
135
90
30
45
60
Answers to Exercises 7, 8, 15–20, 59, and 60 can be found on pp. IA-32 and IA-33.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
484
Chapter 5
1/10/05
1:01 PM
Page 484
• The Trigonometric Functions
Complete the table.
Round the answers to two decimal places.
Arc Length and Central Angles.
DISTANCE, S
(ARC LENGTH)
ANGLE, RADIUS, R
1
ft
2
61.
8 ft
62.
200 cm
254.65 cm
45
63.
16 yd
3.2 yd
5
64.
5.50 in.
4.2 in.
5
12
3
2.29
65. In a circle with a 120-cm radius, an arc 132 cm
long subtends an angle of how many radians? how
many degrees, to the nearest degree? 1.1; 63
71. Linear Speed. A flywheel with a 15-cm diameter
is rotating at a rate of 7 radianssec. What is the
linear speed of a point on its rim, in centimeters
per minute? 3150 cmmin
72. Linear Speed. A wheel with a 30-cm radius is
rotating at a rate of 3 radianssec. What is the
linear speed of a point on its rim, in meters
per minute? 54 mmin
73. Angular Speed of a Printing Press. This text was
printed on a four-color web heatset offset press. A
cylinder on this press has a 13.37-in. diameter. The
linear speed of a point on the cylinder’s surface is
18.33 feet per second. What is the angular speed of
the cylinder, in revolutions per hour? Printers often
refer to the angular speed as impressions per hour
(IPH). (Source: Scott Coulter, Quebecor World,
Taunton, MA) About 18,852 revolutions per hour
66. In a circle with a 10-ft diameter, an arc 20 ft long
subtends an angle of how many radians? how many
degrees, to the nearest degree? 4; 229
67. In a circle with a 2-yd radius, how long is an arc
associated with an angle of 1.6 radians? 3.2 yd
68. In a circle with a 5-m radius, how long is an arc
associated with an angle of 2.1 radians? 10.5 m
69. Angle of Revolution. Through how many radians
does the minute hand of a clock rotate from
12:40 P.M. to 1:30 P.M.? 5, or about 5.24
3
11
12
1
10
2
8
4
9
3
7
6
5
70. Angle of Revolution. A tire on a 2004 Saturn Ion
has an outside diameter of 24.877 in. Through what
angle (in radians) does the tire turn while traveling
1 mi? 5094
24.877 in.
74. Linear Speeds on a Carousel. When Alicia and Zoe
ride the carousel described earlier in this section,
Alicia always selects a horse on the outside row,
whereas Zoe prefers the row closest to the center.
These rows are 19 ft 3 in. and 13 ft 11 in. from
the center, respectively. The angular speed of the
carousel is 2.4 revolutions per minute. What is the
difference, in miles per hour, in the linear speeds
of Alicia and Zoe? (Source: The Children’s
Museum, Indianapolis, IN) 0.92 mph
75. Linear Speed at the Equator. The earth has a
4000-mi radius and rotates one revolution every
24 hr. What is the linear speed of a point on the
equator, in miles per hour? 1047 mph
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 485
Section 5. 4
76. Linear Speed of the Earth. The earth is about
93,000,000 mi from the sun and traverses its orbit,
which is nearly circular, every 365.25 days. What is
the linear velocity of the earth in its orbit, in miles
per hour? 66,659 mph
77. Determining the Speed of a River. A water wheel
has a 10-ft radius. To get a good approximation of
the speed of the river, you count the revolutions of
the wheel and find that it makes 14 revolutions per
minute (rpm). What is the speed of the river, in
miles per hour? 10 mph
•
Radians, Arc Length, and Angular Speed
485
79. John Deere Tractor. A rear wheel and tire on a
John Deere 8520 farm tractor has a 39-in. radius.
Find the angle (in radians) through which a wheel
rotates in 12 sec if the tractor is traveling at a speed
of 22 mph. About 119
39 in.
10 ft
Collaborative Discussion and Writing
80. Explain in your own words why it is preferable to
omit the word, or unit, radians in radian measures.
81. In circular motion with a fixed angular speed, the
length of the radius is directly proportional to the
linear speed. Explain why with an example.
78. The Tour de France. Lance Armstrong won the
2004 Tour de France bicycle race. The wheel of
his bicycle had a 67-cm diameter. His overall
average linear speed during the race was
40.560 kmh. What was the angular speed of
the wheel, in revolutions per hour? (Source:
tourdefrancenews.com) About 19,270 revolutionshr
82. Two new cars are each driven at an average speed
of 60 mph for an extended highway test drive of
2000 mi. The diameters of the wheels of the two
cars are 15 in. and 16 in., respectively. If the cars
use tires of equal durability and profile, differing
only by the diameter, which car will probably need
new tires first? Explain your answer.
Skill Maintenance
In each of Exercises 83–90, fill in the blanks with the
correct terms. Some of the given choices will not be used.
inverse
a relation
a horizontal line
vertical asymptote
a vertical line
horizontal asymptote
exponential function
even function
logarithmic function
odd function
natural
sine of common
cosine of logarithm
tangent of one-to-one
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
486
Chapter 5
1/10/05
1:01 PM
Page 486
• The Trigonometric Functions
83. The domain of a(n)
the range of the inverse f 1.
function f is
[4.1] one-to-one
84. The
is the length of the side
adjacent to divided by the length of the
hypotenuse. [5.1] cosine of 95. Angular Speed of a Gear Wheel. One gear wheel
turns another, the teeth being on the rims. The
wheels have 40-cm and 50-cm radii, and the
smaller wheel rotates at 20 rpm. Find the angular
speed of the larger wheel, in radians per second.
1.676 radianssec
85. The function fx a , where x is a real number,
,
a 0 and a 1, is called the
base a. [4.2] exponential function
x
86. The graph of a rational function may or may not
cross a(n)
. [3.5] horizontal
asymptote
87. If the graph of a function f is symmetric with
respect to the origin, we say that it is a(n)
. [1.7] odd function
88. Logarithms, base e, are called
logarithms. [4.3] natural
89. If it is possible for a(n)
to intersect
the graph of a function more than once, then the
function is not one-to-one and its
is not a function. [4.1] horizontal line; inverse
90. A(n)
[4.3] logarithm
50 cm
40 cm
96. Angular Speed of a Pulley. Two pulleys, 50 cm
and 30 cm in diameter, respectively, are connected
by a belt. The larger pulley makes 12 revolutions
per minute. Find the angular speed of the smaller
pulley, in radians per second. 2.094 radianssec
is an exponent.
Synthesis
91. On the earth, one degree of latitude is how many
kilometers? how many miles? (Assume that the
radius of the earth is 6400 km, or 4000 mi,
approximately.) 111.7 km; 69.8 mi
92. A point on the unit circle has y-coordinate
215. What is its x-coordinate? Check using
a calculator. 25
93. A mil is a unit of angle measure. A right angle has a
measure of 1600 mils. Convert each of the
following to degrees, minutes, and seconds.
a) 100 mils 53730
b) 350 mils 194115
94. A grad is a unit of angle measure similar to a
degree. A right angle has a measure of 100 grads.
Convert each of the following to grads.
a) 48 53.33 grads
5
b)
142.86 grads
7
30 cm
50 cm
97. Distance Between Points on the Earth. To find the
distance between two points on the earth when
their latitude and longitude are known, we can
use a right triangle for an excellent approximation
if the points are not too far apart. Point A is at
latitude 382730
N, longitude 825715
W;
and point B is at latitude 382845
N, longitude
825630
W. Find the distance from A to B in
nautical miles. (One minute of latitude is one
nautical mile.) 1.46 nautical miles
98. Hands of a Clock. At what times between
noon and 1:00 P.M. are the hands of a clock
perpendicular? 16.3636 min after 12:00 noon,
or about 12:16:22 P.M.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 487
Section 5.5
5.5
Circular
Functions: Graphs
and Properties
Study Tip
Take advantage of the numerous
detailed art pieces in this text.
They provide a visual image of the
concept being discussed. Taking
the time to study each figure is an
efficient way to learn and retain
the concepts.
•
Circular Functions: Graphs and Properties
487
Given the coordinates of a point on the unit circle, find its reflections
across the x-axis, the y-axis, and the origin.
Determine the six trigonometric function values for a real number when
the coordinates of the point on the unit circle determined by that real
number are given.
Find function values for any real number using a calculator.
Graph the six circular functions and state their properties.
The domains of the trigonometric functions, defined in Sections 5.1 and
5.3, have been sets of angles or rotations measured in a real number of
degree units. We can also consider the domains to be sets of real numbers, or radians, introduced in Section 5.4. Many applications in calculus
that use the trigonometric functions refer only to radians.
Let’s again consider radian measure and the unit circle. We defined
radian measure for as
s
.
r
y
(x, y)
When r 1,
s
,
1
1
or s .
su
u
x
The arc length s on the unit circle is the same as the radian measure of
the angle .
In the figure above, the point x, y is the point where the terminal
side of the angle with radian measure s intersects the unit circle. We can
now extend our definitions of the trigonometric functions using domains composed of real numbers, or radians.
In the definitions, s can be considered the radian measure of an angle
or the measure of an arc length on the unit circle. Either way, s is a real number. To each real number s, there corresponds an arc length s on the unit
circle. Trigonometric functions with domains composed of real numbers
are called circular functions.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
488
Chapter 5
1/10/05
1:01 PM
Page 488
• The Trigonometric Functions
Basic Circular Functions
For a real number s that determines a point x, y on the unit circle:
y
(x, y)
s
sin s second coordinate y ,
cos s first coordinate x ,
second coordinate
y
tan s x 0,
first coordinate
x
1
1
csc s y 0,
second coordinate
y
1
1
sec s x 0,
first coordinate
x
first coordinate
x
cot s y 0.
second coordinate
y
1
x
We can consider the domains of trigonometric functions to be real
numbers rather than angles. We can determine these values for a specific
real number if we know the coordinates of the point on the unit circle
determined by that number. As with degree measure, we can also find
these function values directly using a calculator.
y
Reflections on the Unit Circle
(35 , 45 )
1
x
Let’s consider the unit circle and a few of its points. For any point x, y on
the unit circle, x 2 y 2 1, we know that 1 x 1 and 1 y 1.
If we know the x- or y-coordinate of a point on the unit circle, we can find
3
the other coordinate. If x 5 , then
35 2 y 2 1
(35 , 45 )
y 2 1 259 16
25
y 45 .
3 4
3
4
Thus, 5 , 5 and 5 , 5 are points on the unit circle. There are two
3
points with an x-coordinate of 5 .
Now let’s consider the radian measure 3 and determine the coordinates of the point on the unit circle determined by 3. We construct a right
triangle by dropping a perpendicular segment from the point to the
x-axis.
y
y
30
(x, y)
1
uu
(x, y)
su
x
1
y
60
q
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 489
Section 5.5
•
489
Circular Functions: Graphs and Properties
Since 3 60, we have a 30– 60 right triangle in which the side
opposite the 30 angle is one half of the hypotenuse. The hypotenuse,
1
1
or radius, is 1, so the side opposite the 30 angle is 2 1, or 2 . Using the
Pythagorean theorem, we can find the other side:
1
2
2
y
y2 1
1
3
4
4
3
3
.
4
2
u
y2 1 y
q, 3
2 x
We know that y is positive since the point is in the first quadrant.
Thus the coordinates of the point determined by 3 are x 12 and
y 32, or 12, 32 . We can always check to see if a point is on
the unit circle by substituting into the equation x 2 y 2 1:
1
2
2
3
2
2
1
3
1.
4
4
Because a unit circle is symmetric with respect to the x-axis, the
y-axis, and the origin, we can use the coordinates of one point on the
unit circle to find coordinates of its reflections.
EXAMPLE 1 Each of the following points lies on the unit circle. Find
their reflections across the x-axis, the y-axis, and the origin.
a)
c)
a)
3 4
,
5 5
, 2
2
2
2 E, R
x
E, R
E, R
b)
2 2
,
2
2
1 3
,
2 2
Solution
b)
y
E, R
c)
y
2
, 2 2
2
y
q, 3
2 q, 3
2 x
x
2
, 2 2
2
2
, 2 2
2
q, 3
2 Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
q, 3
2 BBEPMC05_0321279115.QXP
490
Chapter 5
1/10/05
1:01 PM
Page 490
• The Trigonometric Functions
Finding Function Values
Knowing the coordinates of only a few points on the unit circle along
with their reflections allows us to find trigonometric function values of
the most frequently used real numbers, or radians.
y
(0, 1)
q,
q
u
(1, 0) p
3
2
d
2
, 2 2
2
3
2 , q
0
2p (1, 0) x
A
EXAMPLE 2
a) tan
3
3
4
4
d) cos
3
b) cos
c) sin (0, 1)
Find each of the following function values.
6
e) cot f ) csc 7
2
Solution We locate the point on the unit circle determined by the rotation, and then find its coordinates using reflection if necessary.
a) The coordinates of the point determined by 3
are 12, 32 .
b) The reflection of 22, 22 across the y-axis
is 22, 22 .
y
y
q,
3
2
, 2
2
2
2 u
2
, 2 2
2
d
f
x
Thus, tan
y
32
3.
3
x
12
x
3
2
x
.
4
2
d) The reflection of 12, 32 across the origin is
12, 32 .
Thus, cos
c) The reflection of 32, 12 across the x-axis is
32, 12 .
y
y
3
,q
2
q, 3
2 u
A
x
A
x
o
, q
3
2
Thus, sin 6
y
1
.
2
q, 3
2 Thus, cos
1
4
x .
3
2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 491
Section 5.5
e) The coordinates of the point determined by are
1, 0.
•
Circular Functions: Graphs and Properties
491
f ) The coordinates of the point determined by
72 are 0, 1.
y
y
(0, 1)
t
(1, 0) p
x
x
x
1
, which is not defined.
y
0
We can also think of cot as the reciprocal of
tan . Since tan yx 01 0 and the
reciprocal of 0 is not defined, we know that
cot is not defined.
Thus, cot Normal Sci Eng
Float 0123456789
Radian Degree
Func Par Pol Seq
Connected Dot
Sequential Simul
Real abi reˆθ i
Full Horiz G –T
Thus, csc 7
1
1
1.
2
y
1
Using a calculator, we can find trigonometric function values of any
real number without knowing the coordinates of the point that it determines on the unit circle. Most calculators have both degree and radian
modes. When finding function values of radian measures, or real numbers, we must set the calculator in RADIAN mode. (See the window at left.)
EXAMPLE 3 Find each of the following function values of radian measures using a calculator. Round the answers to four decimal places.
a) cos
2
5
b) tan 3
d) sec
c) sin 24.9
Solution
cos(2π/5)
tan(3)
.3090169944
a) cos
Using a calculator set in RADIAN mode, we find the values.
2
0.3090
5
b) tan 3 0.1425
.1425465431
sin(24.9)
.2306457059
7
c) sin 24.9 0.2306
7
1
1.1099
cos
7
Note in part (d) that the secant function value can be found by taking the reciprocal of the cosine value. Thus we can enter cos 7 and use
the reciprocal key.
d) sec
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
492
Chapter 5
1/10/05
1:01 PM
Page 492
• The Trigonometric Functions
We can graph the unit circle using a
graphing calculator. We use PARAMETRIC mode with the following window
and let X1T cos T and Y1T sin T. Here we use DEGREE mode.
EXPLORING WITH TECHNOLOGY
WINDOW
Tmin 0
Tmax 360
Tstep 15
Xmin 1.5
Xmax 1.5
Xscl 1
Ymin 1
Ymax 1
Yscl 1
1
X1Tcos(T)
Y1Tsin(T)
1.5
1.5
T 30
X .8660254
1
Y .5
Using the trace key and an arrow key to move the cursor around the unit
circle, we see the T, X, and Y values appear on the screen. What do they represent? Repeat this exercise in RADIAN mode. What do the T, X, and Y values
represent? (For more on parametric equations, see Section 9.7.)
From the definitions on p. 488, we can relabel any point x, y on the
unit circle as cos s, sin s, where s is any real number.
y
(x, y) (cos s, sin s)
s
(1, 0) x
Graphs of the Sine and Cosine Functions
y
, 2
2
2
2 (0, 1)
f
q,
3
2
2 2
2, 2
qu
3
d
, q
A 2
0 (1, 0)
p
2p
x
(1, 0)
A 3 , q
f
d
2
q u
2
2
2
2
2 , 2 2 , 2 (0, 1)
3
q, 2 Properties of functions can be observed from their graphs. We begin by
graphing the sine and cosine functions. We make a table of values, plot
the points, and then connect those points with a smooth curve. It is
helpful to first draw a unit circle and label a few points with coordinates.
We can either use the coordinates as the function values or find approximate sine and cosine values directly with a calculator.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 493
Section 5.5
•
493
Circular Functions: Graphs and Properties
s
sin s
cos s
s
sin s
cos s
0
6
4
3
2
34
54
32
74
2
0
0.5
0.7071
0.8660
1
0.7071
0
0.7071
1
0.7071
0
1
0.8660
0.7071
0.5
0
0.7071
1
0.7071
0
0.7071
1
0
6
4
3
2
34
54
32
74
2
0
0.5
0.7071
0.8660
1
0.7071
0
0.7071
1
0.7071
0
1
0.8660
0.7071
0.5
0
0.7071
1
0.7071
0
0.7071
1
The graphs are as follows.
y
y sin s
2
1
2p
w
p
q
q
p
w
2p
s
2p
s
1
2
The sine function
y
2
2p
w
p
q
y cos s
q
p
w
1
2
The cosine function
We can check these graphs using a graphing calculator.
y cos x
y sin x
2
2
2 π
2π
2
Xscl 2 π
π
2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2π
2
Xscl π
2
BBEPMC05_0321279115.QXP
494
Chapter 5
1/10/05
1:01 PM
Page 494
• The Trigonometric Functions
The sine and cosine functions are continuous functions. Note in the
graph of the sine function that function values increase from 0 at s 0
to 1 at s 2, then decrease to 0 at s , decrease further to 1 at
s 32, and increase to 0 at 2. The reverse pattern follows when s
decreases from 0 to 2. Note in the graph of the cosine function that
function values start at 1 when s 0, and decrease to 0 at s 2. They
decrease further to 1 at s , then increase to 0 at s 32, and
increase further to 1 at s 2. An identical pattern follows when s decreases from 0 to 2.
From the unit circle and the graphs of the functions, we know that
the domain of both the sine and cosine functions is the entire set of real
numbers, , . The range of each function is the set of all real numbers from 1 to 1, 1, 1.
Domain and Range of Sine and Cosine Functions
The domain of the sine and cosine functions is , .
The range of the sine and cosine functions is 1, 1.
3
2
2π
3
3
2
2π
3
EXPLORING WITH TECHNOLOGY Another way to construct the sine and
cosine graphs is by considering the unit circle and transferring vertical distances for the sine function and horizontal distances for the cosine function.
Using a graphing calculator, we can visualize the transfer of these distances.
We use the calculator set in PARAMETRIC and RADIAN modes and let
X1T cos T 1 and Y1T sin T for the unit circle centered at 1, 0 and
X2T T and Y2T sin T for the sine curve. Use the following window
settings.
Tmin 0
Tmax 2
Tstep .1
Xmin 2
Xmax 2
Xscl 2
Ymin 3
Ymax 3
Yscl 1
With the calculator set in SIMULTANEOUS mode, we can actually watch the
sine function (in red) “unwind” from the unit circle (in blue). In the two
screens at left, we partially illustrate this animated procedure.
Consult your calculator’s instruction manual for specific keystrokes and
graph both the sine curve and the cosine curve in this manner. (For more on
parametric equations, see Section 9.7.)
A function with a repeating pattern is called periodic. The sine
and cosine functions are examples of periodic functions. The values of
these functions repeat themselves every 2 units. In other words, for
any s, we have
sin s 2 sin s
and
cos s 2 cos s .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 495
Section 5.5
•
Circular Functions: Graphs and Properties
495
To see this another way, think of the part of the graph between 0 and
2 and note that the rest of the graph consists of copies of it. If we translate the graph of y sin x or y cos x to the left or right 2 units, we
will obtain the original graph. We say that each of these functions has a
period of 2.
Periodic Function
A function f is said to be periodic if there exists a positive constant p
such that
fs p fs
for all s in the domain of f. The smallest such positive number p is
called the period of the function.
y
s
s 2p
T
x
The period p can be thought of as the length of the shortest recurring
interval.
We can also use the unit circle to verify that the period of the sine
and cosine functions is 2. Consider any real number s and the point T
that it determines on a unit circle, as shown at left. If we increase s by
2, the point determined by s 2 is again the point T. Hence for any
real number s,
sin s 2 sin s
cos s 2 cos s .
and
It is also true that sin s 4 sin s , sin s 6 sin s , and so on.
In fact, for any integer k, the following equations are identities:
sin s k2 sin s
and cos s k2 cos s ,
or
sin s sin s 2k and
cos s cos s 2k .
The amplitude of a periodic function is defined as one half of the
distance between its maximum and minimum function values. It is always positive. Both the graphs and the unit circle verify that the maximum value of the sine and cosine functions is 1, whereas the minimum
value of each is 1. Thus,
1
the amplitude of the sine function 2 1 1 1
y
y sin x
2p
1
p
Amplitude: 1
p
1
Period: 2p
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2p x
BBEPMC05_0321279115.QXP
496
Chapter 5
1/10/05
1:01 PM
Page 496
• The Trigonometric Functions
and
the amplitude of the cosine function is 12 1 1 1.
y
y cos x
1
Amplitude: 1
p
2p
p
2p x
1
Period: 2p
Using the TABLE feature on a graphing
calculator, compare the y-values for y1 sin x and y2 sin x and for
y3 cos x and y4 cos x. We set TblMin 0 and Tbl 12.
EXPLORING WITH TECHNOLOGY
X
Y1
Y2
X
Y3
Y4
0
.2618
.5236
.7854
1.0472
1.309
1.5708
0
.25882
.5
.70711
.86603
.96593
1
0
.2588
.5
.7071
.866
.9659
1
0
.2618
.5236
.7854
1.0472
1.309
1.5708
1
.96593
.86603
.70711
.5
.25882
0
1
.96593
.86603
.70711
.5
.25882
0
X0
X0
What appears to be the relationship between sin x and sin x and
between cos x and cos x?
Consider any real number s and its opposite, s. These numbers determine points T and T1 on a unit circle that are symmetric with respect
to the x-axis.
y
y
s
s
T(x, y)
T(x, y)
sin s
cos s
x
sin (s)
x
T1(x, y)
T1(x, y)
cos (s)
s
s
Because their second coordinates are opposites of each other, we
know that for any number s,
sin s sin s .
Because their first coordinates are the same, we know that for any
number s,
cos s cos s .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 497
Section 5.5
•
Circular Functions: Graphs and Properties
497
Thus we have shown that the sine function is odd and the cosine function is even.
The following is a summary of the properties of the sine and cosine
functions.
even and odd functions
review section 1.7.
CONNECTING THE CONCEPTS
COMPARING THE SINE AND COSINE
FUNCTIONS
SINE FUNCTION
COSINE FUNCTION
y sin x
2p
1.
2.
3.
4.
5.
6.
p
y
1
y
y cos x
1
1
2p
p
2p
x
Continuous
Period: 2
Domain: All real numbers
Range: 1, 1
Amplitude: 1
Odd: sin s sin s
1.
2.
3.
4.
5.
6.
p
1
p
2p x
Continuous
Period: 2
Domain: All real numbers
Range: 1, 1
Amplitude: 1
Even: cos s cos s
Graphs of the Tangent, Cotangent,
Cosecant, and Secant Functions
To graph the tangent function, we could make a table of values using a
calculator, but in this case it is easier to begin with the definition of tangent and the coordinates of a few points on the unit circle. We recall that
tan s sin s
y
.
x
cos s
y
y
(x, y) (cos s, sin s)
y sin s
s tan s x cos s
(1, 0)
x
(0, 1)
2
, 2 2
2
2
, 2 2
2
(1, 0)
x
(1, 0)
2
, 2 2
2
(0, 1)
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2
, 2 2
2
BBEPMC05_0321279115.QXP
498
Chapter 5
1/10/05
1:01 PM
Page 498
• The Trigonometric Functions
The tangent function is not defined when x, the first coordinate, is 0.
That is, it is not defined for any number s whose cosine is 0:
s
3
5
, , ,....
2
2
2
We draw vertical asymptotes at these locations (see Fig. 1 below).
y
y
2
2
1
1
2pw p q
q
p
w 2p
s
2pw p q
q
p
w 2p s
1
2
2
Figure 1
Figure 2
We also note that
tan s 0 at s 0, , 2, 3, . . . ,
7
3 5 9
tan s 1 at s . . . , , ,
,...,
,
4
4 4 4 4
9
5
3 7
tan s 1 at s . . . , , ,
,....
,
4
4
4 4 4
We can add these ordered pairs to the graph (see Fig. 2 above) and investigate the values in 2, 2 using a calculator. Note that the function value is 0 when s 0, and the values increase without bound as s
increases toward 2. The graph gets closer and closer to an asymptote as
s gets closer to 2, but it never touches the line. As s decreases from 0 to
2, the values decrease without bound. Again the graph gets closer
and closer to an asymptote, but it never touches it. We now complete
the graph.
y tan s
y
y tan x
2
2
1
2 π
2π
2p
w
p
q
q
p
w
1
DOT Mode
2
2
The tangent function
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2p
s
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 499
Section 5.5
•
499
Circular Functions: Graphs and Properties
From the graph, we see that the tangent function is continuous
except where it is not defined. The period of the tangent function is .
Note that although there is a period, there is no amplitude because there
are no maximum and minimum values. When cos s 0, tan s is not defined tan s sin scos s. Thus the domain of the tangent function is
the set of all real numbers except 2 k, where k is an integer. The
range of the function is the set of all real numbers.
The cotangent function cot s cos ssin s is not defined when y,
the second coordinate, is 0 — that is, it is not defined for any number s
whose sine is 0. Thus the cotangent is not defined for s 0, , 2,
3, . . . . The graph of the function is shown below.
y cot s
y
y cot x 1/tan x
2
2
1
2 π
2π
2p
w
p
q
q
p
w
2p
s
1
DOT Mode
2
The cotangent function
The cosecant and sine functions are reciprocal functions, as are the
secant and cosine functions. The graphs of the cosecant and secant functions can be constructed by finding the reciprocals of the values of the
sine and cosine functions, respectively. Thus the functions will be positive together and negative together. The cosecant function is not defined
for those numbers s whose sine is 0. The secant function is not defined
for those numbers s whose cosine is 0. In the graphs below, the sine and
cosine functions are shown by the gray curves for reference.
y csc s
y csc x 1/sin x
y
y sin s 2
2
1
2 π
2π
2p
w
p
q
q
p
1
DOT Mode
2
The cosecant function
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
w
2p
s
BBEPMC05_0321279115.QXP
500
Chapter 5
1/10/05
1:01 PM
Page 500
• The Trigonometric Functions
y
y sec s
y sec x 1/cos x
y cos s
2
2
2 π
2π
2p
w
p
q
q
p
w
2p
s
1
DOT Mode
2
2
The secant function
The following is a summary of the basic properties of the tangent,
cotangent, cosecant, and secant functions. These functions are continuous except where they are not defined.
CONNECTING THE CONCEPTS
COMPARING THE TANGENT, COTANGENT, COSECANT, AND SECANT FUNCTIONS
TANGENT FUNCTION
COTANGENT FUNCTION
1. Period: 2. Domain: All real numbers except 2 k,
where k is an integer
3. Range: All real numbers
COSECANT FUNCTION
1. Period: 2. Domain: All real numbers except k, where k
is an integer
3. Range: All real numbers
SECANT FUNCTION
1. Period: 2
2. Domain: All real numbers except k, where k
is an integer
3. Range: , 1 1, 1. Period: 2
2. Domain: All real numbers except 2 k,
where k is an integer
3. Range: , 1 1, In this chapter, we have used the letter s for arc length and have
avoided the letters x and y, which generally represent first and second
coordinates. Nevertheless, we can represent the arc length on a unit circle
by any variable, such as s, t, x, or . Each arc length determines a point
that can be labeled with an ordered pair. The first coordinate of that
ordered pair is the cosine of the arc length, and the second coordinate is
the sine of the arc length. The identities we have developed hold no matter what symbols are used for variables — for example, cos s cos s ,
cos x cos x , cos cos , and cos t cos t .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 501
Section 5.5
5.5
3.
2
21
,
5
5
4. 1
3
,
2
2
5. The number 4 determines a point on the unit
circle with coordinates 22, 22 . What are the
coordinates of the point determined by 4? 6. A number determines a point on the unit circle
with coordinates 23, 53 . What are the
coordinates of the point determined by ? Find the function value using coordinates of points on
the unit circle. Give exact answers.
1
7. sin 0
8. cos 3
2
9. cot
7
6
501
3
10. tan
14. tan 15. sec
2
Not defined
16. cos 10
17. cos
6
3
2
18. sin
2
3
19. sin
5
4
20. cos
11
6
1
2
4
1
1
3
2
3
2
21. sin 5 0
22. tan
3
2
Not defined
5
2
24. tan
5
3
3
0
1.3065
28. sin 11.7 0.7620
29. cot 342
2.1599
30. tan 1.3
31. cos 6
1
32. sin
0.3090
10
34. sec
10
7
33. csc 4.16 1.1747
35. tan
7
4
1
41. tan
5
6
2
2
27. sec 37
11
4
13. cos
11. sin 3 0
3
2
Find the function value using a calculator set in
RADIAN mode. Round the answer to four decimal
places, where appropriate.
2
25. tan
0.4816
26. cos 0.3090
7
5
37. sin 3
12. csc
4
23. cot
Circular Functions: Graphs and Properties
Exercise Set
The following points are on the unit circle. Find the
coordinates of their reflections across (a) the x-axis,
(b) the y-axis, and (c) the origin.
3 7
2 5
1. ,
2.
,
4 4
3 3
•
39. sin 0
2
9
4
3.6021
4.4940
36. cos 2000 0.3675
0.7071
38. cot 7
Not defined
40. cos 29 0.7481
0
0.8391
42. sin
8
3
0.8660
In Exercises 43 – 48, make hand-drawn graphs.
43. a) Sketch a graph of y sin x . b) By reflecting the graph in part (a), sketch a graph
of y sin x. c) By reflecting the graph in part (a), sketch a graph
of y sin x . Same as (b)
d) How do the graphs in parts (b) and (c) compare?
The same
44. a) Sketch a graph of y cos x . b) By reflecting the graph in part (a), sketch a graph
of y cos x. Same as (a)
c) By reflecting the graph in part (a), sketch a graph
of y cos x . d) How do the graphs in parts (a) and (b) compare?
The same
45. a) Sketch a graph of y sin x . See Exercise 43(a).
b) By translating, sketch a graph of
y sin x . c) By reflecting the graph of part (a), sketch a graph
of y sin x . Same as (b)
d) How do the graphs of parts (b) and (c) compare?
The same
Answers to Exercises 1–6, 43(a), 43(b), 44(a), 44(c), and 45(b) can be found on p. IA-33.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
502
Chapter 5
1/10/05
1:01 PM
Page 502
• The Trigonometric Functions
46. a) Sketch a graph of y sin x . See Exercise 47(a).
b) By translating, sketch a graph of
y sin x . c) By reflecting the graph of part (a), sketch a graph
of y sin x . Same as (b)
d) How do the graphs of parts (b) and (c) compare?
Skill Maintenance
47. a) Sketch a graph of y cos x . b) By translating, sketch a graph of
y cos x . c) By reflecting the graph of part (a), sketch a graph
of y cos x . Same as (b)
d) How do the graphs of parts (b) and (c) compare?
59. fx x, gx 12 x 4 1 The same
The same
48. a) Sketch a graph of y cos x . See Exercise 47(a).
b) By translating, sketch a graph of
y cos x . c) By reflecting the graph of part (a), sketch a graph
of y cos x . Same as (b)
d) How do the graphs of parts (b) and (c) compare?
The same
Graph both functions in the same viewing window and
describe how g is a transformation of f.
57. fx x 2, gx 2x 2 3 58. fx x 2, gx x 22 60. fx x , gx x
3
3
Write an equation for a function that has a graph with
the given characteristics. Check using a graphing
calculator.
61. The shape of y x 3, but reflected across the x-axis,
shifted right 2 units, and shifted down 1 unit
3
[1.7] y x 2 1
62. The shape of y 1x , but shrunk vertically by a
factor of 14 and shifted up 3 units
[1.7] y Synthesis
1
3
4x
49. Of the six circular functions, which are even?
Which are odd? Complete. (For example, sin x 2 sin x .)
63. cos x cos x
50. Of the six circular functions, which have period ?
Which have period 2 ? 64. sin x 65. sin x 2k , k sin x
Consider the coordinates on the unit circle for
Exercises 51–54.
51. In which quadrants is the tangent function positive?
negative? Positive: I, III; negative: II, IV
66. cos x 2k , k cos x
sin x
67. sin x sin x
68. cos x cos x
52. In which quadrants is the sine function positive?
negative? Positive: I, II; negative: III, IV
69. cos x cos x
70. cos x cos x
53. In which quadrants is the cosine function positive?
negative? Positive: I, IV; negative: II, III
71. sin x sin x
72. sin x sin x
54. In which quadrants is the cosecant function positive?
negative? Positive: I, II; negative: III, IV
Collaborative Discussion and Writing
73. Find all numbers x that satisfy the following. Check
using a graphing calculator.
a) sin x 1 2 2k, k b) cos x 1 2k, k c) sin x 0 k, k 55. Describe how the graphs of the sine and cosine
functions are related.
74. Find f g and g f , where fx x 2 2x and
gx cos x . 56. Explain why both the sine and cosine functions are
continuous, but the tangent function, defined as
sinecosine, is not continuous.
Use a graphing calculator to determine the domain, the
range, the period, and the amplitude of the function.
75. y sin x2 76. y cos x 1 Answers to Exercises 46(b), 47(a), 47(b), 48(b), 49, 50, 57–60, and 74–76 can be found on p. IA-33.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:01 PM
Page 503
Section 5.5
•
Circular Functions: Graphs and Properties
y
Determine the domain of the function.
77. fx cos x
79. fx sin x
cos x
78. gx C
1
sin x
83. y sin x cos x
1
D
u
80. gx log sin x Graph.
81. y 3 sin x E
P
x x k, k an integer.
503
O
A B
x
82. y sin x 84. y cos x 85. One of the motivations for developing trigonometry
with a unit circle is that you can actually “see” sin and cos on the circle. Note in the figure at right
that AP sin and OA cos . It turns out that
you can also “see” the other four trigonometric
functions. Prove each of the following.
a) BD tan b) OD sec c) OE csc d) CE cot 86. Using graphs, determine all numbers x that satisfy
sin x cos x . 87. Using a calculator, consider sin xx , where x is
between 0 and 2. As x approaches 0, this function
approaches a limiting value. What is it? 1
Answers to Exercises 77 and 79 – 86 can be found on pp. IA-33 and IA-34.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
522
Chapter 5
1/10/05
1:02 PM
Page 522
• The Trigonometric Functions
Chapter 5 Summary and Review
Important Properties and Formulas
Trigonometric Function Values of an Acute Angle Let be an acute angle of a right triangle. The six
trigonometric functions of are as follows:
opp
,
hyp
hyp
csc ,
opp
adj
,
hyp
hyp
sec ,
adj
sin cos Hypotenuse
Opposite u
u
Adjacent to u
opp
,
adj
adj
cot .
opp
tan Reciprocal Functions
csc 1
,
sin sec 1
,
cos cot 1
tan Function Values of Special Angles
0
30
45
60
90
sin
0
12
22
32
1
cos
1
32
22
12
0
tan
0
33
1
3
Not defined
Cofunction Identities
90 u
u
sin cos 90 ,
tan cot 90 ,
sec csc 90 ,
cos sin 90 ,
cot tan 90 ,
csc sec 90 (continued)
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:02 PM
Page 523
Chapter 5
•
Summary and Review
Trigonometric Functions of Any Angle If Px, y is any point on the terminal side of any angle in standard position,
and r is the distance from the origin to Px, y, where r x 2 y 2, then
y
,
r
r
csc ,
y
x
,
r
r
sec ,
x
sin cos Positive: sin
Negative: cos, tan
y
u
x
x
Basic Circular Functions
For a real number s that determines a point x, y
on the unit circle:
sin s y ,
cos s x ,
y
tan s .
x
y
(x, y)
Positive: All
Negative: None
s
I
x
(, )
x
(, )
(, )
III
IV
Positive: tan
Negative: sin, cos
Positive: cos
Negative: sin, tan
Radian–Degree Equivalents
y
135
f
180 p
h
225
P(x, y)
r
y
y
,
x
x
cot .
y
1
II (, )
y
tan Signs of Function Values
The signs of the function values depend only on
the coordinates of the point P on the terminal side
of an angle.
523
90 60
45
qu
30
d
A 0
w
2p 360 x
j
270
315
Sine is an odd function: sin s sin s .
Cosine is an even function: cos s cos s .
Transformations of Sine and Cosine
Functions
To graph y A sin Bx C D and
y A cos Bx C D :
1. Stretch or shrink the graph horizontally
2
according to B. Period B
2. Stretch or shrink the graph vertically according
to A. (Amplitude A)
3. Translate the graph horizontally according to
C
CB. Phase shift B
4. Translate the graph vertically according to D.
Linear Speed in Terms of Angular Speed
v r
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
524
Chapter 5
1/10/05
1:02 PM
Page 524
• The Trigonometric Functions
Review Exercises
1. Find the six trigonometric function values of . 8
u
3
27. a 30.5, B 51.17 [5.2] A 38.83, b 37.9,
73
c 48.6
91
, find the other five
10
trigonometric function values. 2. Given that sin Find the exact function value, if it exists.
3. cos 45 [5.1] 22
4. cot 60 [5.1] 33
5. cos 495 [5.3] 22
6. sin 150 [5.3] 12
7. sec 270
8. tan 600 [5.3] 3
[5.3] Not defined
9. csc 60 [5.1] 233
Solve each of the following right triangles. Standard
lettering has been used.
26. a 7.3, c 8.6 [5.2] b 4.5, A 58.1, B 31.9
28. One leg of a right triangle bears east. The
hypotenuse is 734 m long and bears N5723E.
Find the perimeter of the triangle. [5.2] 1748 m
29. An observer’s eye is 6 ft above the floor. A mural is
being viewed. The bottom of the mural is at floor
level. The observer looks down 13 to see the bottom
and up 17 to see the top. How tall is the mural?
[5.2] 14 ft
10. cot 45 [5.1] 1
11. Convert 22.27 to degrees, minutes, and seconds.
Round to the nearest second. [5.1] 221612
17°
13°
6 ft
12. Convert 473327
to decimal degree notation.
Round to two decimal places. [5.1] 47.56
Find the function value. Round to four decimal places.
13. tan 2184 [5.3] 0.4452 14. sec 27.9 [5.3] 1.1315
15. cos 181342
16. sin 24524
17. cot 33.2
18. sin 556.13
[5.3] 0.9092
[5.3] 0.9498
[5.3] 1.5292
[5.3] 0.2778
Find in the interval indicated. Round the answer to
the nearest tenth of a degree.
19. cos 0.9041, 180, 270 [5.3] 205.3
20. tan 1.0799, 0, 90 [5.3] 47.2
Find the exact acute angle , in degrees, given the
function value.
3
21. sin [5.1] 60 22. tan 3 [5.1] 60
2
23. cos 2
[5.1] 45
2
24. sec For angles of the following measures, state in which
quadrant the terminal side lies.
30. 142115
[5.3] II
31. 635.2 [5.3] I
32. 392 [5.3] IV
Find a positive angle and a negative angle that are
coterminal with the given angle. Answers may vary.
7
5
33. 65 [5.3] 425, 295 34.
[5.4] , 3
3
3
Find the complement and the supplement.
35. 13.4 36.
6
37. Find the six trigonometric function values for the
angle shown. 23
[5.1] 30
3
25. Given that sin 59.1 0.8581, cos 59.1 0.5135, and
tan 59.1 1.6709, find the six function values for
30.9. y
(2, 3)
3
r
2
u
x
Answers to Exercises 1, 2, 25, and 35–37 can be found on pp. IA-35 and IA-36.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:02 PM
Page 525
Chapter 5
38. Given that tan 25 and that the terminal side
is in quadrant III, find the other five function
values. 54. sin
5
3
56. tan
6
[5.5] •
3
2
55. sin 3
3
525
Review Exercises
7
6
1
[5.5] 2
57. cos 13 [5.5] 1
39. An airplane travels at 530 mph for 3 12 hr in a
direction of 160 from Minneapolis, Minnesota. At
the end of that time, how far south of Minneapolis
is the airplane? [5.3] About 1743 mi
Find the function value. Round to four decimal places.
58. sin 24 [5.5] 0.9056
59. cos 75 [5.5] 0.9218
40. On a unit circle, mark and label the points
determined by 76, 34, 3, and 94.
60. cot 16
For angles of the following measures, convert to radian
measure in terms of , and convert to radian measure
not in terms of . Round the answer to two decimal
places.
41. 145.2 [5.4] 121
, 2.53 42. 30 [5.4] ,
150
0.52
6
Convert to degree measure. Round the answer to two
decimal places.
3
43.
[5.4] 270
44. 3 [5.4] 171.89
2
45. 4.5 [5.4] 257.83
46. 11
[5.4] 1980
47. Find the length of an arc of a circle, given a central
angle of 4 and a radius of 7 cm.
[5.4] 74, or 5.5 cm
48. An arc 18 m long on a circle of radius 8 m subtends
an angle of how many radians? how many degrees,
to the nearest degree? [5.4] 2.25, 129
49. At one time, inside La Madeleine French Bakery and
Cafe in Houston, Texas, there was one of the few
remaining working watermills in the world. The
300-yr-old French-built waterwheel had a radius of
7 ft and made one complete revolution in 70 sec.
What was the linear speed in feet per minute of a
point on the rim? (Source: La Madeleine French
Bakery and Cafe, Houston, TX)
[5.4] About 37.9 ftmin
50. An automobile wheel has a diameter of 14 in. If the
car travels at a speed of 55 mph, what is the angular
velocity, in radians per hour, of a point on the edge
of the wheel? [5.4] 497,829 radianshr
51. The point is on a unit circle. Find the
coordinates of its reflections across the x-axis, the
y-axis, and the origin. [5.5] 35 , 45 , 35 , 45 , 35 , 45 3
5,
45
Find the exact function value, if it exists.
5
52. cos [5.5] 1
53. tan
[5.5] 1
4
[5.5]
61. tan
[5.5] Not defined
3
7
[5.5] 6.1685 63. cos 62. sec 14.3
[5.5] 4.3813
5
[5.5] 0.8090
64. Graph by hand each of the six trigonometric
functions from 2 to 2. 65. What is the period of each of the six trigonometric
functions? [5.5] Period of sin, cos, sec, csc: 2 ;
period of tan, cot: 66. Complete the following table.
FUNCTION
DOMAIN
RANGE
sine
, 1, 1
cosine
, 1, 1
, tangent
Domain of tangent: x x 2 k, k 67. Complete the following table with the sign of the
specified trigonometric function value in each of the
four quadrants.
FUNCTION
I
II
III
IV
sine
cosine
tangent
Determine the amplitude, the period, and the phase
shift of the function, and sketch the graph of the function. Then check the graph using a graphing calculator.
68. y sin x 2
69. y 3 1
cos 2x 2
2
Answers to Exercises 38, 40, 64, 68, and 69 can be found on p. IA-36.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
526
Chapter 5
1/10/05
1:02 PM
Page 526
• The Trigonometric Functions
In Exercises 70–73, without a graphing calculator,
match the function with one of the graphs (a)– (d),
which follow. Then check your work using a graphing
calculator.
y
y
a)
b)
2
2
2
1
1
2 x
2
1
2
2
77. Does 5 sin x 7 have a solution for x ? Why or
why not? 78. Explain the disadvantage of a graphing calculator
when graphing a function like
1
76. Describe the shape of the graph of the cosine
function. How many maximum values are there of
the cosine function? Where do they occur? 2 x
fx sin x
. x
Synthesis
y
c)
All values
80. Graph y 3 sin x2, and determine the domain,
the range, and the period. 2
2
2
79. For what values of x in 0, 2 is sin x x true?
y
d)
1
2 x
2
2
4
1
6
2
81. In the graph below, y1 sin x is shown and y2 is
shown in red. Express y2 as a transformation of the
graph of y1.
2 x
[5.6] y2 2 sin x y1 sin x, y2 ?
2
2
4
70. y cos 2x
[5.6] (d)
71. y 1
sin x 1
2
[5.6] (a)
72. y 2 sin
[5.6] (c)
1
x3
2
73. y cos x [5.6] (b)
2 π
2
74. Sketch a graph of y 3 cos x sin x for values of x
between 0 and 2. Collaborative Discussion and Writing
75. Compare the terms radian and degree.
2π
4
82. Find the domain of y log cos x. 83. Given that sin x 0.6144 and that the terminal side
is in quadrant II, find the other basic circular
function values. [5.3] cos x 0.7890,
tan x 0.7787, cot x 1.2842,
sec x 1.2674, csc x 1.6276
Answers to Exercises 74–78, 80, and 82 can be found on p. IA-36.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
526
Chapter 5
1/10/05
1:02 PM
Page 526
• The Trigonometric Functions
Chapter 5 Test
1. Find the six trigonometric function values of . 65
Find the exact function value, if it exists.
2. sin 120 [5.3] 32
3. tan 45 [5.3] 1
4. cos 3
4
7
[5.4] 1
5. sec
5
4
[5.4] 2
6. Convert 382756
to decimal degree notation.
Round to two decimal places. [5.1] 38.47
Answer to Exercise 1 can be found on p. IA-36.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC05_0321279115.QXP
1/10/05
1:02 PM
Page 527
Chapter 5
Find the function values. Round to four decimal places.
7. tan 526.4
8. sin 12
[5.3] 0.2419
9. sec
5
9
[5.3] 0.2079
[5.4] 5.7588
23. Which is the graph of the function?
y
a)
b)
10. cos 76.07 [5.4] 0.7827
11. Find the exact acute angle , in degrees, for which
sin 12 . [5.1] 30
12. Given that sin 28.4 0.4756, cos 28.4 0.8796,
and tan 28.4 0.5407, find the six trigonometric
function values for 61.6. 2
[5.2] B 54.1, a 32.6, c 55.7
2
15. Find the supplement of
1
2 x
5
. [5.4]
6
6
2
1
1
2
2
2
1
1
2 x
2
2 x
2 x
y
d)
2
[5.3] Answers may vary; 472, 248
y
1
y
c)
[5.6] (c)
2
527
Test
2
13. Solve the right triangle with b 45.1 and A 35.9.
Standard lettering has been used.
14. Find a positive angle and a negative angle coterminal
with a 112 angle.
•
1
1
2
2
16. Given that sin 441 and that the terminal
side is in quadrant IV, find the other five
trigonometric function values. 24. Height of a Kite. The angle of elevation of a kite is
65 with 490 ft of string out. Assuming the string is
taut, how high is the kite? [5.2] About 444 ft
17. Convert 210 to radian measure in terms of .
[5.4] 76
3
18. Convert
to degree measure. [5.4] 135
4
25. Location. A pickup-truck camper travels at
50 mph for 6 hr in a direction of 115 from Buffalo,
Wyoming. At the end of that time, how far east of
Buffalo is the camper? [5.2] About 272 mi
19. Find the length of an arc of a circle given a central
angle of 3 and a radius of 16 cm.
[5.4] 163 16.755 cm
26. Linear Speed. A ferris wheel has a radius of 6 m
and revolves at 1.5 rpm. What is the linear speed,
in meters per minute? [5.4] 18 56.55 mmin
Consider the function y sin x 2 1 for
Exercises 20–23.
20. Find the amplitude. [5.5] 1
21. Find the period.
[5.5] 2
Synthesis
27. Determine the domain of fx 22. Find the phase shift. [5.5] 2
Answers to Exercises 12, 16, and 27 can be found on p. IA-37.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3
cos x
.
BBEPMC06_0321279115.QXP
1/10/05
1:29 PM
Page 529
Trigonometric Identities,
Inverse Functions, and
Equations
6.1
6.2
6.3
6.4
6.5
Identities: Pythagorean and
Sum and Difference
Identities: Cofunction, Double-Angle,
and Half-Angle
Proving Trigonometric Identities
Inverses of the Trigonometric Functions
Solving Trigonometric Equations
6
SUMMARY AND REVIEW
TEST
A P P L I C A T I O N
T
he number of daylight hours in Fairbanks,
Alaska, varies from about 3.9 hr to 20.6 hr
(Source: Astronomical Applications Department, U.S. Naval Observatory, Washington, DC).
The function
Hd 8.3578 sin 0.0166d 1.2711 12.2153
can be used to approximate the number of daylight
hours H on a certain day d in Fairbanks. We can use
this function to determine on which day of the year
there will be about 10.5 hr of daylight.
This problem appears as Exercise 52 in Exercise Set 6.5.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
530
Chapter 6
1/10/05
1:30 PM
Page 530
• Trigonometric Identities, Inverse Functions, and Equations
2.1
6.1
Polynomial
Identities:
Functions
Pythagorean
and
andModeling
Sum and
Difference
State the Pythagorean identities.
Simplify and manipulate expressions containing trigonometric
expressions.
Use the sum and difference identities to find function values.
An identity is an equation that is true for all possible replacements of the
variables. The following is a list of the identities studied in Chapter 5.
Basic Identities
1
1
,
,
csc x sin x csc x
sin x
1
,
sec x
1
tan x ,
cot x
cos x 1
,
cos x
1
cot x ,
tan x
sec x sin x sin x ,
cos x cos x ,
tan x tan x ,
sin x
,
cos x
cos x
cot x sin x
tan x In this section, we will develop some other important identities.
Pythagorean Identities
We now consider three other identities that are fundamental to a study of
trigonometry. They are called the Pythagorean identities. Recall that the
equation of a unit circle in the xy-plane is
y
(x, y), or
(cos s, sin s)
s
x 2 y 2 1.
(1, 0) x
x2 y2 1
For any point on the unit circle, the coordinates x and y satisfy this equation. Suppose that a real number s determines a point on the unit circle
with coordinates x, y, or cos s, sin s. Then x cos s and y sin s .
Substituting cos s for x and sin s for y in the equation of the unit circle
gives us the identity
cos s2 sin s2 1,
Substituting cos s for x
and sin s for y
which can be expressed as
sin2 s cos2 s 1.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 531
Section 6.1
531
• Identities: Pythagorean and Sum and Difference
It is conventional in trigonometry to use the notation sin2 s rather than
sin s2. Note that sin2 s sin s 2.
y
2
y
y (sin x)2 (sin x)(sin x)
1
2
y sin x 2 sin (x ⋅ x)
2
1
2 x
2
1
2 x
1
The identity sin2 s cos2 s 1 gives a relationship between the
sine and the cosine of any real number s. It is an important Pythagorean
identity.
EXPLORING WITH TECHNOLOGY Addition of y-values provides a unique
way of developing the identity sin2 x cos2 x 1. First, graph y1 sin2 x
and y2 cos2 x. By visually adding the y-values, sketch the graph of the
sum, y3 sin2 x cos2 x. Then graph y3 using a graphing calculator and
check your sketch. The resulting graph appears to be the line y4 1, which
is the graph of sin2 x cos2 x. These graphs do not prove the identity, but
they do provide a check in the interval shown.
y1 sin2 x
y2 cos2 x
2
2
2π
2π
2π
2π
2
2
y3 sin2 x cos2 x, y4 1
2
2π
2π
2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
532
Chapter 6
1/10/05
1:30 PM
Page 532
• Trigonometric Identities, Inverse Functions, and Equations
We can divide by sin2 s on both sides of the preceding identity:
sin2 s cos2 s
1
2 .
sin2 s
sin2 s
sin s
Dividing by sin2 s
Simplifying gives us a second identity:
1 cot2 s csc2 s .
This equation is true for any replacement of s with a real number for
which sin2 s 0, since we divided by sin2 s . But the numbers for which
sin2 s 0 (or sin s 0) are exactly the ones for which the cotangent
and cosecant functions are not defined. Hence our new equation holds
for all real numbers s for which cot s and csc s are defined and is thus
an identity.
The third Pythagorean identity can be obtained by dividing by cos2 s on
both sides of the first Pythagorean identity:
sin2 s
cos2 s
1
2
2 cos s cos s cos2 s
tan2 s 1 sec2 s .
Dividing by cos2 s
Simplifying
The identities we have developed hold no matter what symbols are
used for the variables. For example, we could write sin2 s cos2 s 1,
sin2 cos2 1, or sin2 x cos2 x 1.
Pythagorean Identities
sin2 x cos2 x 1,
1 cot 2 x csc2 x ,
1 tan2 x sec2 x
It is often helpful to express the Pythagorean identities in equivalent
forms.
PYTHAGOREAN IDENTITIES
sin2 x cos2 x 1
1 cot 2 x csc2 x
1 tan2 x sec2 x
EQUIVALENT FORMS
sin2 x 1 cos2 x
cos2 x 1 sin2 x
1 csc2 x cot 2 x
cot 2 x csc2 x 1
1 sec2 x tan2 x
tan2 x sec2 x 1
Simplifying Trigonometric Expressions
We can factor, simplify, and manipulate trigonometric expressions in the
same way that we manipulate strictly algebraic expressions.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 533
Section 6.1
Study Tip
EXAMPLE 1
The examples in each section were
chosen to prepare you for success
with the exercise set. Study the
step-by-step annotated solutions
of the examples, noting that
substitutions are highlighted
in red. The time you spend
understanding the examples will
save you valuable time when
you do your assignment.
Solution
• Identities: Pythagorean and Sum and Difference
Multiply and simplify: cos x tan x sec x.
cos x tan x sec x
cos x tan x cos x sec x
1
sin x
cos x
cos x
cos x
cos x
Multiplying
Recalling the identities tan x and sec x sin x 1
533
sin x
cos x
1
and substituting
cos x
Simplifying
There is no general procedure for manipulating trigonometric expressions, but it is often helpful to write everything in terms of sines and
cosines, as we did in Example 1. We also look for the Pythagorean identity, sin2 x cos2 x 1, within a trigonometric expression.
EXAMPLE 2
Factor and simplify: sin2 x cos2 x cos4 x .
Solution
sin2 x cos2 x cos4 x
cos2 x sin2 x cos2 x
cos2 x 1
cos2 x
GCM
y1 cos x [tan x (1/cos x)],
y2 sin x 1
4
Removing a common factor
Using sin2 x cos2 x 1
A graphing calculator can be used to perform a partial check of an identity. First, we graph the expression on the left side of the equals sign. Then
we graph the expression on the right side using the same screen. If the two
graphs are indistinguishable, then we have a partial verification that the
equation is an identity. Of course, we can never see the entire graph, so there
can always be some doubt. Also, the graphs may not overlap precisely, but
you may not be able to tell because the difference between the graphs may be
less than the width of a pixel. However, if the graphs are obviously different,
we know that a mistake has been made.
Consider the identity in Example 1:
cos x tan x sec x sin x 1.
Recalling that sec x 1cos x , we enter
2 π
2π
4
X
Y1
6.283
1
5.498
.2929
4.712
ERROR
3.927
.2929
3.142
1
2.356
1.707
1.571
ERROR
X 6.28318530718
TblStart 2 π
Tbl π/4
Y2
1
.2929
0
.2929
1
1.707
2
y1 cos x tan x 1cos x and
y2 sin x 1.
To graph, we first select SEQUENTIAL mode. Then we select the “line”-graph
style for y1 and the “path”-graph style, denoted by , for y2. The calculator will graph y1 first. Then as it graphs y2, the circular cursor will trace the
leading edge of the graph, allowing us to determine whether the graphs
coincide. As you can see in the first screen on the left, the graphs appear
to be identical. Thus, cos x tan x sec x sin x 1 is most likely an
identity.
The TABLE feature can also be used to check identities. Note in the table
at left that the function values are the same except for those values of x for
which cos x 0. The domain of y1 excludes these values. The domain of y2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
534
Chapter 6
1/10/05
1:30 PM
Page 534
• Trigonometric Identities, Inverse Functions, and Equations
is the set of all real numbers. Thus all real numbers except 2, 32,
52, . . . are possible replacements for x in the identity. Recall that an
identity is an equation that is true for all possible replacements.
Suppose that we had simplified incorrectly in Example 1 and had gotten
cos x 1 on the right instead of sin x 1. Then two different graphs would
have appeared in the window. Thus we would have known that we did not
have an identity and that cos x tan x sec x cos x 1.
y1 cos x (tan x sec x),
y2 cos x 1
4
2π
2π
4
EXAMPLE 3
Simplify each of the following trigonometric expressions.
cot csc 2 sin2 t sin t 3
b)
1 cos2 t sin t
a)
Solution
cos cot sin a)
csc 1
sin cos sin sin cos cos 2 sin t sin t 3
1 cos2 t sin t
2 sin2 t sin t 3
sin2 t sin t
Rewriting in terms of
sines and cosines
Multiplying by the reciprocal
The cosine function is even.
2
b)
factoring
review section R.4.
Substituting sin2 t for 1 cos2 t
2 sin t 3 sin t 1
Factoring in both numerator
and denominator
sin t sin t 1
2 sin t 3
Simplifying
sin t
2 sin t
3
sin t
sin t
3
, or 2 3 csc t
2
sin t
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 535
Section 6.1
• Identities: Pythagorean and Sum and Difference
535
We can add and subtract trigonometric rational expressions in the
same way that we do algebraic expressions.
EXAMPLE 4
Add and simplify:
cos x
tan x .
1 sin x
Solution
rational expressions
review section R.5.
cos x
sin x
cos x
sin x
Using tan x tan x cos x
1 sin x
1 sin x cos x
cos x sin x 1 sin x
cos x
1 sin x cos x cos x 1 sin x
cos2 x sin x sin2 x
cos x 1 sin x
1 sin x
cos x 1 sin x
1
, or sec x
cos x
Multiplying by forms of 1
Adding
Using sin2 x cos2 x 1
Simplifying
When radicals occur, the use of absolute value is sometimes necessary, but it can be difficult to determine when to use it. In Examples 5
and 6, we will assume that all radicands are nonnegative. This means that
the identities are meant to be confined to certain quadrants.
EXAMPLE 5
Multiply and simplify: sin3 x cos x cos x .
Solution
sin3 x cos x cos x sin3 x cos2 x
sin2 x cos2 x sin x
sin x cos x sin x
EXAMPLE 6
Rationalize the denominator:
Solution
2
.
tan x
tan x
2
tan x tan x
2 tan x
tan2 x
2 tan x
tan x
2
tan x
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
536
Chapter 6
1/10/05
1:30 PM
Page 536
• Trigonometric Identities, Inverse Functions, and Equations
Often in calculus, a substitution is a useful manipulation, as we
show in the following example.
EXAMPLE 7 Express 9 x 2 as a trigonometric function of without
using radicals by letting x 3 tan . Assume that 0 2. Then
find sin and cos .
Solution
We have
9 x 2 9 3 tan 2
9 9 tan2 91 tan2 9 sec2 3sec 3 sec .
Substituting 3 tan for x
Factoring
Using 1 tan2 x sec2 x
For 0 2, sec 0,
so sec sec .
We can express 9 x 2 3 sec as
sec 9 x 2
x
u
3
9 x 2
.
3
In a right triangle, we know that sec is hypotenuseadjacent, when
is one of the acute angles. Using the Pythagorean theorem, we can
determine that the side opposite is x. Then from the right triangle, we
see that
sin x
3
and cos .
2
9 x
9 x 2
Sum and Difference Identities
We now develop some important identities involving sums or differences of two numbers (or angles), beginning with an identity for the
cosine of the difference of two numbers. We use the letters u and v for
these numbers.
Let’s consider a real number u in the interval 2, and a real
number v in the interval 0, 2. These determine points A and B on
the unit circle, as shown below. The arc length s is u v , and we know
that 0 s . Recall that the coordinates of A are cos u, sin u, and
the coordinates of B are cos v, sin v.
y
(cos v, sin v)
u
s
v
B
(cos u, sin u) A
(1, 0)
x
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 537
Section 6.1
• Identities: Pythagorean and Sum and Difference
537
Using the distance formula, we can write an expression for the distance AB :
distance formula
review section 1.1.
AB cos u cos v2 sin u sin v2.
This can be simplified as follows:
AB cos2 u 2 cos u cos v cos2 v sin2 u 2 sin u sin v sin2 v
sin2 u cos2 u sin2 v cos2 v 2cos u cos v sin u sin v
2 2cos u cos v sin u sin v.
Now let’s imagine rotating the circle on page 536 so that point B is
at 1, 0 as shown at left. Although the coordinates of point A are now
cos s, sin s, the distance AB has not changed.
Again we use the distance formula to write an expression for the distance AB:
y
(cos s, sin s)
A
s
B (1, 0)
AB cos s 12 sin s 02.
x
This can be simplified as follows:
AB cos2 s 2 cos s 1 sin2 s
sin2 s cos2 s 1 2 cos s
2 2 cos s .
Equating our two expressions for AB, we obtain
2 2cos u cos v sin u sin v 2 2 cos s .
Solving this equation for cos s gives
cos s cos u cos v sin u sin v .
(1)
But s u v , so we have the equation
cos u v cos u cos v sin u sin v .
(2)
Formula (1) above holds when s is the length of the shortest arc from
A to B. Given any real numbers u and v, the length of the shortest arc
from A to B is not always u v . In fact, it could be v u. However,
since cos x cos x , we know that cos v u cos u v. Thus,
cos s is always equal to cos u v. Formula (2) holds for all real numbers u and v. That formula is thus the identity we sought:
y1 cos (x 3),
y2 cos x cos 3 sin x sin 3
2
cos u v
cos u cos v sin u sin v .
Using a graphing calculator, we can graph
2π
2π
y1 cos x 3
and
2
y2 cos x cos 3 sin x sin 3
to illustrate this result.
The cosine sum formula follows easily from the one we have just
derived. Let’s consider cos u v. This is equal to cos u v, and
by the identity above, we have
cos u v cos u v
cos u cos v sin u sin v.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
538
Chapter 6
1/10/05
1:30 PM
Page 538
• Trigonometric Identities, Inverse Functions, and Equations
But cos v cos v and sin v sin v , so the identity we seek is
the following:
y1 cos (x 2),
y2 cos x cos 2 sin x sin 2
2
cos u v
cos u cos v sin u sin v .
Using a graphing calculator, we can graph
2π
2π
y1 cos x 2
and
2
y2 cos x cos 2 sin x sin 2
to illustrate this result.
EXAMPLE 8
Find cos 512 exactly.
Solution We can express 512 as a difference of two numbers whose
sine and cosine values are known:
5 9 4
,
12
12
12
3
.
4
3
or
Then, using cos u v cos u cos v sin u sin v , we have
cos
5
3
cos
12
4
3
cos(5π/12)
.2588190451
3
3
cos
sin
sin
4
3
4
3
2 1
2 3
2
2
2
2
2 6
4
4
6 2
.
4
cos
( (6) (2))/4
.2588190451
We can check using a graphing calculator set in RADIAN mode.
Consider cos 2 . We can use the identity for the cosine of a
difference to simplify as follows:
cos
cos
cos sin
sin 2
2
2
0 cos 1 sin sin .
Thus we have developed the identity
sin cos
.
2
This cofunction identity first
appeared in Section 5.1.
(3)
This identity holds for any real number . From it we can obtain an identity for the cosine function. We first let be any real number. Then we
replace in sin cos 2 with 2 . This gives us
sin
cos
2
2
2
cos ,
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 539
Section 6.1
• Identities: Pythagorean and Sum and Difference
which yields the identity
cos sin
.
2
539
(4)
Using identities (3) and (4) and the identity for the cosine of a difference, we can obtain an identity for the sine of a sum. We start with identity (3) and substitute u v for :
sin cos
sin u v cos
cos
cos
2
Identity (3)
u v
2
Substituting u v for u v
2
u cos v sin
2
sin u cos v cos u sin v .
u sin v
2
Using the identity for the cosine of
a difference
Using identities (3) and (4)
Thus the identity we seek is
sin u v
sin u cos v cos u sin v .
To find a formula for the sine of a difference, we can use the identity
just derived, substituting v for v :
sin u v sin u cos v cos u sin v.
Simplifying gives us
sin u v
sin u cos v cos u sin v .
EXAMPLE 9
Solution
Find sin 105 exactly.
We express 105 as the sum of two measures:
105 45 60.
Then
sin 105 sin 45 60
sin 45 cos 60 cos 45 sin 60
Using sin u v
sin u cos v cos u sin v
2 1
2 3
2
2
2
2
2 6
.
4
sin(105)
.9659258263
((2)(6))/4
.9659258263
We can check this result using a graphing calculator set in DEGREE mode.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
540
Chapter 6
1/10/05
1:30 PM
Page 540
• Trigonometric Identities, Inverse Functions, and Equations
Formulas for the tangent of a sum or a difference can be derived using
identities already established. A summary of the sum and difference identities follows.
Sum and Difference Identities
sin u v sin u cos v cos u sin v ,
cos u v cos u cos v sin u sin v ,
tan u tan v
tan u v 1 tan u tan v
There are six identities here, half of them obtained by using the signs
shown in color.
Find tan 15 exactly.
EXAMPLE 10
Solution We rewrite 15 as 45 30 and use the identity for the tangent of a difference:
tan 45 tan 30
1 tan 45 tan 30
1 33
3 3
.
1 1 33 3 3
tan 15 tan 45 30 2
1
EXAMPLE 11 Assume that sin 3 and sin 3 and that and are between 0 and 2. Then evaluate sin .
Solution
Using the identity for the sine of a sum, we have
sin sin cos cos sin 2
1
3 cos 3 cos .
To finish, we need to know the values of cos and cos . Using reference
triangles and the Pythagorean theorem, we can determine these values
from the diagrams:
y
3
2
a
x
5
cos 5
and
3
cos 22
.
3
Cosine values are positive
in the first quadrant.
Substituting these values gives us
y
2 22
1 5
3
3
3
3
1
42 5
4
2 5, or
.
9
9
9
sin 3
b
22
1
x
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 541
Section 6.1
6.1
• Identities: Pythagorean and Sum and Difference
Exercise Set
Multiply and simplify. Check your result using a
graphing calculator.
1. sin x cos x sin x cos x sin2 x cos2 x
2. tan x cos x csc x sin x sec x
3. cos y sin y sec y csc y sin y cos y
4. sin x cos x sec x csc x 2 tan x cot x
5. sin cos 2
1 2 sin cos 23.
sin4 x cos4 x
sin2 x cos2 x
24.
4 cos3 x
sin x
sin2 x
4 cos x
25.
5 cos sin2 sin cos 5 cot sin2 sin2 cos2 sin cos 26.
tan2 y 3 tan3 y
sec y
sec y
27.
1
2
sin2 s cos2 s cos s sin s
6. 1 tan x2 sec2 x 2 tan x
7. sin x csc x sin2 x csc2 x 1 sin3 x csc3 x
Factor and simplify. Check your result using a graphing
calculator.
9. sin x cos x cos2 x cos x sin x cos x
10. tan2 cot 2 tan cot tan cot 11. sin x cos x
sin x cos x sin x cos x
4
4
12. 4 sin2 y 8 sin y 4
13. 2 cos x cos x 3
2
4sin y 12
2 cos x 3 cos x 1
14. 3 cot 2 6 cot 3
15. sin3 x 27
3cot 12
sin x 3 sin2 x 3 sin x 9
16. 1 125 tan3 s
1 5 tan s 1 5 tan s 25 tan2 s
Simplify and check using a graphing calculator.
sin2 x cos x
17.
tan x
cos2 x sin x
3
30 sin x cos x
6 cos2 x sin x
2
5 sin x
, or 5 tan x sin x
cos x
sin2 x 2 sin x 1
19.
sin x 1
sin x 1
cos 1
20.
cos 1
2
21.
4 tan t sec t 2 sec t
6 tan t sec t 2 sec t
22.
csc x
sec x
cot x
2 tan t 1
3 tan t 1
2
cos x
4
1
cot y
3
1 2 sin s 2 cos s
sin2 s cos2 s
29.
sin2 9 10 cos 5 5sin 3
2 cos 1 3 sin 9
3
30.
9 cos2 25 cos2 1
2 cos 2 6 cos 10
sin x
cos x
2
1
cos2 x
1
3 cos 5 cos 1
4
Simplify and check using a graphing calculator. Assume
that all radicands are nonnegative.
31. sin2 x cos x cos x sin x cos x
32. cos2 x sin x sin x
cos x sin x
33. cos sin2 cos3 cos sin cos 34. tan2 x 2 tan x sin x sin2 x
tan x sin x
35. 1 sin y sin y 1 1 sin y
36. cos 2 cos sin cos cos 2 sin Rationalize the denominator.
sin x sin x cos x
37.
38.
cos x
cos x
39.
cos 1
1
28.
8. 1 sin t 1 sin t cos2 t
18.
541
cos2 y
2 sin2 y
2 cot y
2
Rationalize the numerator.
cos x
cos x
41.
sin x sin x cos x
43.
1 sin y
1 sin y
40.
42.
1 sin y
44.
cos y
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
cos x
tan x
sin x
tan x
1 cos sin 1 cos 1 cos sin x
cot x
cos2 x
2 sin2 x
sin x
cos x
cot x
2
BBEPMC06_0321279115.QXP
542
Chapter 6
1/10/05
1:30 PM
Page 542
• Trigonometric Identities, Inverse Functions, and Equations
Use the given substitution to express the given radical
expression as a trigonometric function without radicals.
Assume that a 0 and 0 2. Then find
expressions for the indicated trigonometric functions.
45. Let x a sin in a2 x 2. Then find cos and
tan . 46. Let x 2 tan in 4 x 2. Then find sin and
cos . 47. Let x 3 sec in x 2 9. Then find sin and
cos . 48. Let x a sec in x 2 a2. Then find sin and
cos . Use the given substitution to express the given radical
expression as a trigonometric function without radicals.
Assume that 0 2.
x2
49. Let x sin in
. sin tan 1 x 2
50. Let x 4 sec in
x 2 16 sin cos .
x2
4
Use the sum and difference identities to evaluate
exactly. Then check using a graphing calculator.
6 2
6 2
51. sin
52. cos 75
4
4
12
53. tan 105 55. cos 15
6 2
4
54. tan
5
12
56. sin
7
12
6 2
4
First write each of the following as a trigonometric
function of a single angle; then evaluate.
57. sin 37 cos 22 cos 37 sin 22 sin 59 0.8572
58. cos 83 cos 53 sin 83 sin 53 cos 30 0.8660
59. cos 19 cos 5 sin 19 sin 5 cos 24 0.9135
60. sin 40 cos 15 cos 40 sin 15 sin 25 0.4226
61.
tan 20 tan 32
tan 52 1.2799
1 tan 20 tan 32
62.
tan 35 tan 12
tan 23 0.4245
1 tan 35 tan 12
63. Derive the formula for the tangent of a sum. 64. Derive the formula for the tangent of a difference.
Assuming that sin u 35 and sin v 45 and that u and
v are between 0 and 2, evaluate each of the following
exactly.
65. cos u v 0
66. tan u v 247
7
67. sin u v 25
68. cos u v
24
25
Assuming that sin 0.6249 and cos 0.1102
and that both and are first-quadrant angles,
evaluate each of the following.
69. tan 1.5789 70. sin 0.7071
71. cos 0.7071
72. cos 0.5351
Simplify.
73. sin sin 2 sin cos 74. cos cos 2 sin sin 75. cos u v cos v sin u v sin v
cos u
76. sin u v cos v cos u v sin v
sin u
Collaborative Discussion and Writing
77. What is the difference between a trigonometric
equation that is an identity and a trigonometric
equation that is not an identity? Give an example
of each.
78. Why is it possible to use a graph to disprove that an
equation is an identity but not to prove that one is?
Skill Maintenance
Solve.
3
79. 2x 3 2 x 2 [2.1] All real numbers
80. x 7 x 3.4
[2.1] No solution
Given that sin 31 0.5150 and cos 31 0.8572,
find the specified function value.
81. sec 59 [5.1] 1.9417
82. tan 59 [5.1] 1.6645
Synthesis
One of the identities gives an
easy way to find an angle formed by two lines. Consider
two lines with equations l1: y m1x b1 and
l2: y m2x b2.
Angles Between Lines.
Answers to Exercises 45–48, 53, 54, 63, and 64 can be found on p. IA-37.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 543
Section 6.1
y
88. Given that fx sin x , show that l2
u2 u 1
or f
l1
u2
The slopes m1 and m2 are the tangents of the angles
1 and 2 that the lines form with the positive direction
of the x-axis. Thus we have m1 tan 1 and
m2 tan 2. To find the measure of 2 1, or ,
we proceed as follows:
tan tan 2 1 tan 2 tan 1
1 tan 2 tan 1
m2 m1
.
1 m2m1
6, or 30
86. l1: 2x y 4 0,
3
, or 135° l2: y 2x 5 0
0; the lines are parallel
126.87
4
87. Circus Guy Wire. In a circus, a guy wire A is
attached to the top of a 30-ft pole. Wire B is used
for performers to walk up to the tight wire, 10 ft
above the ground. Find the angle between the
wires if they are attached to the ground 40 ft from
the pole. 22.83
92. sin x sin x
φ
B
10 ft
94. tan cot 1
2
2
93. cos 2 2 cos 95.
cos 6x
6
cos x
Find the slope of line l1, where m2 is the slope of line l2
and is the smallest positive angle from l1 to l2.
96. m2 43 , 45 17
6 33
0.0645
9 23
98. Line l1 contains the points 2, 4 and 5, 1.
Find the slope of line l2 such that the angle from l1
to l2 is 45. 16
99. Line l1 contains the points 3, 7 and 3, 2.
Line l2 contains 0, 4 and 2, 6. Find the
smallest positive angle from l1 to l2. 168.7
100. Find an identity for sin 2. (Hint: 2 .)
sin 2 2 sin cos 101. Find an identity for cos 2. (Hint: 2 .) Derive the identity. Check using a graphing calculator.
3
102. sin x cos x 2
103. tan x 104.
A
Show that each of the following is not an identity by
finding a replacement or replacements for which the
sides of the equation do not name the same number.
Then use a graphing calculator to show that the
equation is not an identity.
sin 5x
90. sin2 sin 91.
sin 5 x
97. m2 23 , 30
This formula also holds when the lines are taken in the
reverse order. When is acute, tan will be positive.
When is obtuse, tan will be negative.
Find the measure of the angle from l1 to l2.
84. l1: 3y 3x 3,
83. l1: 2x 3 2y ,
l2: y 3x 2
l2: x y 5
85. l1: y 3,
l2: x y 5
fx h fx
cos h 1
sin h
.
cos x
sin x
h
h
h
x
u1
u1
fx h fx
cos h 1
sin h
.
sin x
cos x
h
h
h
89. Given that fx cos x , show that u2
543
• Identities: Pythagorean and Sum and Difference
4
1 tan x
1 tan x
sin tan tan cos 1 tan tan 105. sin sin 2 sin cos sin sin sin cos cos sin sin cos cos sin 2 sin cos 40 ft
Answers to Exercises 88–95 and 101–104 can be found on pp. IA-37 and IA-38.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
544
Chapter 6
1/10/05
1:30 PM
Page 544
• Trigonometric Identities, Inverse Functions, and Equations
6.2
Identities:
Cofunction,
Double-Angle,
and Half-Angle
Use cofunction identities to derive other identities.
Use the double-angle identities to find function values of twice an
angle when one function value is known for that angle.
Use the half-angle identities to find function values of half an
angle when one function value is known for that angle.
Simplify trigonometric expressions using the double-angle and
half-angle identities.
Cofunction Identities
Each of the identities listed below yields a conversion to a cofunction. For
this reason, we call them cofunction identities.
Cofunction Identities
x cos x ,
sin
2
tan
sec
cos
x cot x ,
2
cot
x csc x ,
2
csc
x sin x ,
2
x tan x ,
2
x sec x
2
We verified the first two of these identities in Section 6.1. The other
four can be proved using the first two and the definitions of the trigonometric functions. These identities hold for all real numbers, and thus, for
all angle measures, but if we restrict to values such that 0 90,
or 0 2, then we have a special application to the acute angles of
a right triangle.
Comparing graphs can lead to possible identities. On the left below,
we see that the graph of y sin x 2 is a translation of the graph
of y sin x to the left 2 units. On the right, we see the graph of
y cos x .
y sin (x 2)
2
y
y
y sin x
y cos x
2
2
1
1
2 x
2
1
1
2
2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2 x
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 545
Section 6.2
• Identities: Cofunction, Double-Angle, and Half-Angle
545
Comparing the graphs, we observe a possible identity:
sin x 2
cos x .
The identity can be proved using the identity for the sine of a sum developed in Section 6.1.
Prove the identity sin x 2 cos x .
EXAMPLE 1
Solution
sin x 2
sin x cos
cos x sin
2
2
Using sin u v
sin u cos v cos u sin v
sin x 0 cos x 1
cos x
We now state four more cofunction identities. These new identities
that involve the sine and cosine functions can be verified using previously established identities as seen in Example 1.
Cofunction Identities for the Sine and Cosine
sin x
cos x ,
cos x
sin x
2
2
Find an identity for each of the following.
EXAMPLE 2
a) tan x Solution
a) We have
2
b) sec x 90
tan x 2
sin x cos x cos x
sin x
cot x .
2
2
Using tan x Using cofunction identities
Thus the identity we seek is
tan x 2
sin x
cos x
cot x .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
546
Chapter 6
1/10/05
1:30 PM
Page 546
• Trigonometric Identities, Inverse Functions, and Equations
b) We have
sec x 90 1
1
csc x .
cos x 90 sin x
Thus, sec x 90 csc x .
Double-Angle Identities
If we double an angle of measure x, the new angle will have measure 2x.
Double-angle identities give trigonometric function values of 2x in
terms of function values of x. To develop these identities, we will use the
sum formulas from the preceding section. We first develop a formula for
sin 2x . Recall that
sin u v sin u cos v cos u sin v .
We will consider a number x and substitute it for both u and v in this
identity. Doing so gives us
sin x x sin 2x
sin x cos x cos x sin x
2 sin x cos x .
Our first double-angle identity is thus
sin 2x 2 sin x cos x .
We can graph
y1 sin 2x , and y2 2 sin x cos x
using the “line”-graph style for y1 and the “path”-graph style for y2 and see
that they appear to have the same graph. We can also use the TABLE feature.
y1 sin 2x, y2 2 sin x cos x
2
2π
2π
2
X
6.283
5.498
4.712
3.927
3.142
2.356
1.571
Y1
2E13
1
0
1
0
1
0
Y2
0
1
0
1
0
1
0
X 1.57079632679
Double-angle identities for the cosine and tangent functions can be
derived in much the same way as the identity above:
cos 2x cos2 x sin2 x ,
tan 2x 2 tan x
.
1 tan2 x
EXAMPLE 3 Given that tan 34 and is in quadrant II, find each
of the following.
a) sin 2
c) tan 2
b) cos 2
d) The quadrant in which 2 lies
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 547
Section 6.2
Solution
547
• Identities: Cofunction, Double-Angle, and Half-Angle
By drawing a reference triangle as shown, we find that
sin y
3
5
(4, 3)
and
5
u
3
cos 4
.
5
4
Thus we have the following.
a) sin 2 2 sin cos 2 3
4
5
5
2 tan 1 tan 2
2
4
5
b) cos 2 cos2 sin2 c) tan 2 2 4 3
x
1
2
3
5
24
25
9
7
16
25 25 25
3
3
4 2
2
1
9
16
3
2
16
7
24
7
Note that tan 2 could have been found more easily in this case by
simply dividing:
tan 2 sin 2
cos 2
24
25
7
25
24
7
.
d) Since sin 2 is negative and cos 2 is positive, we know that 2 is in
quadrant IV.
Two other useful identities for cos 2x can be derived easily, as follows.
cos 2x cos2 x sin2 x
1 sin2 x sin2 x
1 2 sin2 x
Double-Angle Identities
sin 2x 2 sin x cos x ,
2 tan x
tan 2x 1 tan2 x
cos 2x cos2 x sin2 x
cos2 x 1 cos2 x
2 cos2 x 1
cos 2x cos2 x sin2 x
1 2 sin2 x
2 cos2 x 1
Solving the last two cosine double-angle identities for sin2 x and cos2 x ,
respectively, we obtain two more identities:
sin2 x 1 cos 2x
1 cos 2x
and cos2 x .
2
2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
548
Chapter 6
1/10/05
1:30 PM
Page 548
• Trigonometric Identities, Inverse Functions, and Equations
Using division and these two identities gives us the following useful
identity:
tan2 x EXAMPLE 4
1 cos 2x
.
1 cos 2x
Find an equivalent expression for each of the following.
a) sin 3 in terms of function values of b) cos3 x in terms of function values of x or 2x, raised only to the first
power
Solution
a) sin 3 sin 2 sin 2 cos cos 2 sin 2 sin cos cos 2 cos2 1 sin Using sin 2 2 sin cos and cos 2 2 cos2 1
2 sin cos2 2 sin cos2 sin 4 sin cos2 sin We could also substitute cos2 sin2 or 1 2 sin2 for cos 2.
Each substitution leads to a different result, but all results are
equivalent.
b) cos3 x cos2 x cos x
1 cos 2x
cos x
2
cos x cos x cos 2x
2
Half-Angle Identities
If we take half of an angle of measure x, the new angle will have measure
x2. Half-angle identities give trigonometric function values of x2 in
terms of function values of x. To develop these identities, we replace x
with x2 and take square roots. For example,
sin2 x 1 cos 2x
2
1 cos 2 x
2
2
x
1
cos
x
sin2
2
2
sin2
sin
x
2
Solving the identity
cos 2x 1 2 sin2 x for sin2 x
x
2
1 cos x
.
2
Substituting
x
for x
2
Taking square roots
The formula is called a half-angle formula. The use of and depends
on the quadrant in which the angle x2 lies. Half-angle identities for the
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 549
Section 6.2
• Identities: Cofunction, Double-Angle, and Half-Angle
549
cosine and tangent functions can be derived in a similar manner. Two
additional formulas for the half-angle tangent identity are listed below.
Half-Angle Identities
sin
x
2
1 cos x
,
2
cos
x
2
1 cos x
,
2
tan
x
2
1 cos x
1 cos x
1 cos x
sin x
1 cos x
sin x
EXAMPLE 5 Find tan 8 exactly. Then check the answer using a
graphing calculator in RADIAN mode.
Solution
4
tan
tan
8
2
4
2
2
2 2 2
1 cos
1
4
2
2
2 2
2
2
2 2 2 2 2 2
2 1
tan(π/8)
.4142135624
((2)1
.4142135624
sin
2
2
The identities that we have developed are also useful for simplifying
trigonometric expressions.
EXAMPLE 6
a)
sin x cos x
1
2 cos 2x
Simplify each of the following.
b) 2 sin2
x
cos x
2
Solution
a) We can obtain 2 sin x cos x in the numerator by multiplying the ex2
pression by 2 :
sin x cos x
2 sin x cos x 2 sin x cos x
1
1
2
cos 2x
2 cos 2x
2 cos 2x
sin 2x
Using sin 2x 2 sin x cos x
cos 2x
tan 2x .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
550
Chapter 6
1/10/05
1:30 PM
Page 550
• Trigonometric Identities, Inverse Functions, and Equations
b) We have
2 sin2
x
1 cos x
cos x 2
2
2
Using sin
cos x
x
2
1 cos x
1 cos x
x
, or sin2
2
2
2
1 cos x cos x
1.
We can check this result using a graph or a table.
y1 2 sin2
x
cos x, y2 1
2
2
2π
2π
2
Y1
X
6.283
1
5.498
1
4.712
1
3.927
1
3.142
1
2.356
1
1.571
1
X 6.28318530718
Y2
1
1
1
1
1
1
1
Tbl π/4
Answers to Exercises 1–8 can be found on p. IA-38.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
550
Chapter 6
6.2
1/10/05
1:30 PM
Page 550
• Trigonometric Identities, Inverse Functions, and Equations
Exercise Set
1. Given that sin 310 0.8090 and
cos 310 0.5878, find each of the following.
a) The other four function values for 310 b) The six function values for 5 2. Given that
sin
2 3
12
2
and cos
2 3
,
12
2
find exact answers for each of the following.
a) The other four function values for 12 b) The six function values for 512 1
3
3. Given that sin and that the terminal side is in
quadrant II, find exact answers for each of the
following.
a) The other function values for b) The six function values for 2 c) The six function values for 2 4
4. Given that cos 5 and that the terminal side is
in quadrant IV, find exact answers for each of the
following.
a) The other function values for b) The six function values for 2 c) The six function values for 2 Find an equivalent expression for each of the following.
5. sec x 6. cot x 2
2
7. tan x 2
8. csc x 2
Find the exact value of sin 2, cos 2, tan 2, and the
quadrant in which 2 lies.
4
9. sin , in quadrant I
5
24
7
24
sin 2 25, cos 2 25, tan 2 7 ; II
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Answers to Exercises 1–8 can be found on p. IA-38.
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 551
Section 6.2
• Identities: Cofunction, Double-Angle, and Half-Angle
5
, in quadrant I
13
sin 2 120
, cos 2 119
, tan 2 120
; II
169
169
119
3
11. cos , in quadrant III
5
7
, cos 2 25
, tan 2 24
; II
sin 2 24
25
7
15
12. tan , in quadrant II
8
sin 2 240
, cos 2 161
, tan 2 240
; III
289
289
161
5
13. tan , in quadrant II
12
, cos 2 119
, tan 2 120
; IV
sin 2 120
169
169
119
10
14. sin , in quadrant IV
10
sin 2 35 , cos 2 45 , tan 2 34 ; IV
15. Find an equivalent expression for cos 4x in terms of
function values of x. c) cot x
10. cos 2
19. sin 112.5
2 2
8
2
5 2 3
22. sin
12
2
21. tan 75
2 3
Given that sin 0.3416 and is in quadrant I, find
each of the following using identities.
23. sin 2 0.6421
24. cos
0.9848
2
25. sin
2
26. sin 4 0.9845
0.1735
y 1 cos x
y cos x sin x
4
2π
4
2π
2π
4
2π
2π
2π
4
4
x
2
a) sin x csc x tan x
c) 2cos2 x sin2 x
b) sin x 2 cos x
d) 1 cos x
sin 2x
2 cos x
a) cos x
c) cos x sin x
b) tan x
d) sin x
28. 2 cos2
29.
30. 2 sin
cos
2
2
2
d) sin cos a) cos2 b) sin
c) sin Simplify. Check your results using a graphing calculator.
x
1 cos x
31. 2 cos2
2
32. cos4 x sin4 x
cos 2x
33. sin x cos x2 sin 2x
1
1 sin 2x
35.
2 sec2 x
sec2 x
36.
1 sin 2x cos 2x
1 sin 2x cos 2x
cos 2x
cot x
37. 4 cos x sin x 2 cos 2x2 2 cos 2x 4 sin x cos x2 8
38. 2 sin x cos3 x 2 sin3 x cos x
4
2π
4
2π
34. sin x cos x2
In Exercises 27– 30, use a graphing calculator to
determine which of the following expressions asserts an
identity. Then derive the identity algebraically.
cos 2x
27.
cos x sin x
a) 1 cos x
b) cos x sin x
y sin x (cot x 1)
4
2 2 20. cos
2
d) sin x cot x 1
y cot x
16. Find an equivalent expression for sin4 in terms of
function values of , 2, or 4, raised only to the
first power. Use the half-angle identities to evaluate exactly.
17. cos 15 2 3
18. tan 67.5 2 1
551
1
2
sin 4x
Collaborative Discussion and Writing
39. Discuss and compare the graphs of y sin x ,
y sin 2x , and y sin x2.
Answers to Exercises 15, 16, and 27–30 can be found on pp. IA-38 and IA-39.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
552
Chapter 6
1/10/05
1:30 PM
Page 552
• Trigonometric Identities, Inverse Functions, and Equations
40. Find all errors in the following:
2 sin 2x cos 4x
22 sin x cos x2 2 cos 2x
8 sin2 x cos2 x 2cos2 x sin2 x
8 sin2 x cos2 x 2.
2
In Exercises 41–48, answer “True” or “False.”
41. 1 cos2 x sin2 x
42. sec2 x tan2 x 1
[6.1] True
44. 1 cot 2 x csc2 x
45. csc2 x cot 2 x 1
46. 1 tan2 x sec2 x
47. 1 sin2 x cos2 x
48. sec2 x 1 tan2 x
[6.1] False
[6.1] False
[6.1] False
[6.1] True
[6.1] True
[6.1] True
Consider the following functions (a)– (f ). Without
graphing them, answer questions 49– 52 below.
1
x
2
2
1
cos 2x 2
4
2
1
d) fx sin x 2
e) fx 2 cos 4x f) fx cos 2 x 8
c) fx sin 2 x 2
49. Which functions have a graph with an amplitude
of 2? [5.5] (a), (e)
2
cos x1 cot x
x sin x
2
1 sin x
56.
cos x cos x tan x 1 tan x
cos2 y sin y 57.
sin2 y sin
2
y
2
cot 2 y
Find sin , cos , and tan under the given conditions.
7 3
58. cos 2 ,
2 2 12 2
59. tan
2
55. cos x cot x sin x [6.1] False
43. sin2 x 1 cos2 x
b) fx cos x sin
Skill Maintenance
a) fx 2 sin
Simplify. Check your results using a graphing calculator.
54. sin
x sec x cos x sin2 x
2
5
3
, 2
3
2
60. Nautical Mile. Latitude is used to measure
north–south location on the earth between the
equator and the poles. For example, Chicago has
latitude 42N. (See the figure.) In Great Britain, the
nautical mile is defined as the length of a minute of
arc of the earth’s radius. Since the earth is flattened
slightly at the poles, a British nautical mile varies
with latitude. In fact, it is given, in feet, by the
function
N 6066 31 cos 2,
where is the latitude in degrees.
90 N
50. Which functions have a graph with a period of ?
42 N
[5.5] (b), (c), (f)
51. Which functions have a graph with a period of 2 ?
[5.5] (d)
52. Which functions have a graph with a phase shift
of ? [5.5] (e)
4
Synthesis
53. Given that cos 51 0.6293, find the six function
values for 141. Equator
0
90 S
a) What is the length of a British nautical mile at
Chicago? 6062.76 ft
b) What is the length of a British nautical mile at the
North Pole? 6097 ft
Answers to Exercises 53, 58, and 59 can be found on p. IA-39.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1:30 PM
Page 553
Section 6.2
553
• Identities: Cofunction, Double-Angle, and Half-Angle
c) Express N in terms of cos only. That is, do
not use the double angle. N 6097 62 cos2 M
72°
I
WISCONSIN
42°
AN
44°
IG
where g is measured in meters per second per second
at sea level.
a) Chicago has latitude 42N. Find g. 9.80359 msec2
b) Philadelphia has latitude 40N. Find g.
46°
H
g 9.780491 0.005288 sin2 0.000006 sin2 2,
g 9.780491 0.005264 sin 0.000024 sin C
61. Acceleration Due to Gravity. The acceleration due
to gravity is often denoted by g in a formula such as
S 12 gt 2, where S is the distance that an object falls
in time t. The number g relates to motion near the
earth’s surface and is usually considered constant.
In fact, however, g is not constant, but varies slightly
with latitude. If stands for latitude, in degrees, g is
given with good approximation by the formula
c) Express g in terms of sin only. That is, eliminate
the double angle.
2
4
N
1/10/05
A
AD
VT.
CA
BBEPMC06_0321279115.QXP
NEW YORK
MASS.
CONN.
Chicago
PENNSYLVANIA
Philadelphia
40°
INDIANA
ILLINOIS
OHIO
N.J.
MD.
DEL.
W.VA.
38°
VIRGINIA
KENTUCKY
90°
88°
86°
84°
82°
80°
9.80180 msec2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
78°
76°
74°
BBEPMC06_0321279115.QXP
572
Chapter 6
1/10/05
1:30 PM
Page 572
• Trigonometric Identities, Inverse Functions, and Equations
Solve trigonometric equations.
6.5
Solving
Trigonometric
Equations
When an equation contains a trigonometric expression with a variable,
such as cos x , it is called a trigonometric equation. Some trigonometric
equations are identities, such as sin2 x cos2 x 1. Now we consider
equations, such as 2 cos x 1, that are usually not identities. As we
have done for other types of equations, we will solve such equations by
finding all values for x that make the equation true.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Answers to Exercises 81–86 can be found on p. IA-44.
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 573
Section 6.5
EXAMPLE 1
Solution
• Solving Trigonometric Equations
573
Solve: 2 cos x 1.
We first solve for cos x :
2 cos x 1
1
cos x .
2
1
The solutions are numbers that have a cosine of 2 . To find them, we use
the unit circle (see Section 5.5).
1
There are just two points on the unit circle for which the cosine is 2 ,
as shown in the figure at left. They are the points corresponding to 23 and
43. These numbers, plus any multiple of 2, are the solutions:
q , 3
2 o
i
(1, 0)
2
2k
3
and
4
2k,
3
where k is any integer. In degrees, the solutions are
q ,
3
2
120 k 360 and 240 k 360,
where k is any integer.
To check the solution to 2 cos x 1, we can graph y1 2 cos x and
y2 1 on the same set of axes and find the first coordinates of the points
of intersection. Using 3 as the Xscl facilitates our reading of the solutions.
First, let’s graph these equations on the interval from 0 to 2, as shown in
the figure on the left below. The only solutions in 0, 2 are 23 and
43.
y1 2 cos x, y2 1
y1 2 cos x, y2 1
4
4
4π
3
2π
0
(23,π 1)
4
(43,π 1)
π
Xscl 3
2π
3
3π
3π
8π
3
2π
3
4
Yscl 1
4π 8π
3 3
π
Xscl 3
Next, let’s change the viewing window to 3, 3, 4, 4 and graph
again. Since the cosine function is periodic, there is an infinite number of solutions. A few of these appear in the graph on the right above. From the
graph, we see that the solutions are 23 2k and 43 2k, where k
is any integer.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
574
Chapter 6
1/10/05
1:30 PM
Page 574
• Trigonometric Identities, Inverse Functions, and Equations
EXAMPLE 2
Solution
Solve: 4 sin2 x 1.
We begin by solving for sin x :
4 sin2 x 1
1
sin2 x 4
sin x 1
.
2
Again, we use the unit circle to find those numbers having a sine of 21 or 12 .
The solutions are
S
23 , q 23 , q F
A
23 , q (1, 0)
G
23 ,
q 11
5
7
2k,
2k,
2k, and
2k,
6
6
6
6
where k is any integer. In degrees, the solutions are
30 k 360, 150 k 360,
210 k 360, and 330 k 360,
where k is any integer.
The general solutions listed above could be condensed using odd as well
as even multiples of :
5
k and
k,
6
6
or, in degrees,
30 k 180 and 150 k 180,
where k is any integer.
Let’s do a partial check using a graphing calculator, checking only the solutions in 0, 2 . We graph y1 4 sin2 x and y2 1 and note that the solutions in 0, 2 are 6, 56, 76, and 116.
y1 4 sin2 x, y2 1
4
0
(56,π 1)
π
, 1
6
( )
4
(76,π 1)
(
11π
, 1
6
)
2π
π
Xscl 6
In most applications, it is sufficient to find just the solutions from 0
to 2 or from 0 to 360. We then remember that any multiple of 2,
or 360, can be added to obtain the rest of the solutions.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 575
Section 6.5
• Solving Trigonometric Equations
575
We must be careful to find all solutions in 0, 2 when solving
trigonometric equations involving double angles.
EXAMPLE 3
Solution
Solve 3 tan 2x 3 in the interval 0, 2 .
We first solve for tan 2x:
3 tan 2x 3
tan 2x 1.
We are looking for solutions x to the equation for which
0 x 2.
Multiplying by 2, we get
0 2x 4,
f,
which is the interval we use when solving tan 2x 1.
Using the unit circle, we find points 2x in 0, 4 for which
tan 2x 1. These values of 2x are as follows:
11p
4
2x , 2
2
2
2 (1, 0)
3 7 11
,
,
,
4
4
4
and
15
.
4
Thus the desired values of x in 0, 2 are each of these values divided
2
, 2 by 2. Therefore,
2
2
j,
15p
4
x
3 7 11
,
,
,
8
8
8
and
15
.
8
Calculators are needed to solve some trigonometric equations. Answers can be found in radians or degrees, depending on the mode setting.
EXAMPLE 4
Solution
cos1(0.4216)
65.06435654
65.06
294.94
65.06 (1, 0)
Solve
1
cos 1 1.2108 in 0, 360.
2
We have
1
cos 1 1.2108
2
1
cos 0.2108
2
cos 0.4216.
Using a calculator set in DEGREE mode (see window at left), we find that
the reference angle, cos1 0.4216, is
65.06.
Since cos is positive, the solutions are in quadrants I and IV. The solutions in 0, 360 are
65.06 and 360 65.06 294.94.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
576
Chapter 6
1/10/05
1:30 PM
Page 576
• Trigonometric Identities, Inverse Functions, and Equations
EXAMPLE 5
Solve 2 cos2 u 1 cos u in 0, 360.
Algebraic Solution
We use the principle of zero products:
2 cos2 u 1 cos u
2 cos u cos u 1 0
2 cos u 1 cos u 1 0
2 cos u 1 0 or cos u 1 0
2 cos u 1 or
cos u 1
1
cos u or
cos u 1.
2
2
Thus,
u 60, 300 or
u 180.
The solutions in 0, 360 are 60, 180, and 300.
Graphical Solution
We can use either the Intersect method or the Zero method to solve
trigonometric equations. Here we illustrate by solving the equation
using both methods. We set the calculator in DEGREE mode.
Intersect Method. We graph the equations
y1 2 cos2 x
and
y2 1 cos x
and use the INTERSECT feature to find the first coordinates of the points
of intersection.
y1 2 cos2 x, y2 1 cos x
3
0
Intersection
X 60
3
360
Y .5
Xscl 60
The leftmost solution is 60. Using the INTERSECT feature two more
times, we find the other solutions, 180 and 300.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 577
Section 6.5
• Solving Trigonometric Equations
577
Zero Method. We write the equation in the form
2 cos2 u cos u 1 0.
Then we graph
y 2 cos2 x cos x 1
and use the ZERO feature to determine the zeros of the function.
y 2 cos2 x cos x 1
3
0
Zero
X 60
3
360
Y0
Xscl 60
The leftmost zero is 60. Using the ZERO feature two more times, we
find the other zeros, 180 and 300. The solutions in 0, 360 are 60,
180, and 300.
EXAMPLE 6
Solution
y sin2 x sin x
2
2π
0
1
π
Xscl 4
Solve sin2 sin 0 in 0, 2 .
We factor and use the principle of zero products:
sin2 sin 0
sin sin 1 0
Factoring
sin 0
or sin 1 0
sin 0
or
sin 1
0, or
.
2
The solutions in 0, 2 are 0, 2, and .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
578
Chapter 6
1/10/05
1:30 PM
Page 578
• Trigonometric Identities, Inverse Functions, and Equations
If a trigonometric equation is quadratic but difficult or impossible to
factor, we use the quadratic formula.
Solve 10 sin2 x 12 sin x 7 0 in 0, 360.
EXAMPLE 7
Solution This equation is quadratic in sin x with a 10, b 12, and
c 7. Substituting into the quadratic formula, we get
sin x sin1(0.4296)
25.44217782
b b 2 4ac
2a
Using the
quadratic formula
12 122 410 7
2 10
Substituting
12 144 280 12 424
20
20
12 20.5913
20
sin x 1.6296 or sin x 0.4296.
25.44
205.44
25.44
Since sine values are never greater than 1, the first of the equations has
no solution. Using the other equation, we find the reference angle to be
25.44. Since sin x is negative, the solutions are in quadrants III and IV.
Thus the solutions in 0, 360 are
334.56
180 25.44 205.44 and 360 25.44 334.56.
Trigonometric equations can involve more than one function.
EXAMPLE 8
y1 2 cos2 x tan x, y2 tan x
2
Solution Using a graphing calculator, we can determine that there are six
solutions. If we let Xscl 4, the solutions are read more easily. In the figures at left, we show the Intersect and Zero methods of solving graphically.
Each illustrates that the solutions in 0, 2 are
2π
0
0,
2
y1 2 cos2 x tan x tan x
2
2
7
3
5
,
, ,
, and
.
4
4
4
4
We can verify these solutions algebraically, as follows:
π
Xscl 4
2π
0
Solve 2 cos2 x tan x tan x in 0, 2 .
2 cos2 x tan x tan x
2 cos2 x tan x tan x 0
tan x 2 cos2 x 1 0
tan x 0
or 2 cos2 x 1 0
1
cos2 x 2
2
2
3 5 7
x
,
,
,
.
4 4 4 4
cos x π
Xscl 4
x 0, or
Thus, x 0, 4, 34, , 54, and 74.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 579
Section 6.5
• Solving Trigonometric Equations
579
When a trigonometric equation involves more than one function, it
is sometimes helpful to use identities to rewrite the equation in terms of
a single function.
EXAMPLE 9
Solve sin x cos x 1 in 0, 2 .
Algebraic Solution
We have
sin x cos x 1
sin x cos x2 12
Squaring both sides
2
sin x 2 sin x cos x cos2 x 1
Using sin2 x cos2 x 1
2 sin x cos x 1 1
2 sin x cos x 0
Using 2 sin x cos x sin 2x
sin 2x 0.
We are looking for solutions x to the equation for which
0 x 2. Multiplying by 2, we get 0 2x 4, which is
the interval we consider to solve sin 2x 0. These values of 2x
are 0, , 2, and 3. Thus the desired values of x in 0, 2 satisfying this equation are 0, 2, , and 32. Now we
check these in the original equation sin x cos x 1:
sin 0 cos 0 0 1 1,
sin
cos
1 0 1,
2
2
sin cos 0 1 1,
3
3
cos
1 0 1.
sin
2
2
We find that and 32 do not check, but the other values
do. Thus the solutions in 0, 2 are
0
and
.
2
When the solution process involves squaring both sides, values are sometimes obtained that are not solutions of the original
equation. As we saw in this example, it is important to check the
possible solutions.
Graphical Solution
We can graph the left side and then the
right side of the equation as seen
in the first window below. Then we
look for points of intersection. We
could also rewrite the equation as
sin x cos x 1 0, graph the left
side, and look for the zeros of the function, as illustrated in the second window
below. In each window, we see the solutions in 0, 2 as 0 and 2.
y1 sin x cos x, y2 1
3
2π
0
3
Xscl π
2
y sin x cos x 1
3
2π
0
3
Xscl π
2
This example illustrates a valuable
advantage of the calculator— that is, with
a graphing calculator, extraneous solutions do not appear.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
580
Chapter 6
1/10/05
1:30 PM
Page 580
• Trigonometric Identities, Inverse Functions, and Equations
EXAMPLE 10
Solve cos 2x sin x 1 in 0, 2 .
Algebraic Solution
Graphical Solution
We have
We graph y1 cos 2x sin x 1 and
look for the zeros of the function.
cos 2x sin x 1
1 2 sin2 x sin x 1
Using the identity
cos 2x 1 2 sin2 x
2 sin x sin x 0
Factoring
sin x 2 sin x 1 0
sin x 0
or 2 sin x 1 0
2
sin x 0
x 0, or
or
y cos 2x sin x 1
2
Principle of
zero products
1
2
5
,
.
x
6 6
sin x 2
All four values check. The solutions in 0, 2 are 0, 6,
56, and .
EXAMPLE 11
Study Tip
2π
0
π
Xscl 6
The solutions in 0, 2 are 0, 6, 56,
and .
Solve tan2 x sec x 1 0 in 0, 2 .
Algebraic Solution
Check your solutions to the oddnumbered exercises in the exercise
sets with the step-by-step
annotated solutions in the
Student’s Solutions Manual. If you
are still having difficulty with the
concepts of this section, make time
to view the content video that
corresponds to the section.
We have
tan2 x sec x 1 0
sec2 x 1 sec x 1 0
Using the identity
1 tan2 x sec2 x , or
tan2 x sec2 x 1
sec2 x sec x 2 0
Factoring
sec x 2 sec x 1 0
Principle of zero
sec x 2
or sec x 1
products
1
cos x 2
2 4
x
,
3 3
or cos x 1
or
Using the identity
cos x 1sec x
x 0.
All these values check. The solutions in 0, 2 are 0, 23, and 43.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 581
Section 6.5
• Solving Trigonometric Equations
581
Graphical Solution
We graph y tan2 x sec x 1, but we enter this equation in the
form
y1 tan2 x 1
1.
cos x
We use the ZERO feature to find zeros of the function.
y tan2 x sec x 1
3
2π
0
3
π
Xscl 3
The solutions in 0, 2 are 0, 23, and 43.
Sometimes we cannot find solutions algebraically, but we can approximate them with a graphing calculator.
EXAMPLE 12
Solve each of the following in 0, 2 .
a) x 1.5 cos x
b) sin x cos x cot x
2
Solution
a) In the screen on the left below, we graph y1 x 2 1.5 and y2 cos x
and look for points of intersection. In the screen on the right, we graph
y1 x 2 1.5 cos x and look for the zeros of the function.
y1 x 2 1.5, y2 cos x
y1 x 2 1.5 cos x
2
2
2π
0
Intersection
X 1.3215996 Y .24662556
2
π
Xscl 4
2π
0
Zero
X 1.3215996 Y 0
2
π
Xscl 4
We determine the solution in 0, 2 to be approximately 1.32.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
582
Chapter 6
1/10/05
1:30 PM
Page 582
• Trigonometric Identities, Inverse Functions, and Equations
b) In the screen on the left, we graph y1 sin x cos x and y2 cot x and
determine the points of intersection. In the screen on the right, we graph
the function y1 sin x cos x cot x and determine the zeros.
y1 sin x cos x, y2 1/tan x
y1 sin x cos x 1/tan x
3
3
2π
0
3
Intersection
X 1.1276527
X 5.661357
Y .47462662
Y 1.395337
2π
0
3
Zero
X 1.1276527
X 5.661357
Y0
Y0
Each method leads to the approximate solutions 1.13 and 5.66 in
0, 2 .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
584
Chapter 6
6.5
1/10/05
1:30 PM
Page 584
• Trigonometric Identities, Inverse Functions, and Equations
Exercise Set
Solve, finding all solutions. Express the solutions in both
radians and degrees.
3
2
1. cos x 2. sin x 2
2
3. tan x 3 5. sin x 1
2
4. cos x 1
2
6. tan x 1 2
7. cos x 2
3
8. sin x 2
Solve, finding all solutions in 0, 2 or 0, 360.
Verify your answer using a graphing calculator.
9. 2 cos x 1 1.2814 98.09, 261.91
10. sin x 3 2.0816 246.69, 293.31
4 5
11. 2 sin x 3 0
,
3 3
19. 6 cos2 5 cos 1 0 20. 2 sin t cos t 2 sin t cos t 1 0 21. sin 2x cos x sin x 0 22. 5 sin2 x 8 sin x 3 198.28, 341.72
23. cos2 x 6 cos x 4 0 139.81, 220.19
24. 2 tan2 x 3 tan x 7 70.12, 250.12
25. 7 cot 2 x 4 cot x
37.22, 169.35, 217.22,
349.35
26. 3 sin2 x 3 sin x 2 207.22, 332.78
Solve, finding all solutions in 0, 2 .
27. cos 2x sin x 1 0, , 7 , 11
6
6
y1 cos 2x sin x,
y2 1
4
0
2p
y1 2 sin x 3
4
4
0
2p
28. 2 sin x cos x sin x 0 0,
2
4
, ,
3
3
y 2 sin x cos x sin x
4
4
12. 2 tan x 4 1 68.20, 248.20
0
y1 2 tan x 4,
y2 1
4
4
0
2p
29. sin 4x 2 sin 2x 0 0,
30. tan x sin x tan x 0
4
13. 2 cos2 x 1
2p
14. csc2 x 4 0 3
, ,
2
2
0, 31. sin 2x cos x sin x 0
0, 32. cos 2x sin x sin x 0
0, 2, , 3 2
15. 2 sin2 x sin x 1
33. 2 sec x tan x 2 sec x tan x 1 0
3 4, 7 4
34. sin 2x sin x cos 2x cos x cos x 0, 2, , 3 2
18. 2 sin2 7 sin 4
5
35. sin 2x sin x 2 cos x 1 0 2 3, 4 3, 3 2
16. cos2 x 2 cos x 3 0
3 11
,
,
,
17. 2 cos2 x 3 cos x 0
6 2 2
6
6
,
6
Answers to Exercises 1–8, 13–15, and 19–21 can be found on p. IA-44.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:30 PM
Page 585
Section 6.5
36. tan2 x 4 2 sec2 x tan x
2.034, 5.176, 4, 5 4
37. sec2 x 2 tan2 x 0 4, 3 4, 5 4, 7 4
• Solving Trigonometric Equations
585
52. Daylight Hours. The data in the following table
give the number of daylight hours for certain days
in Fairbanks, Alaska.
38. cot x tan 2x 3 6, 5 6, 7 6, 11 6
39. 2 cos x 2 sin x 6 12, 5 12
40. 3 cos x sin x 1
6, 3 2
41. sec2 x 2 tan x 6 0.967, 1.853, 4.109, 4.994
42. 5 cos 2x sin x 4 0.379, 2.763, 3.416, 6.009
43. cos x sin x 44.
2
1
sin2 x 1
2
1
2
cos
x 1
2
2 4
,
3 3
3
,
4 4
Solve using a calculator, finding all solutions in 0, 2 .
45. x sin x 1 1.114, 2.773 46. x 2 2 sin x
No solutions in 0, 2 47. 2 cos2 x x 1 0.515 48. x cos x 2 0 5.114
49. cos x 2 x 2 3x
0.422, 1.756
GCM
50. sin x tan
0, 2, 3 2
x
2
Some graphing calculators can use regression to fit a
trigonometric function to a set of data.
51. Sales. Sales of certain products fluctuate in cycles.
The data in the following table show the total sales
of skis per month for a business in a northern
climate.
MONTH, x
August,
8
November, 11
February,
2
May,
5
August,
8
TOTAL SALES, y
(IN THOUSANDS)
$ 0
7
14
7
0
y 7 sin 2.6180x 0.5236 7
a) Using the SINE REGRESSION feature on a
graphing calculator, fit a sine function of the
form y A sin Bx C D to this set of data.
b) Approximate the total sales for December and
for July. $10,500; $13,062
NUMBER OF DAYLIGHT
HOURS, y
DAY, x
January 10,
February 19,
March 3,
April 28,
May 14,
June 11,
July 17,
August 22,
September 19,
October 1,
November 14,
December 28,
10
50
62
118
134
162
198
234
262
274
318
362
4.7
9.0
10.3
16.7
18.5
21.4
19.9
15.8
12.7
11.3
6.4
3.8
Source: Astronomical Applications Department; U.S.
Naval Observatory, Washington, DC
a) Using the SINE REGRESSION feature on a graphing
calculator, model these data with an equation of
the form y A sin Bx C D. b) Approximate the number of daylight hours in
Fairbanks for April 22 (x 112), July 4 (x 185),
and December 15 (x 349). c) Determine on which day of the year there will be
about 10.5 hr of daylight. 64th day (March 5th)
Answers to Exercises 52(a) and 52(b) can be found on p. IA-44.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
586
Chapter 6
1/10/05
1:30 PM
Page 586
• Trigonometric Identities, Inverse Functions, and Equations
Collaborative Discussion and Writing
53. Jan lists her answer to a problem as 6 k,
for any integer k, while Jacob lists his answer as
6 2k and 76 2k , for any integer k.
Are their answers equivalent? Why or why not?
54. Under what circumstances will a graphing calculator
give exact solutions of a trigonometric equation?
Skill Maintenance
Solve the right triangle.
55. C
201
b
[5.2] B 35,
b 140.7,
c 245.4
B
67. Temperature During an Illness. The temperature T,
in degrees Fahrenheit, of a patient t days into a
12-day illness is given by
Tt 101.6 3 sin
t .
8
Find the times t during the illness at which the
patient’s temperature was 103. 1.24 days, 6.76 days
68. Satellite Location. A satellite circles the earth in
such a manner that it is y miles from the equator
(north or south, height from the surface not
considered) t minutes after its launch, where
y 5000 cos
t 10 .
45
At what times t in the interval 0, 240, the first 4 hr,
is the satellite 3000 mi north of the equator?
c
55
23.28 min, 113.28 min, 203.28 min
69. Nautical Mile. (See Exercise 60 in Exercise Set 6.2.)
In Great Britain, the nautical mile is defined as the
length of a minute of arc of the earth’s radius. Since
the earth is flattened at the poles, a British nautical
mile varies with latitude. In fact, it is given, in feet,
by the function
A
56.
T
3.8
14.2
[5.2] R 15.5,
T 74.5,
t 13.7
S
N 6066 31 cos 2,
t
where is the latitude in degrees. At what latitude
north is the length of a British nautical mile found
to be 6040 ft? 16.5N
R
Solve.
x
4
57.
27
3
[2.1] 36
58.
0.01 0.2
0.7
h
[2.1] 14
Synthesis
Solve in 0, 2 .
3
59. sin x 2
60. cos x 2 4 5
, , ,
3 3 3 3
g 9.780491 0.005288 sin2 0.000006 sin2 2 ,
2 4 5
, , ,
3 3 3 3
1
2
where g is measured in meters per second per second
at sea level. At what latitude north does g 9.8?
4
61. tan x 3 3, 4 3
37.95615N
62. 12 sin x 7sin x 1 0 0.063, 0.111, 3.031,
63. ln cos x 0 0
3.079
64. e sin x 1 0, 65. sin ln x 1 e 3 /22k, where k (an integer) 1
66. e ln sin x 1 2
70. Acceleration Due to Gravity. (See Exercise 61 in
Exercise Set 6.2.) The acceleration due to gravity is
often denoted by g in a formula such as S 12 gt 2,
where S is the distance that an object falls in
t seconds. The number g is generally considered
constant, but in fact it varies slightly with latitude.
If stands for latitude, in degrees, an excellent
approximation of g is given by the formula
Solve.
71. cos1 x cos1 35 sin1
1
72. sin
1 1
x tan
3
4
5
tan1 12
1
22
73. Suppose that sin x 5 cos x . Find sin x cos x .
0.1923
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:31 PM
Page 587
Chapter 6
• Summary and Review
Chapter 6 Summary and Review
Important Properties and Formulas
Double-Angle Identities
sin 2x 2 sin x cos x ,
cos 2x cos2 x sin2 x
1 2 sin2 x
2 cos2 x 1,
2 tan x
tan 2x 1 tan2 x
Basic Identities
sin x
1
tan x ,
,
sin x csc x
cos x
cos x
1
cot x cos x ,
,
sec x
sin x
1
tan x ,
cot x
sin x sin x ,
cos x cos x ,
tan x tan x
Half-Angle Identities
Pythagorean Identities
sin2 x cos2 x 1,
1 cot 2 x csc2 x ,
1 tan2 x sec2 x
Sum and Difference Identities
sin u v sin u cos v cos u sin v ,
cos u v cos u cos v sin u sin v ,
tan u tan v
tan u v 1 tan u tan v
Cofunction Identities
sin
x cos x ,
2
tan
sec
x cot x ,
2
sin x
2
cos x ,
cos x
2
sin x
sin
x
2
1 cos x
,
2
cos
x
2
1 cos x
,
2
tan
x
2
1 cos x
1 cos x
sin x
1 cos x
1 cos x
sin x
x csc x
2
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
587
BBEPMC06_0321279115.QXP
588
Chapter 6
1/10/05
1:31 PM
Page 588
• Trigonometric Identities, Inverse Functions, and Equations
Inverse Trigonometric Functions
FUNCTION
DOMAIN
y sin1 x
1, 1
y cos1 x
1, 1
y tan1 x
, RANGE
,
2 2
0, ,
2 2
Composition of Trigonometric Functions
The following are true for any x in the domain of
the inverse function:
sin sin1 x x ,
cos cos1 x x ,
tan tan1 x x .
The following are true for any x in the range of the
inverse function:
sin1 sin x x ,
cos1 cos x x ,
tan1 tan x x .
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
588
Chapter 6
1/10/05
1:31 PM
Page 588
• Trigonometric Identities, Inverse Functions, and Equations
Review Exercises
Complete the Pythagorean identity.
1. 1 cot 2 x [6.1] csc2 x
11.
2. sin2 x cos2 x [6.1] 1
Multiply and simplify. Check using a graphing
calculator.
3. tan y cot y tan y cot y [6.1] tan2 y cot 2 y
4. cos x sec x2
[6.1]
cos2 x 12
cos2 x
3
2
cos y sin y sin2 y cos2 y
12.
13.
4 sin x cos2 x
16 sin2 x cos x
cot x
csc x
2
1
csc2 x
[6.1] 1
[6.1]
1
cot x
4
14. Simplify. Assume the radicand is nonnegative.
Factor and simplify. Check using a graphing calculator.
5. sec x csc x csc2 x [6.1] csc x sec x csc x
6. 3 sin2 y 7 sin y 20 [6.1] 3 sin y 5 sin y 4
sin2 x 2 cos x sin x cos2 x
[6.1] sin x cos x
7. 1000 cos3 u
[6.1] 10 cos u 100 10 cos u cos2 u
Simplify and check using a graphing calculator.
sec4 x tan4 x
2 sin2 x
cos x
8. 2
[6.1] 1 9.
2
sec x tan x
cos31 x
2 sin x
2
[6.1] 2 sec x
3 sin x cos2 x cos x sin x
3 tan x
10.
[6.1]
sin x cos x
cos2 x
sin2 x cos2 x
15. Rationalize the denominator:
16. Rationalize the numerator:
1 sin x
. 1 sin x
cos x
. tan x
17. Given that x 3 tan , express 9 x 2 as a
trigonometric function without radicals. Assume
that 0 2. [6.1] 3 sec Answers to Exercises 11, 15, and 16 can be found on p. IA-44.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:31 PM
Page 589
Chapter 6
Use the sum and difference formulas to write equivalent
expressions. You need not simplify.
3
18. cos x 19. tan 45 30 2
Prove the identity.
1 sin x
cos x
36.
cos x
1 sin x
20. Simplify: cos 27 cos 16 sin 27 sin 16. 37.
1 cos 2
cot sin 2
38.
tan y sin y
y
cos2
2 tan 2
39.
sin x cos x
tan2 x 1
cos2 x
sin x cos x
21. Find cos 165 exactly. 22. Given that tan 3 and sin 22 and that
and are between 0 and 2, evaluate
tan exactly. [6.1] 2 3
23. Assume that sin 0.5812 and cos 0.2341 and
that both and are first-quadrant angles. Evaluate
cos . [6.1] 0.3745
Complete the cofunction identity.
24. cos x 25. cos
2
[6.2] sin x
26. sin x 2
x 2
[6.2] sin x
[6.2] cos x
27. Given that cos 35 and that the terminal side is
in quadrant III:
a) Find the other function values for . b) Find the six function values for 2 . c) Find the six function values for 2. 28. Find an equivalent expression for csc x [6.2] sec x
.
2
29. Find tan 2, cos 2, and sin 2 and the quadrant in
which 2 lies, where cos 45 and is in
quadrant III. 30. Find sin
2 2
exactly. [6.2]
2
8
31. Given that sin 0.2183 and is in quadrant I,
find sin 2, cos
, and cos 4. 2
Simplify and check using a graphing calculator.
x
32. 1 2 sin2
[6.2] cos x
2
33. sin x cos x2 sin 2x
[6.2] 1
34. 2 sin x cos x 2 sin x cos x
3
35.
3
2 cot x
[6.2] tan 2x
cot 2 x 1
[6.2] sin 2x
589
• Review Exercises
In Exercises 40 – 43, use a graphing calculator to
determine which expression (A)–(D) on the right can
be used to complete the identity. Then prove the identity
algebraically.
csc x
40. csc x cos x cot x A.
sec x
1
cos x
41.
B. sin x
sin x cos x sin x
2
cot x 1
C.
42.
sin x
1 tan x
sin x cos x
cos x 1
sin x
D.
43.
1 sin2 x
sin x
cos x 1
Find each of the following exactly in both radians
and degrees.
1
3
44. sin1 45. cos1
2
2
[6.4] 6, 30
46. tan1 1
[6.4] 4, 45
[6.4] 6, 30
47. sin1 0 [6.4] 0, 0
Use a calculator to find each of the following in radians,
rounded to four decimal places, and in degrees, rounded
to the nearest tenth of a degree.
48. cos1 0.2194
49. cot 1 2.381
[6.4] 1.7920, 102.7
Evaluate.
[6.4] 0.3976, 22.8
2
53. cos sin1
7
2
[6.4]
3
3
[6.4]
2
2
[6.4]
257
50. cos cos1
1
2
[6.4] 12
52. sin1 sin
7
[6.4]
Find.
54. cos tan1
Answers to Exercises 18–21, 27, 29, 31, and 36 –43 can be found on pp. IA-44 and IA-45.
b
3
[6.4]
51. tan1 tan
3
b 2
3
3
55. cos 2 sin1
9
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
4
5
BBEPMC06_0321279115.QXP
590
Chapter 6
1/10/05
1:31 PM
Page 590
• Trigonometric Identities, Inverse Functions, and Equations
Solve, finding all solutions. Express the solutions in both
radians and degrees.
2
56. cos x 57. tan x 3 2
Solve, finding all solutions in 0, 2 .
58. 4 sin2 x 1 y
4
y 1
3
2
1
2 x
1
y 4 sin2 x
68. Prove the identity 2 cos2 x 1 cos4 x sin4 x in
three ways:
a) Start with the left side and deduce the right
(method 1).
b) Start with the right side and deduce the left
(method 1).
c) Work with each side separately until you deduce
the same expression (method 2).
Then determine the most efficient method and
explain why you chose that method. 69. Why are the ranges of the inverse trigonometric
functions restricted? Synthesis
70. Find the measure of the angle from l1 to l2:
59. sin 2x sin x cos x 0 l1: x y 3
y
2
Collaborative Discussion and Writing
y sin 2x sin x − cos x
71. Find an identity for cos u v involving only
cosines. 1
2 x
1
y 0
2
60. 2 cos2 x 3 cos x 1
[6.5] 23, , 43
61. sin2 x 7 sin x 0 [6.5] 0, 62. csc2 x 2 cot 2 x 0 [6.5] 4, 34, 54, 74
72. Simplify: cos
2
[6.2] cos x
x csc x sin x.
2
73. Find sin , cos , and tan under the given
conditions:
sin 2 1 ,
2 . 5 2
74. Prove the following equation to be an identity:
ln e sin t sin t . 63. sin 4x 2 sin 2x 0 [6.5] 0, 2, , 32
75. Graph: y sec1 x . 64. 2 cos x 2 sin x 2
76. Show that
[6.5] 712, 2312
65. 6 tan2 x 5 tan x sec2 x
[6.5] 0.864, 2.972, 4.006, 6.114
Solve using a graphing calculator, finding all solutions
in 0, 2 .
66. x cos x 1 [6.5] 4.917
l2: 2x y 5.
[6.1]
108.4
tan1 x is not an identity.
sin1 x
cos1 x
77. Solve e cos x 1 in 0, 2.
[6.5] 2, 32
67. 2 sin2 x x 1 [6.5] No solution in 0, 2 Answers to Exercises 56–59, 68, 69, 71, and 73–76 can be found on pp. IA-45 and IA-46.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC06_0321279115.QXP
1/10/05
1:31 PM
Page 591
Chapter 6
• Test
591
Chapter 6 Test
Simplify.
2 cos2 x cos x 1
1.
cos x 1
2.
sec x
tan x
2
1
tan2 x
Prove each of the following identities.
13. csc x cos x cot x sin x [6.1] 2 cos x 1
14. sin x cos x2 1 sin 2x
15. csc cot 2 [6.1] 1
3. Rationalize the denominator:
1 sin cos . [6.1]
1 sin 1 sin 4. Given that x 2 sin , express 4 x 2 as a
trigonometric function without radicals. Assume
0 2. [6.1] 2 cos Use the sum or difference identities to evaluate exactly.
5. sin 75 6. tan
12
7. Assuming that cos u 135 and cos v 12
13 and that u
and v are between 0 and 2, evaluate cos u v
exactly. [6.1] 120
169
23
8. Given that cos and that the terminal side is
in quadrant II, find cos 2 . [6.2] 53
9. Given that sin 45 and is in quadrant III, find
, II
sin 2 and the quadrant in which 2 lies. [6.2] 24
25
10. Use a half-angle identity to evaluate cos
exactly. 12
11. Given that sin 0.6820 and that is in
quadrant I, find cos 2. [6.2] 0.9304
12. Simplify: sin x cos x2 1 2 sin 2x . [6.2] 3 sin 2x
16.
1 cos 1 cos 1 sin tan 1 csc sec [6.4] 45
17. Find sin1 2
2
exactly in degrees.
18. Find tan1 3 exactly in radians.
[6.4] 3
19. Use a calculator to find cos1 0.6716 in radians,
rounded to four decimal places. [6.4] 2.3072
20. Evaluate cos sin1
21. Find tan sin1
1
3
. [6.4]
2
2
5
5
. [6.4]
2
x
x 25
22. Evaluate cos sin1 12 cos1
1
2
.
[6.4] 0
Solve, finding all solutions in 0, 2 .
23. 4 cos2 x 3 [6.5] 6, 56, 76, 116
24. 2 sin2 x 2 sin x
[6.5] 0, 4, 34, 11
25. 3 cos x sin x 1 [6.5] ,
2 6
Synthesis
26. Find cos , given that cos 2 [6.2] 11
12
Answers to Exercises 5, 6, 10, and 13–16 can be found on p. IA-46.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5 3
,
2.
6 2
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 593
Applications of
Trigonometry
7.1
7.2
7.3
7.4
7.5
7.6
7
The Law of Sines
The Law of Cosines
Complex Numbers: Trigonometric Form
Polar Coordinates and Graphs
Vectors and Applications
Vector Operations
SUMMARY AND REVIEW
TEST
A P P L I C A T I O N
A
n eagle flies from its nest 7 mi in the
direction northeast, where it stops to rest
on a cliff. It then flies 8 mi in the direction
S30°W to land on top of a tree. Place an xy-coordinate
system so that the origin is the bird’s nest, the x-axis
points east, and the y-axis points north. (a) At what
point is the cliff located? (b) At what point is the tree
located?
This problem appears as Exercise 55 in Exercise Set 7.5.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
594
Chapter 7
1/10/05
1:35 PM
Page 594
• Applications of Trigonometry
2.1
7.1
The Law
Polynomial
of Sines
Functions and
Modeling
Use the law of sines to solve triangles.
Find the area of any triangle given the lengths of two sides and the
measure of the included angle.
To solve a triangle means to find the lengths of all its sides and the
measures of all its angles. We solved right triangles in Section 5.2. For
review, let’s solve the right triangle shown below. We begin by listing the
known measures.
Q 37.1
W 90
Z?
Z
q
W
w
37.1
6.3
q?
w?
z 6.3
Q
Since the sum of the three angle measures of any triangle is 180, we
can immediately find the measure of the third angle:
Z 180 90 37.1
52.9.
Then using the tangent and cosine ratios, respectively, we can find q
and w :
q
, or
6.3
q 6.3 tan 37.1 4.8,
tan 37.1 and
6.3
, or
w
6.3
w
7.9.
cos 37.1
cos 37.1 Now all six measures are known and we have solved triangle QWZ.
Q 37.1
W 90
Z 52.9
q 4.8
w 7.9
z 6.3
Solving Oblique Triangles
The trigonometric functions can also be used to solve triangles that are
not right triangles. Such triangles are called oblique. Any triangle, right
or oblique, can be solved if at least one side and any other two measures
are known. The five possible situations are illustrated on the next page.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 595
Section 7.1
1. AAS: Two angles of a triangle
and a side opposite one of them
are known.
• The Law of Sines
595
224
100
2. ASA: Two angles of a triangle
and the included side are
known.
25
AAS
31
51
37.5
ASA
3. SSA: Two sides of a triangle
and an angle opposite one of
them are known. (In this case,
there may be no solution, one
solution, or two solutions. The
latter is known as the
ambiguous case.)
38 q
115.7
20 ~
SSA
4. SAS: Two sides of a triangle
and the included angle are
known.
82.14
58
19.05
SAS
5. SSS: All three sides of the
triangle are known.
210
75
172
SSS
The list above does not include the situation in which only the three
angle measures are given. The reason for this lies in the fact that the
angle measures determine only the shape of the triangle and not the size,
as shown with the following triangles. Thus we cannot solve a triangle
when only the three angle measures are given.
10
10
30
140
10
30 140
30
140
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
596
Chapter 7
1/10/05
1:35 PM
Page 596
• Applications of Trigonometry
In order to solve oblique triangles, we need to derive the law of sines
and the law of cosines. The law of sines applies to the first three situations listed above. The law of cosines, which we develop in Section 7.2,
applies to the last two situations.
The Law of Sines
We consider any oblique triangle. It may or may not have an obtuse
angle. Although we look at only the acute-triangle case, the derivation of
the obtuse-triangle case is essentially the same.
In acute ABC at left, we have drawn an altitude from vertex C. It
has length h. From ADC , we have
C
b
A
c
h
D
a
sin A B
h
,
b
or
h b sin A.
or
h a sin B.
From BDC , we have
sin B h
,
a
With h b sin A and h a sin B, we now have
a sin B b sin A
a sin B
b sin A
sin A sin B sin A sin B
a
b
.
sin A sin B
Dividing by sin A sin B
Simplifying
There is no danger of dividing by 0 here because we are dealing with triangles whose angles are never 0 or 180. Thus the sine value will never
be 0.
If we were to consider altitudes from vertex A and vertex B in the triangle shown above, the same argument would give us
b
c
sin B sin C
and
a
c
.
sin A sin C
We combine these results to obtain the law of sines.
The Law of Sines
In any triangle ABC,
b
c
a
.
sin A sin B sin C
Thus in any triangle, the sides are
proportional to the sines of the
opposite angles.
B
a
c
A
b
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
C
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 597
Section 7.1
• The Law of Sines
597
Study Tip
Solving Triangles (AAS and ASA)
Maximize the learning that you
accomplish during a lecture by
preparing for the class. Your
time is valuable; let each lecture
become a positive learning
experience. Review the lesson from
the previous class and read the
section that will be covered in the
next lecture. Write down questions
you want answered and take an
active part in class discussion.
When two angles and a side of any triangle are known, the law of sines
can be used to solve the triangle.
EXAMPLE 1
triangle.
Solution
In EFG , e 4.56, E 43, and G 57. Solve the
We first make a drawing. We know three of the six measures.
E 43
F?
G 57
F
g
E
4.56
43
57
f
e 4.56
f?
g?
G
From the figure, we see that we have the AAS situation. We begin by
finding F :
F 180 43 57 80.
We can now find the other two sides, using the law of sines:
f
e
sin F sin E
f
4.56
sin 80 sin 43
4.56 sin 80
f
sin 43
f 6.58;
g
e
sin G sin E
g
4.56
sin 57 sin 43
4.56 sin 57
g
sin 43
g 5.61.
Substituting
Solving for f
Substituting
Solving for g
Thus, we have solved the triangle:
E 43,
F 80,
G 57,
e 4.56,
f 6.58,
g 5.61.
The law of sines is frequently used in determining distances.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
598
Chapter 7
1/10/05
1:35 PM
Page 598
• Applications of Trigonometry
EXAMPLE 2 Rescue Mission. During a rescue mission, a Marine fighter
pilot receives data on an unidentified aircraft from an AWACS plane and
is instructed to intercept the aircraft. The diagram shown below appears
on the screen, but before the distance to the point of interception appears
on the screen, communications are jammed. Fortunately, the pilot remembers the law of sines. How far must the pilot fly?
Z
y
115
X
(Unidentified
aircraft)
x
27
500 km
Y
(Pilot)
Solution We let x represent the distance that the pilot must fly in order
to intercept the aircraft and Z represent the point of interception. We
first find angle Z :
Z 180 115 27
38.
Because this application involves the ASA situation, we use the law of
sines to determine x :
x
z
sin X sin Z
x
500
sin 115 sin 38
500 sin 115
x
sin 38
x 736.
Substituting
Solving for x
Thus the pilot must fly approximately 736 km in order to intercept the
unidentified aircraft.
Solving Triangles (SSA)
When two sides of a triangle and an angle opposite one of them are
known, the law of sines can be used to solve the triangle.
Suppose for ABC that b, c, and B are given. The various possibilities are as shown in the eight cases below: five cases when B is acute and
three cases when B is obtuse. Note that b c in cases 1, 2, 3, and 6; b c
in cases 4 and 7; and b c in cases 5 and 8.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 599
Section 7.1
Angle B Is Acute
Case 1: No solution
b c ; side b is too short to
reach the base. No triangle is
formed.
Case 2 : One solution
b c ; side b just reaches the
base and is perpendicular to it.
A
599
• The Law of Sines
Case 3 : Two solutions
b c ; an arc of radius b
meets the base at two points.
(This case is called the
ambiguous case.)
A
A
b
c
c
B
B
Case 4 : One solution
b c ; an arc of radius b
meets the base at just one
point, other than B.
c
Angle B Is Obtuse
Case 6 : No solution
b c ; side b is too short to
reach the base. No triangle
is formed.
b
B
C
Case 7 : No solution
b c ; an arc of radius b meets
the base only at point B. No
triangle is formed.
b
c
B
Case 8: One solution
b c ; an arc of radius b meets
the base at just one point.
A
A
b
c
C
A
C
A
b
Case 5 : One solution
b c ; an arc of radius b meets
the base at just one point.
b
B
b
B
C
A
c
c
b
b
c
B
B
C
The eight cases above lead us to three possibilities in the SSA
situation: no solution, one solution, or two solutions. Let’s investigate
these possibilities further, looking for ways to recognize the number of
solutions.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
600
Chapter 7
1/10/05
1:35 PM
Page 600
• Applications of Trigonometry
EXAMPLE 3 No solution.
Solve the triangle.
Solution
In QRS , q 15, r 28, and Q 43.6.
We make a drawing and list the known measures.
Q 43.6
R?
S?
S
15
28
q 15
r 28
s?
43.6
Q
s
?R?
We observe the SSA situation and use the law of sines to find R:
q
r
sin Q sin R
15
28
sin 43.6 sin R
28 sin 43.6
sin R 15
sin R 1.2873.
Substituting
Solving for sin R
Since there is no angle with a sine greater than 1, there is no solution.
EXAMPLE 4 One solution.
39.7. Solve the triangle.
Solution
In XYZ , x 23.5, y 9.8, and X We make a drawing and organize the given information.
X
z
Y
23.5
39.7
9.8
X 39.7
Y?
Z?
x 23.5
y 9.8
z?
Z
We see the SSA situation and begin by finding Y with the law of sines:
y
15.4°
164.6°
x
x
y
sin X sin Y
23.5
9.8
sin 39.7 sin Y
9.8 sin 39.7
sin Y 23.5
sin Y 0.2664.
Substituting
Solving for sin Y
There are two angles less than 180 with a sine of 0.2664. They are 15.4
and 164.6, to the nearest tenth of a degree. An angle of 164.6 cannot be
an angle of this triangle because it already has an angle of 39.7 and these
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 601
Section 7.1
• The Law of Sines
601
two angles would total more than 180. Thus, 15.4 is the only possibility
for Y. Therefore,
Z 180 39.7 15.4 124.9.
We now find z:
z
x
sin Z sin X
z
23.5
sin 124.9 sin 39.7
23.5 sin 124.9
z
sin 39.7
z 30.2.
Substituting
Solving for z
We now have solved the triangle:
X 39.7,
Y 15.4,
Z 124.9,
x 23.5,
y 9.8,
z 30.2.
The next example illustrates the ambiguous case in which there are
two possible solutions.
EXAMPLE 5 Two solutions.
Solve the triangle.
In ABC , b 15, c 20, and B 29.
Solution We make a drawing, list the known measures, and see that we
again have the SSA situation.
A?
B 29
C?
A
20
B
15
29
a
a?
b 15
c 20
C
We first find C:
y
sin 40 sin 140
140
40
x
b
c
sin B sin C
15
20
sin 29 sin C
20 sin 29
0.6464.
sin C 15
Substituting
Solving for sin C
There are two angles less than 180 with a sine of 0.6464. They are 40
and 140, to the nearest degree. This gives us two possible solutions.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
602
Chapter 7
1/10/05
1:35 PM
Page 602
• Applications of Trigonometry
Possible Solution I.
Possible Solution II.
If C 40, then
If C 140, then
A 180 29 40 111.
A 180 29 140 11.
Then we find a:
Then we find a:
a
b
sin A sin B
a
15
sin 111 sin 29
15 sin 111
a
29.
sin 29
a
b
sin A sin B
a
15
sin 11 sin 29
15 sin 11
a
6.
sin 29
These measures make a triangle as shown
below; thus we have a solution.
These measures make a triangle as shown
below; thus we have a second solution.
A
20
B
111
A
15
40
29
29
20
29
B 6
C
C
15
140
11
Examples 3 – 5 illustrate the SSA situation. Note that we need not
memorize the eight cases or the procedures in finding no solution, one
solution, or two solutions. When we are using the law of sines, the sine
value leads us directly to the correct solution or solutions.
The Area of a Triangle
The familiar formula for the area of a triangle, A 12 bh, can be used
only when h is known. However, we can use the method used to derive
the law of sines to derive an area formula that does not involve the height.
Consider a general triangle ABC , with area K, as shown below.
B
c
A
B
a
h
h
C
D
D
A is acute.
a
c
A
A is obtuse.
C
Note that in the triangle on the right, sin A sin 180 A. Then in
each ADB ,
sin A h
,
c
or h c sin A.
1
Substituting into the formula K 2 bh, we get
1
K 2 bc sin A.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 603
Section 7.1
• The Law of Sines
603
Any pair of sides and the included angle could have been used. Thus we
also have
K 12 ab sin C
and
1
K 2 ac sin B.
The Area of a Triangle
The area K of any ABC is one half the product of the lengths of
two sides and the sine of the included angle:
K
1
1
1
bc sin A ab sin C ac sin B.
2
2
2
EXAMPLE 6 Area of the Peace Monument. Through the Mentoring in
the City Program sponsored by Marian College, in Indianapolis, Indiana,
children have turned a vacant downtown lot into a monument for
peace.* This community project brought together neighborhood volunteers, businesses, and government in hopes of showing children how to
develop positive, nonviolent ways of dealing with conflict. A landscape
architect† used the children’s drawings and ideas to design a triangularshaped peace garden. Two sides of the property, formed by Indiana
Avenue and Senate Avenue, measure 182 ft and 230 ft, respectively, and
together form a 44.7 angle. The third side of the garden, formed by an
apartment building, measures 163 ft. What is the area of this property?
163 ft
182 ft
44.7 Solution
formula:
230 ft
Since we do not know a height of the triangle, we use the area
K 12 bc sin A
K 12 182 ft 230 ft sin 44.7
K 14,722 ft 2.
The area of the property is approximately 14,722 ft 2.
*The Indianapolis Star, August 6, 1995, p. J8.
†Alan Day, a landscape architect with Browning Day Mullins Dierdorf, Inc., donated his
time to this project.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
604
Chapter 7
7.1
1/10/05
1:35 PM
Page 604
• Applications of Trigonometry
Exercise Set
Solve the triangle, if possible.
1. B 38, C 21, b 24 A 121, a 33, c 14
2. A 131, C 23, b 10
or at least approximate, the area of the yard. Find the
area of the yard to the nearest square foot. 787 ft 2
B 26, a 17, c 9
3. A 36.5, a 24, b 34
B 57.4, C 86.1,
c 40, or B 122.6, C 20.9, c 14
4. B 118.3, C 45.6, b 42.1
C
A 16.1, a 13.3, c 34.2
B
42 ft
5. C 6110, c 30.3, b 24.2
135°
53 ft
B 4424, A 7426, a 33.3
6. A 126.5, a 17.2, c 13.5
A
C 39.1, B 14.4, b 5.3
7. c 3 mi, B 37.48, C 32.16
A 110.36, a 5 mi, b 3 mi
8. a 2345 mi, b 2345 mi, A 124.67
No solution
9. b 56.78 yd, c 56.78 yd, C 83.78
B 83.78, A 12.44, a 12.30 yd
10. A 12932, C 1828, b 1204 in.
B 32, a 1752 in., c 720 in.
11. a 20.01 cm, b 10.07 cm, A 30.3
B 14.7, C 135.0, c 28.04 cm
12. b 4.157 km, c 3.446 km, C 5148 13. A 89, a 15.6 in., b 18.4 in.
No solution
14. C 4632, a 56.2 m, c 22.1 m No solution
24. Boarding Stable. A rancher operates a boarding
stable and temporarily needs to make an extra pen.
He has a piece of rope 38 ft long and plans to tie
the rope to one end of the barn (S) and run the
rope around a tree (T) and back to the barn (Q).
The tree is 21 ft from where the rope is first tied,
and the rope from the barn to the tree makes an
angle of 35 with the barn. Does the rancher have
enough rope if he allows 4 12 ft at each end to fasten
the rope? No
15. a 200 m, A 32.76, C 21.97
B 125.27, b 302 m, c 138 m
16. B 115, c 45.6 yd, b 23.8 yd No solution
Find the area of the triangle.
17. B 42, a 7.2 ft, c 3.4 ft 8.2 ft 2
18. A 1712, b 10 in., c 13 in.
19. C 8254, a 4 yd, b 6 yd
Q
2
19 in
12 yd2
S
35°
21 ft
T
20. C 75.16, a 1.5 m, b 2.1 m 1.5 m
2
21. B 135.2, a 46.12 ft, c 36.74 ft 596.98 ft 2
22. A 113, b 18.2 cm, c 23.7 cm 198.5 cm2
Solve.
23. Area of Back Yard. A new homeowner has a
triangular-shaped back yard. Two of the three sides
measure 53 ft and 42 ft and form an included angle
of 135. To determine the amount of fertilizer and
grass seed to be purchased, the owner has to know,
Answer to Exercise 12 can be found on p. IA-46.
25. Rock Concert. In preparation for an outdoor rock
concert, a stage crew must determine how far apart
to place the two large speaker columns on stage
(see the figure at the top of the next page). What
generally works best is to place them at 50 angles
to the center of the front row. The distance from
the center of the front row to each of the speakers is
10 ft. How far apart does the crew need to place the
speakers on stage? About 12.86 ft, or 12 ft 10 in.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 605
Section 7.1
50°
Speaker
605
29. Fire Tower. A ranger in fire tower A spots a fire at a
direction of 295. A ranger in fire tower B, located
45 mi at a direction of 045 from tower A, spots the
same fire at a direction of 255. How far from tower
A is the fire? from tower B? From A: about 35 mi;
50°
from B: about 66 mi
Stage
10 ft
• The Law of Sines
N
10 ft
B
255°
26. Lunar Crater. Points A and B are on opposite sides
of a lunar crater. Point C is 50 m from A. The
measure of BAC is determined to be 112 and the
measure of ACB is determined to be 42. What is
the width of the crater? About 76.3 m
C
Fire
N
45°
45 mi
295° A
27. Length of Pole. A pole leans away from the sun at
an angle of 7 to the vertical. When the angle of
elevation of the sun is 51, the pole casts a shadow
47 ft long on level ground. How long is the pole?
About 51 ft
7°
P
51°
47 ft
In Exercises 28 – 31, keep in mind the two types of
bearing considered in Sections 5.2 and 5.3.
28. Reconnaissance Airplane. A reconnaissance airplane
leaves its airport on the east coast of the United
States and flies in a direction of 085. Because of bad
weather, it returns to another airport 230 km to the
north of its home base. For the return trip, it flies in
a direction of 283. What is the total distance that
the airplane flew? About 1467 km
About 2.4 km or 10.6 km
31. Mackinac Island. Mackinac Island is located
18 mi N3120W of Cheboygan, Michigan, where
the Coast Guard cutter Mackinaw is stationed. A
freighter in distress radios the Coast Guard cutter
for help. It radios its position as S7840E of
Mackinac Island and N6410W of Cheboygan.
How far is the freighter from Cheboygan?
About 22 mi
N
New
airport
30. Lighthouse. A boat leaves lighthouse A and sails
5.1 km. At this time it is sighted from lighthouse B,
7.2 km west of A. The bearing of the boat from B is
N6510E. How far is the boat from B?
Mackinac Island
B
N
230 km
85
283
Original
airport
Cheboygan
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
606
Chapter 7
1/10/05
1:35 PM
Page 606
• Applications of Trigonometry
32. Gears. Three gears are arranged as shown in the
figure below. Find the angle . 89
Synthesis
45. Prove the following area formulas for a general
triangle ABC with area represented by K. a2 sin B sin C
2 sin A
2
c sin A sin B
K
2 sin C
2
b sin C sin A
K
2 sin B
r 28 ft
K
r 22 ft
f
41
46. Area of a Parallelogram. Prove that the area of a
parallelogram is the product of two adjacent sides
and the sine of the included angle. r 36 ft
s1
Collaborative Discussion and Writing
S
33. Explain why the law of sines cannot be used to find
the first angle when solving a triangle given three
sides.
34. We considered eight cases of solving triangles given
two sides and an angle opposite one of them.
Describe the relationship between side b and the
height h in each.
s2
47. Area of a Quadrilateral. Prove that the area of a
quadrilateral is one half the product of the lengths
of its diagonals and the sine of the angle between
the diagonals. c
a
b
d
Skill Maintenance
Find the acute angle A, in both radians and degrees, for
the given function value.
35. cos A 0.2213 [5.1] 1.348, 77.2°
48. Find d. d 18.8 in.
36. cos A 1.5612 [5.1] No angle
Convert to decimal degree notation.
37. 181420 [5.1] 18.24°
d
11 in.
38. 125342 [5.1] 125.06°
39. Find the absolute value: 5. [R.1] 5
Find the values.
3
40. cos
[5.3]
2
6
42. sin 300 [5.3] 3
2
41. sin 45 [5.3]
43. cos 2
3
50
12 in.
2
2
[5.3] 12
44. Multiply: 1 i 1 i. [2.2] 2
Answers to Exercises 45–47 can be found on pp. IA-46 and IA-47.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
15 in.
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 607
Section 7.2
• The Law of Cosines
607
Use the law of cosines to solve triangles.
Determine whether the law of sines or the law of cosines should be
applied to solve a triangle.
7.2
The Law
of Cosines
The law of sines is used to solve triangles given a side and two angles
(AAS and ASA) or given two sides and an angle opposite one of them
(SSA). A second law, called the law of cosines, is needed to solve triangles
given two sides and the included angle (SAS) or given three sides (SSS).
The Law of Cosines
To derive this property, we consider any ABC placed on a coordinate
system. We position the origin at one of the vertices — say, C — and the
positive half of the x-axis along one of the sides — say, CB. Let x, y be
the coordinates of vertex A. Point B has coordinates a, 0 and point C
has coordinates 0, 0.
y
(x, y)
A
c
y
b
(a, 0)
x C (0, 0)
a
Then
cos C x
, so
b
x b cos C
and
sin C y
,
b
y b sin C .
so
B
x
Thus point A has coordinates
b cos C, b sin C.
Next, we use the distance formula to determine c 2:
y
or
(x, y), or
(b cos C, b sin C)
A
y
Now we multiply and simplify:
c
b
(a, 0)
x
c 2 x a2 y 02,
c 2 b cos C a2 b sin C 02.
C (0, 0)
a
B
x
c 2 b 2 cos2 C 2ab cos C a2 b 2 sin2 C
a2 b 2sin2 C cos2 C 2ab cos C
a2 b 2 2ab cos C .
Using the identity
sin2 x cos2 x 1
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
608
Chapter 7
1/10/05
1:35 PM
Page 608
• Applications of Trigonometry
Had we placed the origin at one of the other vertices, we would have
obtained
a2 b 2 c 2 2bc cos A
b 2 a2 c 2 2ac cos B .
or
The Law of Cosines
In any triangle ABC,
or
C
a b c 2bc cos A,
b 2 a2 c 2 2ac cos B,
c 2 a2 b 2 2ab cos C .
2
2
2
b
A
a
B
c
Thus, in any triangle, the square of a side is the sum of the squares
of the other two sides, minus twice the product of those sides and the
cosine of the included angle. When the included angle is 90, the law
of cosines reduces to the Pythagorean theorem.
Solving Triangles (SAS)
When two sides of a triangle and the included angle are known, we can
use the law of cosines to find the third side. The law of cosines or the law
of sines can then be used to finish solving the triangle.
EXAMPLE 1
Solve ABC if a 32, c 48, and B 125.2.
Solution We first label a triangle with the known and unknown
measures.
C
b
A?
B 125.2
C?
a 32
b?
c 48
32
A
48
125.2
B
We can find the third side using the law of cosines, as follows:
b 2 a2 c 2 2ac cos B
b 2 322 48 2 2 32 48 cos 125.2
b 2 5098.8
b 71.
Substituting
We now have a 32, b 71, and c 48, and we need to find the
other two angle measures. At this point, we can find them in two ways.
One way uses the law of sines. The ambiguous case may arise, however,
and we would have to be alert to this possibility. The advantage of using
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 609
Section 7.2
• The Law of Cosines
609
the law of cosines again is that if we solve for the cosine and find that
its value is negative, then we know that the angle is obtuse. If the value
of the cosine is positive, then the angle is acute. Thus we use the law of
cosines to find a second angle.
Let’s find angle A. We select the formula from the law of cosines that
contains cos A and substitute:
a2 b 2 c 2 2bc cos A
322 712 48 2 2 71 48 cos A
1024 5041 2304 6816 cos A
6321 6816 cos A
cos A 0.9273768
A 22.0.
Substituting
The third angle is now easy to find:
C 180 125.2 22.0
32.8.
Thus,
A 22.0,
B 125.2,
C 32.8,
a 32,
b 71,
c 48.
Due to errors created by rounding, answers may vary depending on
the order in which they are found. Had we found the measure of angle C
first in Example 1, the angle measures would have been C 34.1 and
A 20.7. Variances in rounding also change the answers. Had we used
71.4 for b in Example 1, the angle measures would have been A 21.5
and C 33.3.
Suppose we used the law of sines at the outset in Example 1 to find b.
We were given only three measures: a 32, c 48, and B 125.2.
When substituting these measures into the proportions, we see that there
is not enough information to use the law of sines:
a
b
32
b
l
,
sin A sin B
sin A sin 125.2
b
c
b
48
l
,
sin B sin C
sin 125.2 sin C
c
32
48
a
l
.
sin A sin C
sin A sin C
In all three situations, the resulting equation, after the substitutions, still
has two unknowns. Thus we cannot use the law of sines to find b.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
610
Chapter 7
1/10/05
1:35 PM
Page 610
• Applications of Trigonometry
Solving Triangles (SSS)
When all three sides of a triangle are known, the law of cosines can be
used to solve the triangle.
EXAMPLE 2
Solution
Solve RST if r 3.5, s 4.7, and t 2.8.
We sketch a triangle and label it with the given measures.
R?
S?
T?
R
4.7
2.8
S
3.5
r 3.5
s 4.7
t 2.8
T
Since we do not know any of the angle measures, we cannot use the law
of sines. We begin instead by finding an angle with the law of cosines. We
choose to find S first and select the formula that contains cos S :
s 2 r 2 t 2 2rt cos S
4.72 3.52 2.82 23.5 2.8 cos S
3.52 2.82 4.72
cos S 23.5 2.8
cos S 0.1020408
S 95.86.
Substituting
Similarly, we find angle R:
r 2 s 2 t 2 2st cos R
3.52 4.72 2.82 24.7 2.8 cos R
4.72 2.82 3.52
cos R 24.7 2.8
cos R 0.6717325
R 47.80.
Then
T 180 95.86 47.80 36.34.
Thus,
R 47.80,
S 95.86,
T 36.34,
r 3.5,
s 4.7,
t 2.8.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 611
Section 7.2
0.5 cm
B
C
2 cm
2 cm
A
• The Law of Cosines
611
EXAMPLE 3 Knife Bevel. Knifemakers know that the bevel of the blade
(the angle formed at the cutting edge of the blade) determines the cutting characteristics of the knife. A small bevel like that of a straight razor
makes for a keen edge, but is impractical for heavy-duty cutting because
the edge dulls quickly and is prone to chipping. A large bevel is suitable
for heavy-duty work like chopping wood. Survival knives, being universal in application, are a compromise between small and large bevels. The
diagram at left illustrates the blade of a hand-made Randall Model 18 survival knife. What is its bevel? (Source : Randall Made Knives, P.O.
Box 1988, Orlando, FL 32802)
Solution We know three sides of a triangle. We can use the law of
cosines to find the bevel, angle A.
a2 b 2 c 2 2bc cos A
0.52 22 22 2 2 2 cos A
0.25 4 4 8 cos A
4 4 0.25
cos A 8
cos A 0.96875
A 14.36.
Thus the bevel is approximately 14.36.
CONNECTING THE CONCEPTS
CHOOSING THE APPROPRIATE LAW
The following summarizes the situations in which to use the law of sines and the
law of cosines.
To solve an oblique triangle:
Use the law of sines for:
Use the law of cosines for:
AAS
ASA
SSA
SAS
SSS
The law of cosines can also be used for the SSA situation, but since
the process involves solving a quadratic equation, we do not include that
option in the list above.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
612
Chapter 7
1/10/05
1:35 PM
Page 612
• Applications of Trigonometry
EXAMPLE 4 In ABC , three measures are given. Determine which law
to use when solving the triangle. You need not solve the triangle.
a)
b)
c)
d)
e)
f)
a 14, b 23, c 10
a 207, B 43.8, C 57.6
A 112, C 37, a 84.7
B 101, a 960, c 1042
b 17.26, a 27.29, A 39
A 61, B 39, C 80
Solution It is helpful to make a drawing of a triangle with the given information. The triangle need not be drawn to scale. The given parts are
shown in color.
FIGURE
a)
SITUATION
C
A
b)
A
Law of Cosines
ASA
Law of Sines
AAS
Law of Sines
SAS
Law of Cosines
SSA
Law of Sines
AAA
Cannot be solved
B
C
A
d)
SSS
B
C
c)
LAW TO USE
B
C
Study Tip
The InterAct Math Tutorial
software that accompanies this
text provides practice exercises
that correlate at the objective level
to the odd-numbered exercises in
the text. Each practice exercise is
accompanied by an example and
guided solution designed to
involve students in the solution
process. This software is available
in your campus lab or on CD-ROM.
A
e)
B
C
A
f)
B
C
A
B
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 613
Section 7.2
7.2
• The Law of Cosines
613
Exercise Set
Solve the triangle, if possible.
1. A 30, b 12, c 24 a 15, B 24, C 126
2. B 133, a 12, c 15 b 25, A 20, C 27
3. a 12, b 14, c 20
A 36.18, B 43.53, C 100.29
4. a 22.3, b 22.3, c 36.1
A 35.96, B 35.96, C 108.08
5. B 7240, c 16 m, a 78 m
a rangefinder. She observes some poachers, and the
rangefinder indicates that they are 500 ft from her
position. They are headed toward big game that she
knows to be 375 ft from her position. Using her
compass, she finds that the poachers’ azimuth (the
direction measured as an angle from north) is 355
and that of the big game is 42. What is the distance
between the poachers and the game? About 367 ft
b 75 m, A 9451, C 1229
6. C 22.28, a 25.4 cm, b 73.8 cm
c 51.2 cm, A 10.82, B 146.90
Poachers
N
7. a 16 m, b 20 m, c 32 m
A 24.15, B 30.75, C 125.10
8. B 72.66, a 23.78 km, c 25.74 km
Game
b 29.38 km, A 50.59, C 56.75
9. a 2 ft, b 3 ft, c 8 ft No solution
500 ft
10. A 9613, b 15.8 yd, c 18.4 yd
42°
a 25.5 yd, B 382, C 4545
11. a 26.12 km, b 21.34 km, c 19.25 km
355°
A 79.93, B 53.55, C 46.52
12. C 2843, a 6 mm, b 9 mm
375 ft
Rangers
c 5 mm, A 3857, B 11220
13. a 60.12 mi, b 40.23 mi, C 48.7
c 45.17 mi, A 89.3, B 42.0
14. a 11.2 cm, b 5.4 cm, c 7 cm
A 128.71, B 22.10, C 29.19
15. b 10.2 in., c 17.3 in., A 53.456
a 13.9 in., B 36.127, C 90.417
No solution
16. a 17 yd, b 15.4 yd, c 1.5 yd
Determine which law applies. Then solve the triangle.
17. A 70, B 12, b 21.4
Law of sines; C 98, a 96.7, c 101.9
18. a 15, c 7, B 62
Law of cosines; b 13, A 92, C 26
19. a 3.3, b 2.7, c 2.8
Law of cosines; A 73.71, B 51.75, C 54.54
20. a 1.5, b 2.5, A 58 No solution
26. Circus Highwire Act. A circus highwire act walks up
an approach wire to reach a highwire. The approach
wire is 122 ft long and is currently anchored so that
it forms the maximum allowable angle of 35 with
the ground. A greater approach angle causes the
aerialists to slip. However, the aerialists find that
there is enough room to anchor the approach wire
30 ft back in order to make the approach angle less
severe. When this is done, how much farther will
they have to walk up the approach wire, and what
will the new approach angle be?
26 ft farther, about 28°
21. A 40.2, B 39.8, C 100 Cannot be solved
C
22. a 60, b 40, C 47
Law of cosines; c 44, A 91, B 42
High wire
23. a 3.6, b 6.2, c 4.1
Law of cosines; A 33.71, B 107.08, C 39.21
24. B 11030, C 810, c 0.912
Law of sines; A 6120, a 5.633, b 6.014
Solve.
25. Poachers. A park ranger establishes an observation
post from which to watch for poachers. Despite
losing her map, the ranger does have a compass and
35°
A 30 ft B
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Ground
D
BBEPMC07_0321279115.QXP
614
Chapter 7
1/10/05
1:35 PM
Page 614
• Applications of Trigonometry
27. In-line Skater. An in-line skater skates on a
fitness trail along the Pacific Ocean from point A to
point B. As shown below, two streets intersecting at
point C also intersect the trail at A and B. In his
car, the skater found the lengths of AC and BC to be
approximately 0.5 mi and 1.3 mi, respectively. From
a map, he estimates the included angle at C to be
110. How far did he skate from A to B?
30. Survival Trip. A group of college students is
learning to navigate for an upcoming survival trip.
On a map, they have been given three points at
which they are to check in. The map also shows the
distances between the points. However, to navigate
they need to know the angle measurements.
Calculate the angles for them.
S 112.5, T 27.2, U 40.3
About 1.5 mi
S
C
31.6 km
110°
0.5 mi
22.4 km
1.3 mi
Start
T
A
B
28. Baseball Bunt. A batter in a baseball game drops a
bunt down the first-base line. It rolls 34 ft at an
angle of 25 with the base path. The pitcher’s mound
is 60.5 ft from home plate. How far must the pitcher
travel to pick up the ball? (Hint: A baseball diamond
is a square.) About 30.8 ft
Pitcher
34 ft
25
Batter
29. Ships. Two ships leave harbor at the same time.
The first sails N15W at 25 knots (a knot is one
nautical mile per hour). The second sails N32E at
20 knots. After 2 hr, how far apart are the ships?
U
50
45.2 km
50
31. Airplanes. Two airplanes leave an airport at the
same time. The first flies 150 kmh in a direction
of 320. The second flies 200 kmh in a direction
of 200. After 3 hr, how far apart are the planes?
About 912 km
32. Slow-pitch Softball. A slow-pitch softball diamond
is a square 65 ft on a side. The pitcher’s mound is
46 ft from home plate. How far is it from the
pitcher’s mound to first base? About 46 ft
33. Isosceles Trapezoid. The longer base of an isosceles
trapezoid measures 14 ft. The nonparallel sides
measure 10 ft, and the base angles measure 80.
a) Find the length of a diagonal. About 16 ft
b) Find the area. About 122 ft 2
34. Area of Sail. A sail that is in the shape of an
isosceles triangle has a vertex angle of 38. The
angle is included by two sides, each measuring
20 ft. Find the area of the sail. About 124 ft 2
35. Three circles are arranged as shown in the figure
below. Find the length PQ. About 4.7 cm
About 37 nautical mi
r 1.1 cm
P
d
N15W
109
N32E
r 1.8 cm
Q
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
r 1.4 cm
BBEPMC07_0321279115.QXP
1/10/05
1:35 PM
Page 615
Section 7.2
36. Swimming Pool. A triangular swimming pool
measures 44 ft on one side and 32.8 ft on another
side. These sides form an angle that measures 40.8.
How long is the other side? About 28.8 ft
615
• The Law of Cosines
elevation of the ends of the bridge are 78 and 72.
How deep is the canyon? About 9386 ft
5045 ft
Collaborative Discussion and Writing
37. Try to solve this triangle using the law of cosines.
Then explain why it is easier to solve it using the
law of sines.
h
78°
C
a
11.1
19
A
28.5
B
38. Explain why we cannot solve a triangle given SAS
with the law of sines.
Skill Maintenance
Classify the function as linear, quadratic, cubic, quartic,
rational, exponential, logarithmic, or trigonometric.
39. fx 34 x 4 [3.1] Quartic
40. y 3 17x
[1.3] Linear
41. y sin x 3 sin x
2
42. fx 2x1/2
43. fx [5.5] Trigonometric
[4.2] Exponential
x 2x 3
[3.5] Rational
x1
2
44. fx 27 x 3 [3.1] Cubic
45. y e x e x 4
[4.2] Exponential
50. Heron’s Formula. If a, b, and c are the lengths of the
sides of a triangle, then the area K of the triangle
is given by
K ss a s b s c,
1
where s 2 a b c. The number s is called the
semiperimeter. Prove Heron’s formula. (Hint: Use
1
the area formula K 2 bc sin A developed in
Section 7.1.) Then use Heron’s formula to find the
area of the triangular swimming pool described in
Exercise 36. 51. A 12 a 2 sin ; when 90°
51. Area of Isosceles Triangle. Find a formula for the
area of an isosceles triangle in terms of the
congruent sides and their included angle. Under
what conditions will the area of a triangle with
fixed congruent sides be maximum?
52. Reconnaissance Plane. A reconnaissance plane
patrolling at 5000 ft sights a submarine at bearing
35 and at an angle of depression of 25. A carrier is
at bearing 105 and at an angle of depression of 60.
How far is the submarine from the carrier?
About 10,106 ft
46. y log2 x 2 log2 x 3 [4.3] Logarithmic
105
47. fx cos x 3 [5.5] Trigonometric
48. y 1 2
2x
2x 2 [2.3] Quadratic
72°
60
5000 ft
E
25
35
N
x
Synthesis
y
49. Canyon Depth. A bridge is being built across a
canyon. The length of the bridge is 5045 ft. From
the deepest point in the canyon, the angles of
z
E
Answer to Exercise 50 can be found on p. IA-47.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
N
BBEPMN00_0321279115.QXP
1/7/05
3:53 PM
Page A-37
A-37
Chapters 4 – 5
12. [4.2]
13. [4.3]
y
4
4
4
2
2
2
2
2
4
f (x) 4
x
ex
2
2
2
3
4
4
x
f(x) ln (x 2)
4
17. [4.3] 15
20. [4.3] 2.7726
x 2z
21. [4.3] 0.5331 22. [4.3] 1.2851 23. [4.4] loga
y
1
2
24. [4.4] 5 ln x 5 ln y
25. [4.4] 0.656 26. [4.4] 4t
14. [4.3] 5
15. [4.3] 1
16. [4.3] 0
18. [4.3] x e 4
19. [4.3] x log3 5.4
27. [4.5] 12
28. [4.5] 1
29. [4.5] 1
30. [4.5] 4.174
31. [4.6] 1.54% 32. [4.6] (a) 4.5%;
(b) Pt 1000e 0.045t ; (c) $1433.33; (d) 15.4 yr
27
33. [4.5] 8
Chapter 5
Exercise Set 5.1
1. sin cot 158
15
17 ,
cos 2
3
19. 2
21.
23. 12
25. 1
27. 2
2
3
29. 62.4 m 31. 9.72 33. 35.01 35. 3.03
37. 49.65 39. 0.25 41. 5.01 43. 1736
45. 83130 47. 1145 49. 474936 51. 54
53. 3927 55. 0.6293 57. 0.0737 59. 1.2765
61. 0.7621 63. 0.9336 65. 12.4288 67. 1.0000
69. 1.7032 71. 30.8 73. 12.5 75. 64.4
77. 46.5 79. 25.2 81. 38.6 83. 45 85. 60
87. 45 89. 60 91. 30
1
93. cos 20 sin 70 sec 20
1
95. tan 52 cot 38 cot 52
97. sin 25 0.4226, cos 25 0.9063, tan 25 0.4663,
csc 25 2.3662, sec 25 1.1034, cot 25 2.1445
99. sin 184955 0.3228, cos 184955 0.9465,
tan 184955 0.3411, csc 184955 3.0979,
sec 184955 1.0565, cot 184955 2.9317
1
1
101. sin 8 q, cos 8 p, tan 8 , csc 8 ,
r
q
1
sec 8 , cot 8 r
103. Discussion and Writing
p
105. [4.2]
104. [4.2]
17.
y
y
y
8
17 ,
tan 15
8 ,
csc 17
15 ,
sec 17
8 ,
1
23
3
, cos , tan 3, csc ,
2
2
3
3
sec 2, cot 3
765
465
7
5. sin , cos , tan ,
65
65
4
4
65
65
, sec , cot csc 7
4
7
25
35
3
3
2
7. csc , or
; sec ; cot , or
5
2
5
5
5
25
7
9. cos 257 , tan 247 , csc 25
,
,
sec
cot
24
7
24
25
5
5
11. sin , cos , csc , sec 5,
5
5
2
1
cot 2
2
25
35
5
13. sin , cos , tan , sec ,
3
3
5
5
5
cot 2
25
5
15. sin , tan 2, csc , sec 5,
5
2
1
cot 2
5
4
5
4
3
2
3. sin 3 2 1
1
f(x) 2x
f(x) e x/2 3
2
1
2
3
3 2 1
1
x
106. [4.3]
107. [4.3]
y
y
2
3
x
3
2
1
3
2
1
1
1
2
3
1
1
2
3
4
5
x
g(x) log 2 x
1
1
2
3
1
2
3
4
5
h(x) ln x
108. [4.5] 9.21 109. [4.5] 4
110. [4.5] 101
97
111. [4.5] 343 113. 0.6534
115. Area 12 ab. But a c sin A, so Area 12 bc sin A.
Exercise Set 5.2
1. F 60, d 3, f 5.2
3. A 22.7, a 52.7, c 136.6
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x
BBEPMN00_0321279115.QXP
A-38
1/7/05
3:53 PM
Page A-38
Answers
5. P 4738, n 34.4, p 25.4
7. B 217, b 0.39, c 9.74
9. A 77.2, B 12.8, a 439
11. B 42.42, a 35.7, b 32.6
13. B 55, a 28.0, c 48.8
15. A 62.4, B 27.6, a 3.56
17. About 62.2 ft 19. About 2.5 ft 21. About 606 ft
23. About 92.9 cm 25. About 599 ft 27. About 8 km
29. About 275 ft 31. About 24 km
33. Discussion and Writing
35. [1.1] 102, or about 14.142
36. [1.1] 310, or about 9.487
37. [4.3] 103 0.001
38. [4.3] ln t 4
39. 3.3
41. Cut so that 79.38 43. 27
77. Positive: all
79. sin 319 0.6561, cos 319 0.7547,
tan 319 0.8693, csc 319 1.5242,
sec 319 1.3250, cot 319 1.1504
81. sin 115 0.9063, cos 115 0.4226,
tan 115 2.1445, csc 115 1.1034,
sec 115 2.3663, cot 115 0.4663
83. East: about 130 km; south: 75 km
85. About 223 km 87. 1.1585 89. 1.4910
91. 0.8771 93. 0.4352 95. 0.9563 97. 2.9238
99. 275.4 101. 200.1 103. 288.1 105. 72.6
107. Discussion and Writing
y
109. [3.4]
1
Exercise Set 5.3
1. III
3. III
5. I 7. III
9. II
11. II
13. 434, 794, 286, 646
15. 475.3, 835.3, 244.7, 604.7
17. 180, 540, 540, 900 19. 72.89, 162.89
21. 775646, 1675646 23. 44.8, 134.8
5
13
25. sin 135 , cos 12
13 , tan 12 , csc 5 ,
13
12
sec 12 , cot 5
27
23
21
27. sin , cos , tan ,
7
7
3
7
21
3
, sec , cot csc 2
3
2
213
313
2
29. sin , cos , tan 13
13
3
541
441
5
31. sin , cos , tan 41
41
4
22
2
33. cos , tan , csc 3,
3
4
32
, cot 22
sec 4
25
1
5
35. sin , cos , tan ,
5
5
2
5
csc 5, sec 2
37. sin 45 , tan 43 , csc 54 , sec 53 ,
3
cot 4
3
2
39. 30; 41. 45; 1 43. 0
45. 45; 2
2
3
47. 30; 2
49. 30; 3
51. 30; 3
53. Not defined 55. 1
57. 60; 3
2
59. 45;
61. 45; 2 63. 1
65. 0
67. 0
2
69. 0
71. Positive: cos, sec; negative: sin, csc, tan, cot
73. Positive: tan, cot; negative: sin, csc, cos, sec
75. Positive: sin, csc; negative: cos, sec, tan, cot
6
6
x 5
x
x5
f(x) 1
x 2 25
y
110. [3.1]
25
20
15
10
5
43
1
5
1 2 3 4 x
10
g(x) x3 2x 1
15
20
25
111. [1.2], [3.5] Domain: x x 2; range: x x 1
112. [1.2], [3.1] Domain: x x 32 and x 5 ;
range: all real numbers
113. [2.1] 12
114. [2.3] 2, 3 115. [2.1] 12, 0
116. [2.3] 2, 0, 3, 0
117. 19.625 in.
Exercise Set 5.4
1.
11p
(c) f ; (e) 4
y
3.
y
(a) d
(f)
17p
4
(b) i
(d) p
14p
(e) 6
(a) A
x
x
(c) F
(b) w
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
(f)
10p
(d) 6
23p
4
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-39
A-39
Chapter 5
5. M:
2
4
3
5
3
11
, ; N:
, ; P:
, ; Q:
,
3
3
2
2
4
4
6
6
5.
y
7.
(a) 2.4
2
2
,
2
2
25.
33.
41.
43.
(c) 32
x
(d) 320
17.
9
7
,
4
4
19
5
4
8
,
13.
,
6
6
3
3
2
15. Complement: ; supplement:
6
3
5
17. Complement: ; supplement:
8
8
5
11
19. Complement:
; supplement:
12
12
5
10
214.6
5
21.
23.
25. 27. 29.
12
9
180
72
17
31. 33. 4.19 35. 1.05 37. 2.06
9
39. 0.02 41. 6.02 43. 1.66 45. 135
47. 1440 49. 57.30 51. 134.47 53. 225
55. 5156.62 57. 51.43
59. 0 0 radians, 30 , 45 , 60 ,
6
4
3
3
5
3
, 135 , 180 , 225 , 270 ,
90 2
4
4
2
7
, 360 2
315 4
61. 2.29 63. 3.2 yd
65. 1.1; 63 67. 3.2 yd
5
cm
69.
, or about 5.24
71. 3150
3
min
73. About 18,852 revolutions per hour 75. 1047 mph
77. 10 mph 79. About 202
81. Discussion and Writing
83. [4.1] One-to-one 84. [5.1] Cosine of 85. [4.2] Exponential function
86. [3.5] Horizontal asymptote
87. [1.7] Odd function
88. [4.3] Natural 89. [4.1] Horizontal line; inverse
90. [4.3] Logarithm
91. 111.7 km; 69.8 mi
93. (a) 53730; (b) 194115 95. 1.676 radianssec
97. 1.46 nautical miles
1. (a) 3. (a)
3
7
,
; (b)
4
4
3 7
,
; (c)
4 4
3
7
,
4
4
2 21
2
2 21
21
,
; (b) , ; (c) ,
5
5
5
5
5
5
(b)
q i S p x
y sin x
1
y
1
p f q
d
y sin (x)
d
q
p x
f
1
(c) same as (b); (d) the same
45. (a) See Exercise 43(a);
(b)
y
1
p f q
y sin (x p)
d
d
q
p x
f
1
(c) same as (b); (d) the same
y
47. (a)
y cos x
1
f q
pSi
d
f
u A
Adu
q i S p x
1
(b)
y
y cos (x p)
1
p f q
f
Adu
d
d
1
(c) same as (b); (d) the same
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
q
f
p x
3
2
23. 0
31. 1
39. 0
d
pSiq u A
11.
Exercise Set 5.5
21. 0
1
f
9.
13. 11. 0
3
2
19. 2
2
0.4816 27. 1.3065 29. 2.1599
1.1747 35. 1
37. 0.7071
0.8391
y
(a)
15. Not defined
(b) 7.5
9. 3
7. 0
BBEPMN00_0321279115.QXP
A-40
1/7/05
3:54 PM
Page A-40
Answers
49. Even: cosine, secant; odd: sine, tangent, cosecant,
cotangent
51. Positive: I, III; negative: II, IV
53. Positive: I, IV; negative: II, III
55. Discussion and Writing
y
57. [1.7]
6
f (x) x 2
77.
79.
2k,
2k , k 2
2
xx k , k 2
y
81.
2
4
2
2
2
4
2p w p
x
g(x) 2x 2 3
q
y
83.
g(x) (x 2)2
2p w
2
2
4
1
p
2
4
q
x
85. (a)
4
Shift the graph of f right 2 units.
59. [1.7]
f (x) |x|
y
g(x) 8
1
2
|x 4| 1
4
4
4
8
x
4
8
y
f (x) x3
4
x
2
2
2
2
4
w
p
2p
x
(b) OPA ODB ;
OD OB
OP
OA
1
OD
1
cos OD sec (d) OAP ECO
CE
CO
AO AP
CE
1
cos sin cos CE sin CE cot OE csc 87. 1
4
4
OPA ODB ;
BD
AP
Thus,
OA OB
sin BD
cos 1
tan BD
(c) OAP ECO ;
OE CO
PO AP
1
OE
1
sin Shift the graph of f to the right 4 units, shrink it vertically,
then shift it up 1 unit.
60. [1.7]
q
1
2
2
8
y sin x cos x
2
y
4
x
w
p
3
4
Stretch the graph of f vertically, then shift it down 3 units.
f(x) x 2
2p
q
4
58. [1.7]
y 3 sin x
4
3
2
1
4
g(x) x 3
Reflect the graph of f across the x-axis.
1
61. [1.7] y x 23 1
62. [1.7] y 3
4x
63. cos x
65. sin x
67. sin x
69. cos x
71. sin x
73. (a)
2k, k ; (b) 2k,
2
75. Domain: , ; range: 0, 1;
k ; (c) k, k period: ; amplitude: 12
Visualizing the Graph
1. J
8. A
2. H
9. C
3. E
10. I
4. F
5. B
6. D
7. G
Exercise Set 5.6
1. Amplitude: 1; period: 2 ; phase shift: 0
y
2
1
2p w p
q
1
2
q
p
w
2p
y sin x 1
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-41
A-41
Chapter 5
3. Amplitude: 3; period: 2 ; phase shift: 0
13. Amplitude: 3; period: 2 ; phase shift: y
3
2
1
w
2p
p q
w
q
1
2
2p
p
x
w
2p
p q
y 3 cos x
1
5. Amplitude: 2 ; period: 2 ; phase shift: 0
y 3 cos (x p)
3
2
1
w
q
1
2
p
2p
x
2p
x
15. Amplitude: 13 ; period: 2 ; phase shift: 0
y
y
2
1
1
2p w p
2 1
2
x
q 1
2
3
1
y
2 cos x
2
q
p
w
y a sin x 4
5
17. Amplitude: 1; period: 2 ; phase shift: 0
7. Amplitude: 1; period: ; phase shift: 0
y
y
3
2
2
1
1
2 2
x
2p w p
1
2
y
2
1
2p w p
q
1
q
11. Amplitude:
p
w
2p
x
y 2 sin qx
2
1
q
1
p
w
2p
x
y cos (x ) 2
19. Amplitude: 2; period: 4 ; phase shift: 1
21. Amplitude: ; period: ; phase shift: 2
4
3
23. Amplitude: 3; period: 2; phase shift:
25. Amplitude: 12 ; period: 1; phase shift: 0
27. Amplitude: 1; period: 4 ; phase shift: 29. Amplitude: 1; period: 1; phase shift: 0
4
1
31. Amplitude: ; period: 2; phase shift:
4
33. (b) 35. (h)
37. (a) 39. (f )
1
41. y 2 cos x 1 43. y cos x 2
2
y
45.
1
; period: 2 ; phase shift: 2
2
y
w
q
y sin (2x)
9. Amplitude: 2; period: 4 ; phase shift: 0
2p
y
y q sin x q
4
q
p
q
1
p
w
2p
x
2p
p
1
2
3
p
2p
y 2 cos x cos 2x
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x
BBEPMN00_0321279115.QXP
A-42
1/7/05
3:54 PM
Page A-42
Answers
y
47.
2p
p
49.
2p
p
1
2
y
75.
y sin x cos 2x
2
3
2
1
x
2p w p
y
2
1
2p
77.
p
2p
p
2
4
y sin x cos x
y 3 cos x sin 2x
79.
p
2p
p
3
10
2p
y cos 2x 2x
q
p
w
2p
x
y
3
2
1
2p
2p w p
10
2p
83.
2p
10
Discussion and Writing
[3.5] Rational
64. [4.3] Logarithmic
[3.1] Quartic
66. [1.3] Linear
[5.6] Trigonometric 68. [4.2] Exponential
[1.3] Linear 70. [5.6] Trigonometric
[3.1] Cubic
72. [4.2] Exponential
Maximum: 8; minimum: 4
q 1
y 2 sec (x p)
q
p
w
3
y 4 cos 2x 2 sin x
10
61.
63.
65.
67.
69.
71.
73.
81.
59.
10
q 1
2
3
3
10
2p
y 2 cot x
2
10
10
2p
y tan x
q p w 2p x
2p p q
y cos x x
57.
x
1
y x sin x
2p
2p
y 2 tan qx
2
55.
2p
w
y
x
2
3
4
53.
p
3
2
1
x
1
2p
q
y
2p w p
y
51.
q 1
2
3
y
10
8
6
4
2
r w q
w x
y 2 csc qx f
85. 9.42, 6.28, 3.14, 3.14, 6.28, 9.42
87. 3.14, 0, 3.14
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2p
x
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-43
Chapter 5
89. (a) y 101.6 3 sin p x
8
(b) 104.6, 98.6
40. [5.4]
A-43
y
105
k
0
97
373
873
3
1. [5.1] sin , cos , tan ,
73
73
8
8
73
73
, sec , cot csc 3
8
3
3
1091
91
2. [5.1] cos , tan , csc ,
10
3
91
10
391
sec , cot 3
91
2
3
2
3. [5.1]
4. [5.1]
5. [5.3] 6. [5.3] 12
2
3
2
23
7. [5.3] Not defined 8. [5.3] 3
9. [5.1]
3
10. [5.1] 1
11. [5.1] 221612 12. [5.1] 47.56
13. [5.3] 0.4452 14. [5.3] 1.1315 15. [5.3] 0.9498
16. [5.3] 0.9092 17. [5.3] 1.5282 18. [5.3] 0.2778
19. [5.3] 205.3 20. [5.3] 47.2 21. [5.1] 60
22. [5.1] 60 23. [5.1] 45 24. [5.1] 30
25. [5.1] sin 30.9 0.5135, cos 30.9 0.8581,
tan 30.9 0.5985, csc 30.9 1.9474, sec 30.9 1.1654,
cot 30.9 1.6709
26. [5.2] b 4.5, A 58.1, B 31.9
27. [5.2] A 38.83, b 37.9, c 48.6
28. [5.2] 1748 m
29. [5.2] 14 ft
30. [5.3] II
31. [5.3] I
32. [5.3] IV
5
33. [5.3] 425, 295 34. [5.4] , 3
3
35. [5.3] Complement: 76.6; supplement: 166.6
5
36. [5.4] Complement: ; supplement:
3
6
313
213
3
37. [5.3] sin , cos , tan ,
13
13
2
2
13
13
, sec , cot csc 3
2
3
2
5
5
38. [5.3] sin , cos , cot ,
3
3
2
35
3
, csc 39. [5.3] About 1743 mi
sec 5
2
u
f
91. Amplitude: 3000; period: 90; phase shift: 10
93. 4 in.
Review Exercises: Chapter 5
x
F
12
, 0.52
6
[5.4] 270 44. [5.4] 171.89 45. [5.4] 257.83
7
[5.4] 1980 47. [5.4]
, or 5.5 cm
4
[5.4] 2.25, 129 49. [5.4] About 37.9 ftmin
[5.4] 497,829 radianshr
[5.5] 35 , 45 , 35 , 45 , 35 , 45 3
[5.5] 1
53. [5.5] 1
54. [5.5] 55. [5.5] 12
2
3
[5.5]
57. [5.5] 1
58. [5.5] 0.9056
3
[5.5] 0.9218 60. [5.5] Not defined 61. [5.5] 4.3813
[5.5] 6.1685 63. [5.5] 0.8090
[5.5]
41. [5.4] 121
150 , 2.53
43.
46.
48.
50.
51.
52.
56.
59.
62.
64.
y
2
1
2p
p
42. [5.4] y
y sin x
p
2
2p p
2p x
2
1
2
y cos x
p
y
3
2
1
2p w p
q 1
2
3
y tan x
q
p
y
q 1
2p
x
y cot x
3
2
1
2p w p
w
q
p
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
w
2p
x
2p x
BBEPMN00_0321279115.QXP
A-44
1/7/05
3:54 PM
Page A-44
Answers
y
1
; period: ; phase shift:
2
4
69. [5.6] Amplitude:
3
2
y
4
2p w p
q 1
2
3
q
p
2p w p
3
y sec x
2
w
2p
y 3 qcos 2x q
1
y
q
x
2p w p
y csc x
3
2
1
70. [5.6] (d)
74. [5.6] y
w
q
1
2p
p
q
q
71. [5.6] (a)
p
w
2p
72. [5.6] (c)
x
73. [5.6] (b)
3
x
2
1
FUNCTION
DOMAIN
RANGE
Sine
, 1, 1
Cosine
, 1, 1
xx Tangent
k, k 2
, 67. [5.3]
FUNCTION
I
II
III
IV
Sine
Cosine
Tangent
68. [5.6] Amplitude: 1; period: 2 ; phase shift: y
2
2p p
1
2
y sin x q
p
2p x
2
q
1
65. [5.5] Period of sin, cos, sec, csc: 2 ; period of tan, cot: 66. [5.5]
2p x
p
2
3
y 3 cos x sin x
75. Discussion and Writing [5.1], [5.4] Both degrees and
radians are units of angle measure. A degree is defined to be
1
360 of one complete positive revolution. Degree notation has
been in use since Babylonian times. Radians are defined in
terms of intercepted arc length on a circle, with one radian
being the measure of the angle for which the arc length equals
the radius. There are 2 radians in one complete revolution.
76. Discussion and Writing [5.5] The graph of the cosine
function is shaped like a continuous wave, with “high” points
at y 1 and “low” points at y 1. The maximum value of
the cosine function is 1, and it occurs at all points where
x 2k, k .
77. Discussion and Writing [5.5] No; sin x is never greater
than 1.
78. Discussion and Writing [5.6] When x is very large or very
small, the amplitude of the function becomes small. The
dimensions of the window must be adjusted to be able to see
the shape of the graph. Also, when x is 0, the function is
undefined, but this may not be obvious from the graph.
79. [5.5] All values
80. [5.6] Domain: , ; range: 3, 3; period 4
y
y 3 sin
3
x
2
2
1
1
2p
p
4p
3p
x
2
3
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-45
Chapters 5 – 6
2
2
82. [5.6] The domain consists of the intervals
2k,
2k , k .
2
2
83. [5.3] cos x 0.7890, tan x 0.7787,
cot x 1.2842, sec x 1.2674, csc x 1.6276
81. [5.6] y2 2 sin x Test: Chapter 5
465
4
7
765
, or
; cos , or
;
65
65
65
65
4
7
65
65
; sec ; cot tan ; csc 7
4
7
4
3
2. [5.3]
3. [5.3] 1
4. [5.4] 1
5. [5.4] 2
2
6. [5.1] 38.47 7. [5.3] 0.2419 8. [5.3] 0.2079
9. [5.4] 5.7588 10. [5.4] 0.7827 11. [5.1] 30
12. [5.1] sin 61.6 0.8796; cos 61.6 0.4756;
tan 61.6 1.8495; csc 61.6 1.1369; sec 61.6 2.1026;
cot 61.6 0.5407
13. [5.2] B 54.1, a 32.6, c 55.7
14. [5.3] Answers may vary; 472, 248 15. [5.4]
6
5
4
41
16. [5.3] cos ; tan ; csc ;
5
4
41
7
5
41
sec ; cot 17. [5.4]
18. [5.4] 135
5
4
6
16
16.755 cm
19. [5.4]
20. [5.5] 1
21. [5.5] 2
3
22. [5.5]
23. [5.6] (c)
24. [5.2] About 444 ft
2
25. [5.2] About 272 mi
26. [5.4] 18 56.55 mmin
2k, k an integer
27. [5.5] x 2k x 2
2
1. [5.1] sin Copyright © 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
A-45
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-45
Chapters 5 – 6
Chapter 6
Exercise Set 6.1
1. sin2 x cos2 x
3. sin y cos y
5. 1 2 sin cos 7. sin3 x csc 3 x
9. cos x sin x cos x
11. sin x cos x sin x cos x
13. 2 cos x 3 cos x 1
15. sin x 3 sin2 x 3 sin x 9 17. tan x
2 tan t 1
19. sin x 1 21.
23. 1
3 tan t 1
1 2 sin s 2 cos s
5 cot 25.
27.
sin cos sin2 s cos2 s
A-45
5sin 3
31. sin x cos x
3
33. cos sin cos 35. 1 sin y
cos x
sin x cos x
2 cot y
37.
39.
41.
cos x
2
sin x cos x
1 sin y
x
a 2 x 2
43.
45. cos , tan 2
cos y
a
a x 2
3
x 2 9
47. sin , cos 49. sin tan x
x
6 2
3 1
51.
53.
, or 2 3
4
1 3
6 2
55.
57. sin 59 0.8572
4
59. cos 24 0.9135 61. tan 52 1.2799
sin 63. tan cos sin cos cos sin cos cos sin sin 1
sin cos cos sin cos cos cos cos sin sin 1
cos cos sin sin cos cos sin sin 1
cos cos tan tan 1 tan tan 65. 0
67. 257
69. 1.5789 71. 0.7071
73. 2 sin cos 75. cos u 77. Discussion and Writing
79. [2.1] All real numbers
80. [2.1] No solution
81. [5.1] 1.9417 82. [5.1] 1.6645 83. 0; the lines are
3
parallel
85.
, or 135 87. 22.83
4
cos x h cos x
89.
h
cos x cos h sin x sin h cos x
h
cos x cos h cos x sin x sin h
h
h
cos h 1
sin h
sin x
cos x
h
h
sin 5x sin 91. Let x . Then
0 sin 5.
5
x
5
Answers may vary.
93. Let . Then cos 2 cos
0, but
4
2
2 cos 2 cos
2. Answers may vary.
4
29.
BBEPMN00_0321279115.QXP
A-46
1/7/05
3:54 PM
Page A-46
Answers
1
cos cos 6x
. Then
6.
6
cos x
32
cos
6
6 33
Answers may vary. 97.
0.0645
9 23
99. 168.7 101. cos 2 cos2 sin2 , or 1 2 sin2 ,
or 2 cos2 1
tan x tan
1 tan x
4
103. tan x 4
1 tan x
1 tan x tan
4
105. sin sin sin cos cos sin sin cos cos sin 2 sin cos 95. Let x Exercise Set 6.2
3
3
3
1.3763, csc
1.2361, sec
1.7013,
10
10
10
3
cot
0.7266; (b) sin
0.5878, cos
0.8090,
10
5
5
tan
0.7266, csc
1.7013, sec
1.2361,
5
5
5
1.3763
cot
5
22
2
3. (a) cos , tan , csc 3,
3
4
32
, cot 22;
sec 4
22
1
(b) sin
, cos
,
2
3
2
3
1. (a) tan
tan
sec
22, csc
2
3, cot
2
(c) sin 2
17.
tan 2
22, csc sec 2
3, cot 2
2
2
4
csc x
cot x
7. tan x 2
2
24
7
24
9. sin 2 25 , cos 2 25 , tan 2 7 ; II
5. sec x 24
7
24
11. sin 2 25 , cos 2 25 , tan 2 7 ; II
120
119
120
13. sin 2 169 , cos 2 169 , tan 2 119 ; IV
2
2
15. cos 4x 1 8 sin x cos x , or
cos4 x 6 sin2 x cos2 x sin4 x , or 8 cos4 x 8 cos2 x 1
2
21. 2 3
Exercise Set 6.3
1.
sec x sin x tan x
sin x
1
sin x cos x
cos x
1 sin2 x
cos x
cos2 x
cos x
cos x
1
,
3
32
,
4
2 2
25. 0.1735
cos2 x sin2 x
cos 2x
27. (d);
cos x sin x
cos x sin x
cos x sin x cos x sin x
cos x sin x
cos x sin x
sin x
cos x sin x
sin x
cos x sin x
sin x
sin x sin x
sin x cot x 1
2 sin x cos x
sin 2x
29. (d);
31. cos x
33. 1
sin x
2 cos x
2 cos x
35. cos 2x
37. 8 39. Discussion and Writing
41. [6.1] True
42. [6.1] False
43. [6.1] False
44. [6.1] True
45. [6.1] False
46. [6.1] True
47. [6.1] False
48. [6.1] True
49. [5.5] (a), (e)
50. [5.5] (b), (c), (f )
51. [5.5] (d)
52. [5.5] (e)
53. sin 141 0.6293, cos 141 0.7772,
tan 141 0.8097, csc 141 1.5891, sec 141 1.2867,
55. cos x 1 cot x
57. cot 2 y
cot 141 1.2350
8
15
59. sin 15
,
,
cos
tan
17
17
8
61. (a) 9.80359 msec 2; (b) 9.80180 msec 2;
(c) g 9.780491 0.005264 sin2 0.000024 sin4 2
;
2
4
19.
2
23. 0.6421
32
,
2
4
22
, cos 3
2
2 3
cos x
3.
1 cos x
sin x
sin x
1 cos x
1 cos x
sin x
1 cos x 1 cos x
sin x 1 cos x
1 cos2 x
sin x 1 cos x
sin2 x
1 cos x
sin x
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-47
Chapter 6
5.
A-47
11.
1 tan 1 cot 1 tan 1 cot cos sin 1
1
cos sin sin cos 1
1
cos sin cos sin sin cos cos sin cos sin sin cos cos sin cos cos sin cos cos sin sin sin cos sin sin cos cos sin sin cos cos sin sin cos cos sin cos sin cos sin cos sin 0
1 cos 5 cos 3 sin 5 sin 3
1 cos 5 cos 3 sin 5 sin 3 1 cos 5 3 1 cos 2
0
13.
2 sin cos3 2 sin3 cos 2 sin cos cos2 sin2 2 sin cos 1
sin 1
cos sin 1
cos sin sin cos 1
sin sin cos 1
sin sin cos 1
sin cos 1
cos cos 9.
2 tan 1 tan2 2 tan sec 2 2 sin cos2 cos 1
2 sin cos sin 2
2 sin cos tan x sin x
2 tan x
sin x
sin x
1 cos x
2
sin x
cos x
sin x sin x cos x cos x
cos x
sin x
1 cos x
2
1
2
sin2
x
2
1 cos x
2
17.
sin sin sin cos sin cos cos sin cos sin 2
2
sin cos cos2 sin2 cos2 1 cos2 cos2 1 cos2 cos2 cos2 cos2 cos2 cos2 cos2 cos2 cos2 cos2 tan 1
cos2 tan 1
sin cos2 1
cos sin cos2 1
cos sin cos 1
sin cos 1
sin 2
2 sin cos 15.
7.
cos2 cot cos2 cot cos cos2 sin cos
cos2 sin 1
cos cos sin 2 sin2 1 cos 2
19.
tan tan cot tan2 tan cot tan2 1
sec 2 sec 2 21.
1 cos2 x
sin2 x
cos2 x
1
2 sin x
sin2 x
csc 2 x cot 2 x
csc 2 x csc 2 x 1
2 csc 2 x 1
2 csc 2 x 1
sin2 sin2 1 cos2 1 cos2 cos2 cos2 BBEPMN00_0321279115.QXP
A-48
1/7/05
3:54 PM
Page A-48
Answers
23.
35. C;
1 sin x sin x 1
1 sin x 1 sin x
1 sin x2 1 sin x2
1 sin2 x
2
1 2 sin x sin x 1 2 sin x sin2 x
cos2 x
4 sin x
cos2 x
25.
cos2 cot 2 cot 2 cos2 1 sin2 cot 2 cos2 cot 2 sin2 sin2 cot 2 cos2 27.
4 sec x tan x
1
sin x
cos x cos x
4 sin x
cos2 x
4
1 sin 2 sin cos cos cos2 2 sin2 cos2 1 sin2 1 sin2 cos2 sin2 sin2 cos2 1 sin2 cos2 cos2 sin2 1
29.
1 sin x
sec x tan x2
1 sin x
sin x 2
1
1 sin x 1 sin x
1 sin x 1 sin x
cos x cos x
2
1 sin x2
1 sin x
2
1 sin x
cos2 x
2
1 sin x
cos2 x
31. B;
cos x cot x
cos x
1 csc x
cos x cos x
1
sin x
1
1
sin x
sin x
sin x cos x cos x
sin x
sin x 1
cos x sin x 1
sin x 1
cos x
33. A;
2
2
4
sin x cos x 1
1
cot x sin2 x
1
cos x
sin2 x
sin x
1
cos x sin x
tan x cot x
sin x cos x
cos x sin x
sin2 x cos2 x
cos x sin x
1
cos x sin x
37. Discussion and Writing
y
39. [4.1] (a), (d)
2
2
4
sin3 x cos3 x
sin x cos x
sin x cos x sin2 x sin x cos x cos2 x
sin x cos x
sin2 x sin x cos x cos2 x
sin x cos x 1
f(x) 3x 2
4
f 1(x) (b) yes; (c) f 1x 2
4
2
x2
3
x
x2
3
40. [4.1] (a), (d)
f(x) x 3 1
y
4
3
f 1(x) x
1
2
4 2
2
4
x
2
4
(b) yes; (c) f 1x x 1
y
41. [4.1] (a), (d)
3
f(x) x 2 4, x 0
4
f 1(x) x 4
4 2
4
x
2
4
(b) yes; (c) f 1x x 4
y
42. [4.1] (a), (d)
6
f 1(x) x 2 2, x 0
4
f(x) x 2
2
4
2
(b) yes; (c) f 1x x 2 2, x 0
6
x
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-49
Chapter 6
5
7
43. [2.3] 0, 2
44. [2.3] 4, 3
45. [2.3] 2, 3i
46. [2.3] 5 26
47. [2.5] 27
48. [2.5] 9
49.
ln tan x ln cot x 1
ln
cot x
ln 1 ln cot x 0 ln cot x ln cot x 51. log cos x sin x log cos x sin x
log cos x sin x cos x sin x
log cos2 x sin2 x log cos 2x
1
1
53.
Ctan tan sin sin C
cos cos 1
sin cos sin cos C
cos cos cos cos C sin Exercise Set 6.4
, 45 7. 0, 0
4
9.
, 30 11.
, 30 13. , 30
6
6
6
15. , 30 17.
, 90 19.
, 60
6
2
3
21. 0.3520, 20.2 23. 1.2917, 74.0 25. 2.9463, 168.8
27. 0.1600, 9.2 29. 0.8289, 47.5
31. 0.9600, 55.0
33. sin1: 1, 1; cos1: 1, 1; tan1: , 2000
35. sin1
37. 0.3
39.
41.
d
4
5
11
43. 45. 12
47. 1
49.
51.
3
3
33
a
p
q 2 p 2
53. 55.
57.
59.
2
6
p
3
a 9
2
3
61.
63. 65. xy 1 x 2 1 y 2 2
10
67. 0.9861 69. Discussion and Writing
71. Discussion and Writing
72. [5.5] Periodic
73. [5.4] Radian measure 74. [5.1] Similar
75. [5.2] Angle of depression
76. [5.4] Angular speed
77. [5.3] Supplementary 78. [5.5] Amplitude
79. [5.1] Acute
80. [5.5] Circular
1. , 60
3
3.
, 45
4
5.
A-49
81.
sin1 x cos1 x
sin sin1 x cos1 x
sin sin1 x cos cos1 x cos sin1 x sin cos1 x
x x 1 x 2 1 x 2
x2 1 x2
1
83.
x
sin1 x
tan1
1 x 2
x
sin sin1 x sin tan1
1 x 2
x
x
85.
sin1 x
cos1 1 x 2
1
sin sin x sin cos1 1 x 2 x
x
y
h
y
87. tan1
tan1 ; 38.7
x
x
2
sin
2
1
Visualizing the Graph
1. D
8. J
2. G 3. C
9. F 10. B
4. H
5. I
6. A
7. E
Exercise Set 6.5
11
2k,
2k, or 30 k 360, 330 k 360
6
6
2
3.
k, or 120 k 180
3
5
5.
2k,
2k, or 30 k 360, 150 k 360
6
6
3
5
7.
2k,
2k, or 135 k 360, 225 k 360
4
4
4 5
3 5 7
9. 98.09, 261.91 11.
,
13.
,
,
,
3 3
4 4 4 4
5 3
3 11
15.
,
,
17.
, ,
,
6 6 2
6 2 2
6
19. 109.47, 120, 240, 250.53
3
5 7
21. 0, ,
, ,
,
23. 139.81, 220.19
4 4
4 4
7 11
25. 37.22, 169.35, 217.22, 349.35 27. 0, ,
,
6
6
3
3 7
29. 0, , ,
31. 0, 33.
,
2
2
4 4
2 4 3
3 5 7
5
35.
,
,
37.
,
,
,
39.
,
3 3 2
4 4 4 4
12 12
1.
BBEPMN00_0321279115.QXP
A-50
1/7/05
3:54 PM
Page A-50
Answers
2 4
,
3 3
45. 1.114, 2.773 47. 0.515 49. 0.422, 1.756
51. (a) y 7 sin 2.6180x 0.5236 7;
(b) $10,500, $13,062 53. Discussion and Writing
55. [5.2] B 35, b 140.7, c 245.4
56. [5.2] R 15.5, T 74.5, t 13.7 57. [2.1] 36
2 4 5
4
58. [2.1] 14
59.
,
,
,
61.
,
3 3 3 3
3 3
3/22k
63. 0
65. e
, where k (an integer) 1
67. 1.24 days, 6.76 days
69. 16.5N
71. 1
73. 0.1923
41. 0.967, 1.853, 4.109, 4.994
43.
29. [6.2] tan 2 247 , cos 2 257 , sin 2 24
25 ; I
30. [6.2]
cos
cot
csc
4
, tan
2
5
4
, sec
2
3
3
,
2
4
5
,
2
4
5
; (c) sin 2
3
2
cos 2
cot 2
csc 2
4
, tan 5
2
4
, sec 3
2
5
3
3
,
4
5
,
4
28. [6.2] sec x
3
,
5
2
31. [6.2] sin 2 0.4261, cos
32. [6.2] cos x
35. [6.2] tan 2x
36. [6.3]
33. [6.2] 1
1 sin x
cos x
1 sin x cos x
cos x
cos x
cos x sin x cos x
cos2 x
Review Exercises: Chapter 6
1. [6.1] csc 2 x
2. [6.1] 1
3. [6.1] tan2 y cot 2 y
2
2
cos x 1
4. [6.1]
5. [6.1] csc x sec x csc x
cos2 x
6. [6.1] 3 sin y 5 sin y 4
7. [6.1] 10 cos u 100 10 cos u cos2 u
8. [6.1] 1
3 tan x
1
9. [6.1] 2 sec x
10. [6.1]
sin x cos x
3 cos y 3 sin y 2
11. [6.1]
12. [6.1] 1
cos2 y sin2 y
1
13. [6.1] 4 cot x
14. [6.1] sin x cos x
cos x
cos x
15. [6.1]
16. [6.1]
17. [6.1] 3 sec 1 sin x
sin x
3
3
18. [6.1] cos x cos
sin x sin
2
2
tan 45 tan 30
19. [6.1]
1 tan 45 tan 30
6 2
20. [6.1] cos 27 16, or cos 11 21. [6.1]
4
22. [6.1] 2 3
23. [6.1] 0.3745 24. [6.2] sin x
25. [6.2] sin x
26. [6.2] cos x
4
4
3
27. [6.2] (a) sin , tan , cot ,
5
3
4
5
5
3
sec , csc ; (b) sin
,
3
4
2
5
2 2
0.9940, cos 4 0.6369
2
34. [6.2] sin 2x
cos x
1 sin x
cos x
1 sin x
1 sin x 1 sin x
cos x sin x cos x
1 sin2 x
cos x sin x cos x
cos2 x
37. [6.3]
1 cos 2
sin 2
1 2 cos2 1
2 sin cos cos sin cot cos sin 38. [6.3]
1
2
1
2
tan y sin y
2 tan y
sin y sin y cos y
cos y
sin y
cos y
cos2
y
2
1 cos y
2
sin y 1 cos y cos y
cos y
sin y
1 cos y
2
39. [6.3]
sin x cos x
cos2 x
tan2 x 1
sin x cos x
sin2 x
1
cos2 x
sin x cos x
1
sin2 x cos2 x
cos2 x
sin x cos x
sin x cos x
cos2 x
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-51
Chapter 6
40. [6.3] B;
csc x cos x cot x
cos x
1
cos x
sin x
sin x
1 cos2 x
sin x
sin2 x
sin x
sin x
49. [6.4] 0.3976, 22.8
sin x
cos x
1
sin x cos x sin x
cos2 x
1
sin x cos x sin x cos x
1 cos2 x
sin x cos x
sin2 x
sin x cos x
sin x
cos x
sin x cos x
1 sin2 x
sin x cos x
cos2 x
sin x
cos x
42. [6.3] A;
cot x 1
1 tan x
cos x sin x
sin x sin x
cos x sin x
cos x cos x
cos x
cos x sin x
sin x
cos x sin x
cos x
sin x
csc x
sec x
1
sin x
1
cos x
1
cos x
sin x
1
cos x
sin x
43. [6.3] C;
sin x
2
cos x 1
sin x
cos x 1
sin x
cos x 12 sin2 x
sin x cos x 1
cos2 x 2 cos x 1 sin2 x
sin x cos x 1
2 cos x 2
sin x cos x 1
2cos x 1
sin x cos x 1
2
sin x
44. [6.4] , 30 45. [6.4] , 30
6
6
46. [6.4] , 45 47. [6.4] 0, 0 48. [6.4] 1.7920, 102.7
4
51. [6.4]
3
3
3
2
54. [6.4]
2
b 2 9
3
5
55. [6.4] 257
56. [6.5]
2k,
2k, or
4
4
135 k 360, 225 k 360
57. [6.5]
k, or 60 k 180
3
5 7 11
58. [6.5] ,
,
,
6 6 6
6
3 5 3 7
2
4
59. [6.5] , ,
,
,
,
60. [6.5]
, ,
4 2 4 4 2 4
3
3
3 5 7
61. [6.5] 0, 62. [6.5] ,
,
,
4 4 4 4
3
7 23
63. [6.5] 0, , ,
64. [6.5]
,
2
2
12 12
65. [6.5] 0.864, 2.972, 4.006, 6.114
66. [6.5] 4.917 67. [6.5] No solution in 0, 2 68. [6.3] Discussion and Writing
(a) 2 cos2 x 1 cos 2x cos2 x sin2 x
1 cos2 x sin2 x
cos2 x sin2 x cos2 x sin2 x
cos4 x sin4 x ;
(b) cos4 x sin4 x cos2 x sin2 x cos2 x sin2 x
1 cos2 x sin2 x
cos2 x sin2 x cos 2x
2 cos2 x 1;
(c)
cos4 x sin4 x
2 cos2 x 1
cos 2x cos2 x sin2 x cos2 x sin2 x
1 cos2 x sin2 x
cos2 x sin2 x
cos 2x
Answers may vary. Method 2 may be the more efficient
because it involves straightforward factorization and
simplification. Method 1(a) requires a “trick” such as
multiplying by a particular expression equivalent to 1.
69. Discussion and Writing [6.4] The ranges of the inverse
trigonometric functions are restricted in order that they
might be functions.
70. [6.1] 108.4
71. [6.1] cos u v cos u cos v sin u sin v
u cos
v
cos u cos v cos
2
2
2
72. [6.2] cos x
1
1
6
6
73. [6.2] sin ; cos ;
2
5
2
5
5 26
tan 5 26
74. [6.3] ln e sin t loge e sin t sin t
52. [6.4]
41. [6.3] D;
7
50. [6.4] 12
A-51
53. [6.4]
BBEPMN00_0321279115.QXP
A-52
1/7/05
3:54 PM
Page A-52
Answers
75. [6.4]
16. [6.3]
y
p
y
sec1
1 sin 1 csc x
1 sin 1
1
sin 1 sin sin 1
sin sin q
3 2 1
76. [6.4] Let x sin1
1
cos
1
2
3
x
2
2
. Then tan1
0.6155 and
2
2
4
1.
2
2
4
2
2
77. [6.5]
3
,
2 2
17. [6.4] 45
tan sec sin cos 1
cos sin 18. [6.4]
3
19. [6.4] 2.3072
5
3
21. [6.4]
22. [6.4] 0
2
2
x 25
5 7 11
3
23. [6.5] ,
,
,
24. [6.5] 0, ,
,
6 6 6
6
4 4
11
25. [6.5] ,
26. [6.2] 11
12
2 6
20. [6.4]
Test: Chapter 6
1. [6.1] 2 cos x 1
4. [6.1] 2 cos 7. [6.1] 120
169
10. [6.2]
12.
13.
14.
15.
2. [6.1] 1
5. [6.1]
8. [6.2]
2 6
4
5
3
cos 1 sin 3 3
6. [6.1]
3 3
3. [6.1]
Chapter 7
9. [6.2] 24
25 , II
2 3
11. [6.2] 0.9304
2
[6.2] 3 sin 2x
[6.3]
csc x cos x cot x sin x
cos x
1
cos x sin x
sin x
1 cos2 x
sin x
sin2 x
sin x
sin x
[6.3]
sin x cos x2
2
sin x 2 sin x cos x cos2 x
1 2 sin x cos x
1 sin 2x
[6.3]
1 cos csc cot 2
1 cos 2
1
cos 1 cos sin sin 1 cos Exercise Set 7.1
1 sin 2x
2
1 cos 2
1 cos2 1 cos 2
sin2 1 cos 2
sin2 1 cos sin 1 cos 1 cos 1. A 121°, a 33, c 14 3. B 57.4°, C 86.1°,
c 40, or B 122.6°, C 20.9°, c 14 5. B 44°24,
A 74°26, a 33.3 7. A 110.36°, a 5 mi, b 3 mi
9. B 83.78°, A 12.44°, a 12.30 yd 11. B 14.7°,
C 135.0°, c 28.04 cm 13. No solution
15. B 125.27°, b 302 m, c 138 m 17. 8.2 ft 2
19. 12 yd2 21. 596.98 ft 2 23. 787 ft 2
25. About 12.86 ft, or 12 ft, 10 in. 27. About 51 ft
29. From A: about 35 mi; from B: about 66 mi
31. About 22 mi
33. Discussion and Writing
35. [5.1] 1.348, 77.2° 36. [5.1] No angle
37. [5.1] 18.24° 38. [5.1] 125.06° 39. [R.1] 5
3
3
2
40. [5.3]
41. [5.3]
42. [5.3] 2
2
2
BBEPMN00_0321279115.QXP
1/7/05
3:54 PM
Page A-53
A-53
Chapters 6 – 7
1
44. [2.2] 2
45. Use the formula for the
2
area of a triangle and the law of sines.
43. [5.3] 1
c sin B
bc sin A and b ,
2
sin C
2
c sin A sin B
so K .
2 sin C
1
a sin B
K ab sin C and b ,
2
sin A
2
a sin B sin C
so K .
2 sin A
1
b sin C
K bc sin A and c ,
2
sin B
2
b sin A sin C
so K .
2 sin B
K
47.
A
a
Discussion and Writing
39. [3.1] Quartic
[1.3] Linear
41. [5.5] Trigonometric
[4.2] Exponential 43. [3.4] Rational
[3.1] Cubic
45. [4.2] Exponential
[4.3] Logarithmic
47. [5.5] Trigonometric
[2.3] Quadratic
49. About 9386 ft
1
51. A a 2 sin ; when 90°
2
Exercise Set 7.3
1. 5;
3. 1;
Imaginary
Imaginary
4 3i
4
2
2
4
2
b
4
2
4 2
B
c
u
37.
40.
42.
44.
46.
48.
C
Real
4 2
2
4
Real
4
Real
2
4
d
i
4
5. 17;
7. 3;
Imaginary
Imaginary
D
For the quadrilateral ABCD, we have
1
1
Area bd sin ac sin 2
2
1
1
adsin 180° bc sin 180° 2
2
Exercise Set 7.2
1. a 15, B 24°, C 126° 3. A 36.18°, B 43.53°,
C 100.29° 5. b 75 m, A 94°51, C 12°29
7. A 24.15°, B 30.75°, C 125.10° 9. No solution
11. A 79.93°, B 53.55°, C 46.52°
13. c 45.17 mi, A 89.3°, B 42.0° 15. a 13.9 in.,
B 36.127°, C 90.417° 17. Law of sines; C 98°,
a 96.7, c 101.9 19. Law of cosines; A 73.71°,
B 51.75°, C 54.54° 21. Cannot be solved
23. Law of cosines; A 33.71°, B 107.08°, C 39.21°
25. About 367 ft 27. About 1.5 mi
29. About 37 nautical mi 31. About 912 km
33. (a) About 16 ft; (b) about 122 ft2 35. About 4.7 cm
4
2
2
3
4 2
2
4i
2
Note: sin sin 180° .
1
bd ac ad bc sin 2
1
a b c d sin 2
1
d 1d 2 sin ,
2
where d 1 a b and d 2 c d.
4
Real
4 2
2
2
4
4
7
7
i sin
, or
4
4
32 cos 315° i sin 315°
, or 4cos 90° i sin 90°
i sin
11. 4i; 4 cos
2
2
9. 3 3i; 32 cos
13. 2 cos
15. 3 cos
7
7
i sin
, or 2cos 315° i sin 315°
4
4
3
3
, or 3cos 270° i sin 270°
i sin
2
2
i sin
, or 2cos 30° i sin 30°
6
6
2
2
cos 0 i sin 0, or cos 0° i sin 0°
19.
5
5
17. 2 cos
21. 6 cos
23.
5
5
i sin
, or 6cos 225° i sin 225°
4
4
3
33
i
2
2
25. 10i
27. 2 2i
29. 2i
Summer 2012
L.E.A.D. Ambassador Team 3
(Statistics Course Content)
“Algebra 2 and Trigonometry”
Ann Xavier Gantert
Amsco School Publications, Inc. (2009)
14411C15.pgs
8/14/08
10:36 AM
Page 587
CHAPTER
STATISTICS
Every year the admission officers of colleges
choose, from thousands of applicants, those students
who will be offered a place in the incoming class for
the next year. An attempt is made by the college to
choose students who will be best able to succeed academically and who best fit the profile of the student
body of that college. Although this choice is based not
only on academic standing, the scores on standardized
tests are an important part of the selection. Statistics
establishes the validity of the information obtained
from standardized tests and influence the interpretation of the data obtained from them.
15
CHAPTER
TABLE OF CONTENTS
15-1 Gathering Data
15-2 Measures of Central
Tendency
15-3 Measures of Central
Tendency for Grouped Data
15-4 Measures of Dispersion
15-5 Variance and Standard
Deviation
15-6 Normal Distribution
15-7 Bivariate Statistics
15-8 Correlation Coefficient
15-9 Non-Linear Regression
15-10 Interpolation and
Extrapolation
Chapter Summary
Vocabulary
Review Exercises
Cumulative Review
587
14411C15.pgs
8/14/08
588
10:36 AM
Page 588
Statistics
15-1 GATHERING DATA
Important choices in our lives are often made by evaluating information, but in
order to use information wisely, it is necessary to organize and condense the
multitude of facts and figures that can be collected. Statistics is the science that
deals with the collection, organization, summarization, and interpretation of
related information called data. Univariate statistics consists of one number for
each data value.
Collection of Data
Where do data come from? Individuals, government organizations, businesses,
and other political, scientific, and social groups usually keep records of their
activities. These records provide factual data. In addition to factual data, the outcome of an event such as an election, the sale of a product, or the success of a
movie often depends on the opinion or choices of the public.
Common methods of collecting data include the following:
1. Censuses: every ten years, the government conducts a census to determine
the U.S. population. Each year, almanacs are published that summarize
and update data of general interest.
2. Surveys: written questionnaires, personal interviews, or telephone requests
for information can be used when experience, preference, or opinions are
sought.
3. Controlled experiments: a structured study that usually consists of two
groups: one that makes use of the subject of the study (for example, a new
medicine) and a control group that does not. Comparison of results for the
two groups is used to indicate effectiveness.
4. Observational studies: similar to controlled experiments except that the
researcher does not apply the treatment to the subjects. For example, to
determine if a new drug causes cancer, it would be unethical to give the
drug to patients. A researcher observes the occurrence of cancer among
groups of people who previously took the drug.
When information is gathered, it may include data for all cases to which
the result of the study is to be applied. This source of information is called the
population. When the entire population can be examined, the study is a census.
For example, a study on the age and number of accidents of every driver insured
by an auto insurance company would constitute a census if all of the company’s
records are included. However, when it is not possible to obtain information
from every case, a sample is used to determine data that may then be applied to
every case. For example, in order to determine the quality of their product, the
quality-control department of a business may study a sample of the product
being produced.
14411C15.pgs
8/14/08
10:36 AM
Page 589
Gathering Data
589
In order that the sample reflect the properties of the whole group, the following conditions should exist:
1. The sample must be representative of the group being studied.
2. The sample must be large enough to be effective.
3. The selection should be random or determined in such a way as to eliminate any bias.
If a new medicine being tested is proposed for use by people of all ages, of
different ethnic backgrounds, and for use by both men and women, then the
sample must be made up of people who represent these differences in sufficient
number to be effective. A political survey to be effective must include people of
different cultural, ethnic, financial, geographic, and political backgrounds.
Potential Pitfalls of Surveys
Surveys are a very common way of collecting data. However, if not done correctly, the results of the survey can be invalid. One potential problem is with the
wording of the survey questions. For example, the question, “Do you agree that
teachers should make more money?” will likely lead to a person answering
“Yes.” A more neutral form of this question would be, “Do you believe that
teachers’ salaries are too high, too low, or just about right?”
Questions can also be too vague, “loaded” (that is, use words with unintended connotations), or confusing. For example, for many people, words that
invoke race will likely lead to an emotional response.
Another potential problem with surveys is the way that participants are
selected. For example, a magazine would like to examine the typical teenager’s
opinion on a pop singer in a given city. The magazine editors conduct a survey
by going to a local mall. The problem with this survey is that teenagers who go
to the mall are not necessarily representative of all teenagers in the city. A better survey would be done by visiting the high schools of the city.
Many surveys often rely on volunteers. However, volunteers are likely to
have stronger opinions than the general population. This is why the selection of
participants, if possible, should be random.
EXAMPLE 1
A new medicine intended for use by adults is being tested on five men whose
ages are 22, 24, 25, 27, and 30. Does the sample provide a valid test?
Solution No:
• The sample is too small.
• The sample includes only men.
• The sample does not include adults over 30.
14411C15.pgs
8/14/08
590
10:36 AM
Page 590
Statistics
EXAMPLE 2
The management of a health club has received complaints about the temperature of the water in the swimming pool. They want to sample 50 of the 200 members of the club to determine if the temperature of the pool should be changed.
How should this sample be chosen?
Solution One suggestion might be to poll the first 50 people who use the pool on a
given day. However, this will only include people who use the pool and who
can therefore tolerate the water temperature.
Another suggestion might be to place 50 questionnaires at the entrance
desk and request members to respond. However, this includes only people
who choose to respond and who therefore may be more interested in a
change.
A third suggestion might be to contact by phone every fourth person on
the membership list and ask for a response. This method will produce a random
sample but will include people who have no interest in using the pool. This
sample may be improved by eliminating the responses of those people.
Organization of Data
In order to be more efficiently presented and more easily understood and interpreted, the data collected must be organized and summarized. Charts and
graphs such as the histogram are useful tools. The stem-and-leaf diagram is an
effective way of organizing small sets of data.
For example, the heights of
Heights of Children
the 20 children in a seventhgrade class are shown to the 61 71 58 72 60 53 74 61 68 65
right. To draw a stem-and-leaf 72 67 64 48 70 56 65 67 59 61
diagram, choose the tens digit as
the stem and the units digit as the
leaf.
(1) Draw a vertical line and list
the tens digits, 4, 5, 6, and 7
(or the stem), from bottom
to top to the left of the line:
Stem
(2) Enter each height by writing
the leaf, the units digit, to
the right of the line, following the appropriate stem:
Stem
7
6
5
4
7
6
5
4
Leaf
12420
1018574571
8369
8
14411C15.pgs
8/14/08
10:36 AM
Page 591
Gathering Data
(3) Arrange the leaves in numerical order after
each stem:
(4) Add a key to indicate the meaning of the
numbers in the diagram:
Stem
591
Leaf
7
01224
6
0111455778
5
3689
4
8
Key: 4 8 5 48
For larger sets of data, a frequency distribution
table can be drawn. Information is grouped and the
frequency, the number of times that a particular
value or group of values occurs, is stated for each
group. For example, the table to the right lists the
scores of 250 students on a test administered to all
tenth graders in a school district.
The table tells us that 24 students scored between
91 and 100, that 82 scored between 81 and 90, that 77
scored between 71 and 80. The table also tells us that
the largest number of students scored in the 80’s and
that ten students scored 50 or below. Note that unlike
the stem-and-leaf diagram, the table does not give us
the individual scores in each interval.
Score
Frequency
91–100
24
81–90
82
71–80
77
61–70
36
51–60
21
41–50
8
31–40
2
EXAMPLE 2
The prices of a gallon of milk in 15 stores are listed below.
$3.15 $3.39 $3.28 $2.98 $3.25 $3.45 $3.58 $3.24
$3.35 $3.29 $3.29 $3.30 $3.25 $3.40 $3.29
a. Organize the data in a stem-and-leaf diagram.
b. Display the data in a frequency distribution table.
c. If the 15 stores were chosen at random from the more than 100 stores that sell
milk in Monroe County, does the data set represent a population or a sample?
Solution a. Use the first two
digits of the price
as the stem.
Use the last digit as
the leaf.
Stem
Stem
3.5
3.4
3.3
3.2
3.1
3.0
2.9
3.5
3.4
3.3
3.2
3.1
3.0
2.9
Leaf
8
50
950
8549959
5
8
Write the leaves in
numerical order.
Stem
Leaf
3.5
8
3.4
05
3.3
059
3.2
4558999
3.1
5
3.0
2.9
8
Key: 2.9 8 5 2.98
8/14/08
592
10:36 AM
Page 592
Statistics
b. Divide the data into groups of length $0.10 starting with $2.90. These groups
correspond with the stems of the stem-and-leaf diagram. The frequencies
can be determined by the use of a tally to represent each price.
Stem
Leaf
3.5
8
3.4
05
3.3
059
3.2
4558999
3.1
5
3.0
2.9
8
Key: 2.9 8 5 2.98
Price
Tally
$3.50–$3.59
|
1
$3.40–$3.49
||
2
$3.30–$3.39
|||
3
$3.20–$3.29
|||| ||
7
$3.10–$3.19
|
1
Frequency
$3.00–$3.09
$2.90–$2.99
0
|
1
c. The data set is obtained from a random selection of stores from all of
the stores in the study and is therefore a sample. Answer
Once the data have been organized, a graph can be used to visualize the
intervals and their frequencies. A histogram is a vertical bar graph where each
interval is represented by the width of the bar and the frequency of the interval
is represented by the height of the bar. The bars are placed next to each other
to show that, as one interval ends, the next interval begins. The histogram below
shows the data of Example 2:
PRICE OF A GALLON OF MILK
Frequency
14411C15.pgs
8
7
6
5
4
3
2
1
0
.99
$2
2
–$
0
9
.
.19
.09
$3
.00
3
–$
.10
$3
3
–$
.20
$3
Price
.49
.39
.29
3
–$
.30
$3
3
–$
.40
$3
.59
3
–$
.50
$3
3
–$
14411C15.pgs
8/14/08
10:36 AM
Page 593
Gathering Data
593
A graphing calculator can be used to display a histogram from the data on
a frequency distribution table.
(1) Clear L1 and L2, the lists to be used, of existing data.
ENTER:
STAT
4
2nd
L1
,
2nd
1 to edit the lists. Enter the
(2) Press STAT
minimum value of each interval in L1 and
the frequencies into L2.
L2
ENTER
L1
L2
3.5
1
3.4
2
3.3
3
3.2
7
3.1
1
3
0
2.9
1
L3(1)=
L3
(3) Clear any functions in the Y menu.
(4) Turn on Plot1 from the STAT PLOT menu
and select for histogram. Make
sure to also set Xlist to L1 and Freq to L2.
ENTER:
2nd
STAT PLOT
䉲
䉴
䉴
L1
䉲
2nd
ENTER
Plot1 Plot2 Plot3
On
Off
.
Ty p e : ......
ENTER
1
䉲
2nd
Xlist:L1
Freq:L2
L2
(5) In the WINDOW menu, enter Xmin as 2.8,
the length of one interval less than the
smallest interval value, and Xmax as 3.7, the
length of one interval more than the largest
interval value. Enter Xscl as 0.10, the
length of the interval. The Ymin is 0 and
Ymax is 9 to be greater than the largest
frequency.
(6) Press GRAPH to graph the histogram.
We can view the frequency associated
with each interval by pressing TRACE .
Use the left and right arrows to move
from one interval to the next.
WINDOW
Xmin =2.8
Xmax=3.7
Xscl =.1
Ymin =0
Ymax=9
Yscl =1
Xres=1
3
14411C15.pgs
8/14/08
594
10:36 AM
Page 594
Statistics
Exercises
Writing About Mathematics
1. In a controlled experiment, two groups are formed to determine the effectiveness of a new
cold remedy. One group takes the medicine and one does not. Explain why the two groups
are necessary.
2. In the experiment described in Exercise 1, explain why it is necessary that a participant does
not know to which group he or she belongs.
Developing Skills
In 3–5, organize the data in a stem-and-leaf diagram.
3. The grades on a chemistry test:
95 90 84 85 74 67 78 86 54 82
75 67 92 66 90 68 88 85 76 87
4. The weights of people starting a weight-loss program:
173 210 182 190 175 169 236 192 203 196 201
187 205 195 224 177 195 207 188 184 196 155
5. The heights, in centimeters, of 25 ten-year-old children:
137 134 130 144 131 141 136 140 137 129 139 137 144
127 147 143 132 132 142 142 131 129 138 151 137
In 6–8, organize the data in a frequency distribution table.
6. The numbers of books read during the summer months by each of 25 students:
2 2 5 1 3 0 7 2 4 3 3 1 8
5 7 3 4 1 0 6 3 4 1 1 2
7. The sizes of 26 pairs of jeans sold during a recent sale:
8 12 14 10 12 16 14 6 10 9 8 13 12
8 12 10 12 14 10 12 16 10 11 15 8 14
8. The number of siblings of each of 30 students in a class:
2 1 1 5 1 0 2 2 1 3 4 0 6 2 0
3 1 2 2 1 1 1 0 2 1 0 1 1 2 3
14411C15.pgs
8/14/08
10:36 AM
Page 595
595
Gathering Data
In 9–11, graph the histogram of each set of data.
9.
10.
xi
fi
xi
35–39
13
101–110
30–34
19
25–29
fi
11.
xi
fi
3
$55–$59
20
91–100
6
$50–$54
15
10
81–90
10
$45–$49
12
20–24
13
71–80
13
$40–$44
5
15–19
8
61–70
14
$35–$39
10
10–14
19
51–60
2
$30–$34
12
5–9
15
41–50
2
$25–$29
16
Applying Skills
In 12–18, suggest a method that might be used to collect data for each study. Tell whether your
method uses a population or a sample.
12. Average temperature for each month for a given city
13. Customer satisfaction at a restaurant
14. Temperature of a patient in a hospital over a period of time
15. Grades for students on a test
16. Population of each of the states of the United States
17. Heights of children entering kindergarten
18. Popularity of a new movie
19. The grades on a math test of 25 students are listed below.
86 92 77 84 75 95 66 88 84 53 98 87 83
74 61 82 93 98 87 77 86 58 72 76 89
a. Organize the data in a stem-and-leaf diagram.
b. Organize the data in a frequency distribution table.
c. How many students scored 70 or above on the test?
d. How many students scored 60 or below on the test?
20. The stem-and-leaf diagram at the right shows the ages of 30 people
in an exercise class. Use the diagram to answer the following questions.
a. How many people are 45 years old?
b. How many people are older than 60?
c. How many people are younger than 30?
d. What is the age of the oldest person in the class?
e. What is the age of the youngest person in the class?
Stem
Leaf
7
2
6
015
5
136679
4
2245567
3
9
2
1335788
1
02269
Key: 1 9 5 19
14411C15.pgs
8/14/08
596
10:36 AM
Page 596
Statistics
Hands-On Activity
In this activity, you will take a survey of 25 people. You will need a stopwatch or a clock with a second hand. Perform the following experiment with each participant to determine how each perceives
the length of a minute:
1. Indicate the starting point of the minute.
2. Have the person tell you when he or she believes that a minute has passed.
3. Record the actual number of seconds that have passed.
After surveying all 25 participants, use a stem-and-leaf diagram to record your data. Keep this
data. You will use it throughout your study of this chapter.
15-2 MEASURES OF CENTRAL TENDENCY
After data have been collected, it is often useful to represent the data by a single value that in some way seems to represent all of the data. This number is
called the measure of central tendency. The most frequently used measures of
central tendency are the mean, the median, and the mode.
The Mean
The mean or arithmetic mean is the most common measure of central tendency. The mean is the sum of all of the data values divided by the number of
data values.
For example, nine members of the basketball team played during all or part
of the last game. The number of points scored by each of the players was:
21, 15, 12, 9, 8, 7, 5, 2, 2
Mean 5 21 1 15 1 12 1 9 19 8 1 7 1 5 1 2 1 2 5 81
9 59
Note that if each of the 9 players had scored 9 points, the total number of
points scored in the game would have been the same.
The summation symbol, S, is often used to designate the sum of the data values. We designate a data value as xi and the sum of n data values as
n
c1x
n
a xi 5 x1 1 x2 1 x3 1
i51
For the set of data given above:
n 5 9, x1 5 21, x2 5 15, x3 5 12, x4 5 9, x5 5 8, x6 5 7, x7 5 5, x8 5 2, x9 5 2
9
a xi 5 81
i51
The subscript for each data value indicates its position in a list of data values, not its value, although the value of i and the data value may be the same.
14411C15.pgs
8/14/08
10:36 AM
Page 597
597
Measures of Central Tendency
Procedure
To find the mean of a set of data:
1. Add the data values.
2. Divide the sum by n, the total number of data values.
EXAMPLE 1
An English teacher recorded the number of spelling errors in the 40 essays written by students. The table below shows the number of spelling errors and the
frequency of that number of errors, that is, the number of essays that contained
that number of misspellings. Find the mean number of spelling errors for these
essays.
Errors
0
1
2
3
4
5
6
7
8
9
10
Frequency
1
3
2
2
6
9
7
5
2
1
2
Solution To find the total number of spelling errors, first multiply each number of
errors by the frequency with which that number of errors occurred. For example, since 2 essays each contained 10 errors, there were 20 errors in these
essays. Add the products in the fi xi row to find the total number of errors in
the essays. Divide this total by the total frequency.
Total
Errors (xi)
0
1
2
3
4
5
6
7
8
9
10
Frequency (fi)
1
3
2
2
6
9
7
5
2
1
2
40
Errors Frequency (fixi)
0
3
4
6
24
45
42
35
16
9
20
204
10
a fixi
Mean 5
i50
40
5 204
40 5 5.1
Note that for this set of data, the data value is equal to i for each xi.
Answer 5.1 errors
The Median
The median is the middle number of a data set arranged in numerical order.
When the data are arranged in numerical order, the number of values less than
or equal to the median is equal to the number of values greater than or equal to
8/14/08
Page 598
Statistics
the median. Consider again the nine members of the basketball team who
played during all or part of the last game and scored the following number of
points:
2, 2, 5, 7, 8, 9, 12, 15, 21
We can write 9 5 4 1 1 1 4. Therefore, we think of four scores below the
median, the median, and four scores above the median.
8,
scores less than the median
↑
9, 12, 15, 21
d
2, 2, 5, 7,
d
scores greater than the median
median
Note that the number of data values can be written as 2(4) 1 1. The median
is the (4 1 1) or 5th value from either end of the distribution. The median number of points scored is 8.
When the number of values in a set of data is even, then the median is the
mean of the two middle values. For example, eight members of a basketball
team played in a game and scored the following numbers of points.
4, 6, 6, 7, 11, 12, 18, 20
We can separate the eight data values into two groups of 4 values. Therefore,
we average the largest score of the four lowest scores and the smallest score of
the four highest scores.
4, 6, 6, 7,
four lowest scores
7 1 11
2
↑
,
11, 12, 18, 20
d
598
10:36 AM
d
14411C15.pgs
four highest scores
median
median 5 7 12 11 5 9
Note that the number of data values can be written as 2(4). The median is
the mean of the 4th and 5th value from either end of the distribution. The
median is 9. There are four scores greater than the median and four scores lower
than the median. The median is a middle mark.
Procedure
To find the median of a set of data:
1. Arrange the data in order from largest to smallest or from smallest to
largest.
2. a. If the number of data values is odd, write that number as 2n 1 1.The
median is the score that is the (n 1 1)th score from either end of the
distribution.
b. If the number of data values is even, write that number as 2n.The median
is the score that is the mean of the nth score and the (n 1 1)th score
from either end of the distribution.
10:36 AM
Page 599
Measures of Central Tendency
599
The Mode
The mode is the value or values that occur most frequently in a set of data. For
example, in the set of numbers
3, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 10,
the number that occurs most frequently is 6. Therefore, 6 is the mode for this set
of data.
When a set of numbers is arranged in a
Number of
frequency distribution table, the mode is the
Books Read
Frequency
entry with the highest frequency. The table to
8
1
the right shows, for a given month, the number of books read by each student in a class.
7
1
The largest number of students, 12, each
6
3
read four books. The mode for this distribu5
6
tion is 4.
A data set may have more than one
4
12
mode. For example, in the set of numbers 3, 4,
3
4
5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 10, the numbers 5 and
2
2
6 each occur four times, more frequently than
any other number. Therefore, 5 and 6 are
1
0
modes for this set of data. The set of data is
0
1
said to be bimodal.
Quartiles
When a set of data is listed in numerical order, the median separates the data
into two equal parts. The quartiles separate the data into four equal parts. To
find the quartiles, we first separate the data into two equal parts and then separate each of these parts into two equal parts. For example, the grades of 20 students on a math test are listed below.
58, 60, 65, 70, 72, 75, 76, 80, 80, 81, 83, 84, 85, 87, 88, 88, 90, 93, 95, 98
Since there are 20 or 2(10) grades, the median grade that separates the data
into two equal parts is the average of the 10th and 11th grade.
58, 60, 65, 70, 72, 75, 76, 80, 80, 81,
Lower half
83, 84, 85, 87, 88, 88, 90, 93, 95, 98
↑
82
i
8/14/08
i
14411C15.pgs
Upper half
Median
The 10th grade is 81 and the 11th grade is 83. Therefore, the mean of these
83
two grades, 81 1
or 82, separates the data into two equal parts. This number
2
is the median grade.
8/14/08
Page 600
Statistics
Now separate each half into two equal parts. Find the median of the two
lower quarters and the median of the two upper quarters.
D
D
↑
73.5
D
58, 60, 65, 70, 72 75, 76, 80, 80, 81 83, 84, 85, 87, 88 88, 90, 93, 95, 98
D
↑
82
↑
88
The numbers 73.5, 82, and 88 are the quartiles for this data.
• One quarter of the grades are less than or equal to 73.5. Therefore, 73.5 is
the first quartile or the lower quartile.
• Two quarters of the grades are less than or equal to 82. Therefore, 82 is the
second quartile. The second quartile is always the median.
• Three quarters of the grades are less than or equal to 88. Therefore, 88 is
the third quartile or the upper quartile.
Note: The minimum, first quartile, median, third quartile, and maximum make
up the five statistical summary of a data set.
When the data set has an odd number of values, the median or second quartile will be one of the values. This number is not included in either half of the
data when finding the first and third quartiles.
For example, the heights, in inches, of 19 children are given below:
37, 39, 40, 42, 42, 43, 44, 44, 44, 45, 46, 47, 47, 48, 49, 49, 50, 52, 53
There are 19 5 2(9) 1 1 data values. Therefore, the second quartile is the
10th height or 45.
45 , 46, 47, 47, 48, 49, 49, 50, 52, 53
I
37, 39, 40, 42, 42, 43, 44, 44, 44,
I
There are 9 5 2(4) 1 1 heights in the lower half of the data and also in the
upper half of the data. The middle height of each half is the 5th height.
45 , 46, 47, 47, 48,
49 , 49, 50, 52, 53
D
42 , 43, 44, 44, 44,
D
37, 39, 40, 42,
D
600
10:36 AM
D
14411C15.pgs
↑
↑
↑
Lower quartile
Median
Upper quartile
In this example, the first or lower quartile is 42, the median or second quartile is 45, and the third or upper quartile is 49. Each of these values is one of the
data values, and the remaining values are separated into four groups with the
same number of heights in each group.
Box-and-Whisker Plot
A box-and-whisker plot is a diagram that is used to display the quartile values
and the maximum and minimum values of a distribution. We will use the data
from the set of heights given above.
10:36 AM
Page 601
Measures of Central Tendency
601
49 , 49, 50, 52, 53
D
45 , 46, 47, 47, 48,
D
42 , 43, 44, 44, 44,
37, 39, 40, 42,
D
8/14/08
D
14411C15.pgs
1. Choose a scale that includes the maximum and minimum values of the
data. We will use a scale from 35 to 55.
2. Above the scale, place dots to represent the minimum value, the lower
quartile, the mean, the upper quartile, and the maximum value.
35
42
37
45
49
53
55
3. Draw a box with opposite sides through the lower and upper quartiles and
a vertical line through the median.
35
42
37
45
49
53
55
4. Draw whiskers by drawing a line to join the dot that represents the minimum value to the dot that represents the lower quartile and a line to join
the dot that represents the upper quartile to the dot that represents the
maximum value.
35
42
37
45
49
53
55
A graphing calculator can be used to find the quartiles and to display the
box-and-whisker plot.
(1) Enter the data for this set of heights into L1.
(2) Turn off all plots and enter the required
choices in Plot 1.
ENTER:
䉲
2nd
2nd
STAT PLOT
䉴
L1
䉴
䉴
䉲
ALPHA
On
Off
.
Ty p e : ......
ENTER
1
䉴
Plot1 Plot2 Plot3
ENTER
1
䉲
Xlist:L1
Freq:1
8/14/08
602
10:36 AM
Page 602
Statistics
(3) Now display the plot by entering
9 . You can press TRACE
ZOOM
to
display the five statistical summary.
The five statistical summary of a set of data
can also be displayed on the calculator by using
1-Var Stats.
ENTER:
STAT
䉴
ENTER
ENTER
The first entry under 1-Var Stats is –x, the value of the mean. The next is
a x, the sum of the data values. The next three entries are values that will be
used in the sections that follow. The last entry is the number of data values.
The arrow tells us that there is more information. Scroll down to display the
minimum value, minX 5 37, the lower quartile, Q1 5 42, the median or second quartile, Med 5 45, the upper quartile, Q3 5 49, and the maximum value,
maxX 5 53.
>
1–Var Stats
–
x=45.31578947
x=861
x2=39353
Sx=4.321170515
X=4.205918531
n=19
1–Var Stats
n=19
minX=37
Q1=42
Med=45
Q3=49
maxX=53
>
14411C15.pgs
EXAMPLE 1
Find the mean, the median, and the mode of the following set of grades:
92, 90, 90, 90, 88, 87, 85, 70
n
Solution
Mean 5 n1 a xi 5 18(92 1 90 1 90 1 90 1 88 1 87 1 85 1 70) 5 692
8 5 86.5
i51
88
Median 5 the average of the 4th and 5th grades 5 90 1
5 89
2
Mode 5 the grade that appears most frequently 5 90
EXAMPLE 2
The following list shows the length of time, in minutes, for each of 35 employees
to commute to work.
25 12 20 18 35 25 40 35 27 30 60 22 36 20 18
27 35 42 35 55 27 30 15 22 10 35 27 15 57 18
25 45 24 27 25
14411C15.pgs
8/14/08
10:36 AM
Page 603
Measures of Central Tendency
603
a. Organize the data in a stem-and-leaf diagram.
b. Find the median, lower quartile, and upper quartile.
c. Draw a box-and-whisker plot.
Solution a. (1) Choose the tens digit as the
stem and enter the units digit
as the leaf for each value.
Stem
(2) Write the leaves in numerical
order from smallest to largest.
Stem
b. (1) The median is the middle value
of the 35 data values when the
values are arranged in order.
Stem
35 5 2(17) 1 1
The median is the 18th value.
Separate the data into groups
of 17 from each end of the distribution. The median is 27.
(2) There are 17 5 2(8) 1 1 data
values below the median. The
lower quartile is the 9th data
value from the lower end.
The upper quartile is the 9th
data value from the upper
end.
6
5
4
3
2
1
Leaf
0
57
025
55065505
50572077275475
2885058
Leaf
6
0
5
57
4
025
3
00555556
2
00224555577777
1
0255888
Key: 1 0 5 10
6
5
4
3
2
1
Stem
6
5
4
3
2
1
Leaf
0
57
025
005 5 5556
002 2 455557 7 777
0255888
Leaf
0
57
025
005 5 5 5 56
0 0 2 2 455557 7 777
0255888
The lower quartile is 20, the
median is 27, and the upper quartile is 35.
c.
10
20
27
35
60
14411C15.pgs
8/14/08
604
10:36 AM
Page 604
Statistics
Note: In the stem-and-leaf diagram of Example 2, the list is read from left to
right to find the lower quartile. The list is read from right to left (starting from
the top) to find the upper quartile.
Exercises
Writing About Mathematics
1. Cameron said that the number of data values of any set of data that are less than the lower
quartile or greater than the upper quartile is exactly 50% of the number of data values. Do
you agree with Cameron? Explain why or why not.
2. Carlos said that for a set of 2n data values or of 2n 1 1 data values, the lower quartile is the
median of the smallest n values and the upper quartile is the median of the largest n values.
Do you agree with Carlos? Explain why or why not.
Developing Skills
In 3–8, find the mean, the median, and the mode of each set of data.
3. Grades: 74, 78, 78, 80, 80, 80, 82, 88, 90
4. Heights: 60, 62, 63, 63, 64, 65, 66, 68, 68, 68, 70, 75
5. Weights: 110, 112, 113, 115, 115, 116, 118, 118, 125, 134, 145, 148
6. Number of student absences: 0, 0, 0, 1, 1, 2, 2, 2, 3, 4, 5, 9
7. Hourly wages: $6.90, $7.10, $7.50, $7.50, $8.25, $9.30, $9.50, $10.00
8. Tips: $1.00, $1.50, $2.25, $3.00, $3.30, $3.50, $4.00, $4.75, $5.00, $5.00, $5.00
In 9–14, find the median and the first and third quartiles for each set of data values.
9. 2, 3, 5, 8, 9, 11, 15, 16, 17, 20, 22, 23, 25
10. 34, 35, 35, 36, 38, 40, 42, 43, 43, 43, 44, 46, 48, 50
11. 23, 27, 15, 38, 12, 17, 22, 39, 28, 20, 27, 18, 25, 28, 30, 29
12. 92, 86, 77, 85, 88, 90, 81, 83, 95, 76, 65, 88, 91, 81, 88, 87, 95
13. 75, 72, 69, 68, 66, 65, 64, 63, 63, 61, 60, 59, 59, 58, 56, 54, 52, 50
14. 32, 32, 30, 30, 29, 27, 26, 22, 20, 20, 19, 18, 17
15. A student received the following grades on six tests: 90, 92, 92, 95, 95, x.
a. For what value(s) of x will the set of grades have no mode?
b. For what value(s) of x will the set of grades have only one mode?
c. For what value(s) of x will the set of grades be bimodal?
16. What are the first, second, and third quartiles for the set of integers from 1 to 100?
17. What are the first, second, and third quartiles for the set of integers from 0 to 100?
14411C15.pgs
8/14/08
10:36 AM
Page 605
Measures of Central Tendency for Grouped Data
Applying Skills
18. The grades on a English test are shown in the
stem-and-leaf diagram to the right.
a. Find the mean grade.
b. Find the median grade.
c. Find the first and third quartiles.
d. Draw a box-and-whisker plot for this data.
Stem
605
Leaf
9
001259
8
022555788
7
35667
6
055
5
5
4
7
Key: 4 7 5 47
19. The weights in pounds of the members of the football team are shown below:
181 199 178 203 211 208 209 202 212 194
185 208 223 206 202 213 202 186 189 203
a. Find the mean.
b. Find the median.
c. Find the mode or modes.
d. Find the first and third quartiles.
e. Draw a box-and-whisker plot.
20. Mrs. Gillis gave a test to her two classes of algebra. The mean grade for her class of 20 students was 86 and the mean grade of her class of 15 students was 79. What is the mean grade
when she combines the grades of both classes?
Hands-On Activity
Use the estimates of a minute collected in the Hands-On Activity of the previous section to determine the five statistical summary for your data. Draw a box-and-whisker plot to display the data.
15-3 MEASURES OF CENTRAL TENDENCY FOR GROUPED DATA
Most statistical studies involve much larger numbers of data values than can be
conveniently displayed in a list showing each data value. Large sets of data are
usually organized into a frequency distribution table.
Frequency Distribution Tables for
Individual Data Values
A frequency distribution table records the individual data values and the frequency or number of times that the data value occurs in the data set. The example on page 606 illustrates this method of recording data.
Each of an English teacher’s 100 students recently completed a book report.
The teacher recorded the number of misspelled words in each report. The table
14411C15.pgs
8/14/08
606
10:36 AM
Page 606
Statistics
records the number of reports for each number of misspelled words. Let xi represent the number of misspelled words in a report and fi represent the number
of reports that contain xi misspelled words.
Total
xi
0
1
2
3
4
5
6
7
8
9
10
fi
5
7
6
8
19
26
16
7
5
0
1
100
xifi
0
7
12
24
76
130
96
49
40
0
10
444
The mean of this set of data is the total number of misspelled words divided
by the number of reports. To find the total number of misspelled words, we must
first multiply each number of misspelled words, xi, by the number of reports that
contain that number of misspelled words, fi. That is, we must find xi fi for each
number of misspelled words.
10
For this set of data, when we sum the fi row, a fi 5 100, and when we sum
i50
10
the xi fi row a xifi 5 444.
i50
10
a xi fi
Mean 5
i50
10
a fi
5 444
100 5 4.44
i50
To find the median and the quartiles for this set of data, we will find the cumulative frequency for each number of misspelled words. The cumulative frequency
is the accumulation or the sum of all frequencies less than or equal to a given
frequency. For example, the cumulative frequency for 3 misspelled words on a
report is the sum of the frequencies for 3 or fewer misspelled words, and the
cumulative frequency for 6 misspelled words on a report is the sum of the frequencies for 6 or fewer misspelled words. The third row of the table shows that
0 misspelled words occur 5 times, 1 or fewer occur 7 1 5 or 12 times, 2 or fewer
occur 6 1 12 or 18 times. In each case, the cumulative frequency for xi is the frequency for xi plus the cumulative frequency for xi21. The cumulative frequency
for the largest data value is always equal to the total number of data values.
xi
0
1
2
3
4
5
6
7
8
9
10
fi
5
7
6
8
19
26
16
7
5
0
1
Cumulative
Frequency
5
12
18
26
45
71
87
94
99
99
100
↑
3
Lower
quartile
↑
5
↑
6
Median Upper
quartile
14411C15.pgs
8/14/08
10:36 AM
Page 607
Measures of Central Tendency for Grouped Data
607
Since there are 100 data values, the median is the average of the 50th and
51st values. To find these values, look at the cumulative frequency column. There
are 45 values less than or equal to 4 and 71 less than or equal to 5. Therefore,
the 50th and 51st values are both 5 and the median is 5 misspelled words.
Similarly, the upper quartile is the average of the 75th and 76th values. Since
there are 71 values less than or equal to 5, the 75th and 76th values are both 6
misspelled words. The lower quartile is the average of the 25th and 26th values.
Since there are 18 values less than or equal to 2, the 25th and 26th values are
both 3 misspelled words.
Percentiles
A percentile is a number that tells us what percent of the total number of data
values lie at or below a given measure.
For example, let us use the data from the previous section. The table records
the number of reports, fi, that contain each number of misspelled words, xi , on
100 essays.
xi
0
1
2
3
4
5
6
7
8
9
10
fi
5
7
6
8
19
26
16
7
5
0
1
Cumulative
Frequency
5
12
18
26
45
71
87
94
99
99
100
To find the percentile rank of 7 misspelled words, first find the number of
essays with fewer than 7 misspelled words, 87. Add to this half of the essays with
7 misspelled words, 72 or 3.5. Add these two numbers and divide the sum by the
number of essays, 100.
87 1 3.5
100
5 90.5
100 5 90.5%
Percentiles are usually not written with fractions. We say that 7 misspelled
words is at the 90.5th or 91st percentile. That is, 91% of the essays had 7 or fewer
misspelled words.
Frequency Distribution Tables for Grouped Data
Often the number of different data values in a set of data is too large to list each
data value separately in a frequency distribution table. In this case, it is useful to
list the data in terms of groups of data values rather than in terms of individual
data values. The following example illustrates this method of recording data.
There are 50 members of a weight-loss program. The weights range from
181 to 285 pounds. It is convenient to arrange these weights in groups of 10
pounds starting with 180–189 and ending with 280–289. The frequency distribution table shows the frequencies of the weights for each interval.
14411C15.pgs
8/14/08
608
10:36 AM
Page 608
Statistics
Weights
Midpoint
xi
Frequency
fi
280–289
270–279
260–269
250–259
240–249
230–239
220–229
210–219
200–209
190–199
180–189
284.5
274.5
264.5
254.5
244.5
234.5
224.5
214.5
204.5
194.5
184.5
1
3
4
8
12
10
5
3
2
1
1
Cumulative
Frequency
xifi
284.5
823.5
1,058.0
2,036.0
2,934.0
2,345.0
1,122.5
643.5
409.0
194.5
184.5
50
49
46
42
34
22
12
7
4
2
1
In order to find the mean, assume that the
weights are evenly distributed throughout the
interval. The mean is found by using the midpoint of the weight intervals as representative of
each value in the interval groupings.
11
a xi fi
Mean 5
i51
11
a fi
12,035
5 50
5 240.7
i51
To find the median for a set of data that is
organized in intervals greater than 1, first find
50
12,035
the interval in which the median lies by using
the cumulative frequency.
There are 50 data values. Therefore, the median is the value between the
25th and the 26th values. The cumulative frequency tells us that there are 22 values less than or equal to 239 and 34 values less than or equal to 249. Therefore,
the 25th and 26th values are in the interval 240–249. Can we give a better
approximation for the median?
The endpoints of an interval are the lowest and highest data values to be
entered in that interval. The boundary values are the values that separate intervals. The lower boundary of an interval is midway between the lower endpoint
of the interval and the upper endpoint of the interval that precedes it. The lower
boundary of the 240–249 is midway between 240 and 239, that is, 239.5. The
upper endpoint of this interval is between 249 and 250, that is 249.5.
Since there are 34 weights less than or equal to 249 and 22 weights less than
240, the weights in the 240–249 interval are the 23rd through the 34th weights.
Think of these 12 weights as being evenly spaced throughout the interval.
median
23
24
25
26
27
29
28
30
31
32
239.5
33
34
249.5
3
12
3
The midpoint between the 25th and 26th weights is 12
of the distance
between the boundaries of the interval, a difference of 10.
3
Median 5 239.5 1 12
(10) 5 239.5 1 2.5 5 242
The estimated median for the weights is 242 pounds. Thus, if we assume that
the data values are evenly distributed within each interval, we can obtain a better approximation for the median.
14411C15.pgs
8/14/08
10:36 AM
Page 609
Measures of Central Tendency for Grouped Data
609
EXAMPLE 1
The numbers of pets owned by the children in a
sixth-grade class are given in the table.
No. of Pets
6
5
4
3
2
1
0
a. Find the mean.
b. Find the median.
c. Find the mode for this set of data.
d. Find the percentile rank of 4 pets.
Frequency
2
1
3
5
8
14
7
Solution a. Add to the table the number of pets times the frequency, and the cumulative frequency for each row of the table.
No. of Pets
xi
Frequency
fi
fi xi
Cumulative
Frequency
6
5
4
3
2
1
0
2
1
3
5
8
14
7
12
5
12
15
16
14
0
40
38
37
34
29
21
7
40
74
Add the numbers in the fixi column and the numbers in the fi column.
6
6
a fi 5 40
a fixi 5 74
i50
i50
6
a xi fi
Mean 5
i50
6
a fi
5 74
40 5 1.85
i50
b. There are 40 data values in this set of data. The median is the average of the
20th and the 21st data values. Since there are 21 students who own 1 or
fewer pets, both the 20th and the 21st data value is 1. Therefore, the median
number of pets is 1.
c. The largest number of students, 14, have 1 pet. The mode is 1.
d. There are 34 students with fewer than 4 pets and 3 students with 4 pets. Add
34 and half of 3 and divide the sum by the total number of students, 40.
34 1 32
40
5 35.5
40 5 0.8875 5 88.75%
Four pets represents the 89th percentile.
8/14/08
610
10:37 AM
Page 610
Statistics
Calculator Clear the lists first if necessary by keying in STAT
Solution for
2nd
ENTER .
L2
a, b, and c
4
2nd
L2
ENTER
L1
,
Enter the number of pets in L1.
Enter the frequency for each number of pets in L2.
Display 1-Var Stats.
ENTER:
DISPLAY:
STAT
䉴
2nd
1
L1
,
2nd
1–Var Stats
–
x=1.85
x=74
x2=236
Sx=1.594059485
x=1.574007624
n=40
1–Var Stats
n=40
minX=0
Q1=1
Med=1
Q3=3
maxX=6
>
14411C15.pgs
>
– , is the mean, 1.85. Use the down-arrow key,
The first entry, x
the median, Med. The median is 1.
6
䉲
, to display
6
Note that a x 5 74 is a fixi and n 5 40 is a fi.
i50
i50
Answers a. mean 5 1.85 b. median 5 1 c. mode 5 1 d. 4 is the 89th percentile
Note: The other information displayed under 1-Var Stats will be used in the
sections that follow.
EXAMPLE 2
A local business made the summary of the ages of 45 employees shown below.
Find the mean and the median age of the employees to the nearest integer.
Age
Frequency
20–24 25–29 30–34 35–39 40–44 45–49
1
2
5
6
7
10
50–54 55–59 60–64
7
5
Solution Add to the table the midpoint, the midpoint times the frequency, and the
cumulative frequency for each interval.
2
14411C15.pgs
8/14/08
10:37 AM
Page 611
Measures of Central Tendency for Grouped Data
Age
Midpoint
(xi)
Frequency
(fi)
xifi
Cumulative
Frequency
60–64
55–59
50–54
45–49
40–44
35–39
30–34
25–29
20–24
62
57
52
47
42
37
32
27
22
2
5
7
10
7
6
5
2
1
124
285
364
470
294
222
160
54
22
45
43
38
31
21
14
8
3
1
45
1,995
611
11
a xi fi
Mean 5
i51
11
a fi
1,995
5 45 5 44.333 . . . 44
i51
The mean is the middle age of 45 ages or the 23rd age. Since there are 21 ages
less than 45, the 23rd age is the second of 10 ages in the 45–49 interval. The
boundaries of the 45–49 interval are 44.5 to 49.5
median
Median 5 44.5 1
112
10
35
5 44.5 1 0.75 5 45.25 45
22
44.5
1 12
23
24
25
26
49.5
10
Answer mean 5 44, median 5 45
Exercises
Writing About Mathematics
1. Adelaide said that since, in Example 2, there are 10 employees whose ages are in the 45–49
interval, there must be two employees of age 45. Do you agree with Adelaide? Explain why
or why not.
2. Gail said that since, in Example 2, there are 10 employees whose ages are in the 45–49 interval, there must at least two employees who are the same age. Do you agree with Gail?
Explain why or why not.
14411C15.pgs
8/14/08
612
10:37 AM
Page 612
Statistics
Developing Skills
In 3–8, find the mean, the median, and the mode for each set of data.
3.
6.
xi
fi
5
4
3
2
1
0
6
10
15
11
2
1
xi
fi
10
9
8
7
6
5
4
1
1
3
7
6
2
2
4.
7.
xi
fi
50
40
30
20
10
8
12
17
10
3
5.
xi
fi
$1.10
$1.20
$1.30
$1.40
$1.50
1
5
8
6
6
xi
fi
12
11
10
9
8
7
6
5
7
15
13
16
14
15
9
2
xi
fi
95
90
85
80
75
70
65
60
2
8
12
10
9
3
0
1
8.
9. Find the percentile rank of 2 for the data in Exercise 3.
10. Find the percentile rank of 20 for the data in Exercise 4.
11. Find the percentile rank of 8 for the data in Exercise 5.
12. Find the percentile rank of 6 for the data in Exercise 6.
In 13–18, find the mean and the median for each set of data to the nearest tenth.
13.
xi
fi
21–25
16–20
11–15
6–10
1–5
2
3
12
6
1
14.
xi
fi
91–100 5
81–90
8
71–80 10
61–70
6
51–60
0
41–50
1
15.
xi
fi
$1.51–$1.60
$1.41–$1.50
$1.31–$1.40
$1.21–$1.30
$1.11–$1.20
$1.01–$1.10
2
5
14
4
2
3
14411C15.pgs
8/14/08
10:37 AM
Page 613
Measures of Central Tendency for Grouped Data
16.
xi
fi
17–19
14–16
11–13
8–10
5–7
20
27
32
39
32
17.
18.
xi
fi
$60–$69
$50–$59
$40–$49
$30–$39
$20–$29
$10–$19
16
5
16
2
5
37
613
xi
fi
0.151–0.160
0.141–0.150
0.131–0.140
0.121–0.130
0.111–0.120
16
5
0
6
0
Applying Skills
19. The table shows the number of correct answers on a test consisting of 15 questions. Find the
mean, the median, and the mode for the number of correct answers.
Correct
Answers
6
7
8
9
10
11
12
13
14
15
Frequency
1
0
1
3
5
8
9
6
5
2
20. The ages of students in a calculus class at a high school
are shown in the table. Find the mean and median age.
Age
Frequency
19
18
17
16
15
2
8
9
1
1
21. Each time Mrs. Taggart fills the tank of her car, she estimates, from the number of miles
driven and the number of gallons of gasoline needed to fill the tank, the fuel efficiency of
her car, that is, the number of miles per gallon. The table shows the result of the last 20
times that she filled the car.
a. Find the mean and the median fuel efficiency (miles per gallon) for her car.
b. Find the percentile rank of 34 miles per gallon.
Miles per
Gallon
32
33
34
35
36
37
38
39
40
Frequency
1
3
2
5
3
3
2
0
1
22. The table shows the initial weights of people enrolled in a weight-loss program. Find the
mean and median weight.
Weight
Frequency
Weight
Frequency
191–200
201–210
211–220
221–230
231–240
241–250
1
1
5
7
10
12
251–260
261–270
271–280
281–290
291–300
13
16
8
5
2
14411C15.pgs
8/14/08
614
10:37 AM
Page 614
Statistics
23. In order to improve customer relations, an auto-insurance company surveyed 100 people to
determine the length of time needed to complete a report form following an auto accident.
The result of the survey is summarized in the following table showing the number of minutes needed to complete the form. Find the mean and median amount of time needed to
complete the form.
Minutes
Frequency
26–30
31–35
36–40
41–45
46–50
51–55
56–60
61–65
66–70
2
8
12
15
10
24
26
1
2
Hands-On Activity
Organize the data from the survey in the Hands-On Activity of Section 15-1 using intervals of five
seconds. Use the table to find the mean number of seconds. Compare this result with the mean
found using the individual data values.
15-4 MEASURES OF DISPERSION
The mean and the median of a set of data help us to describe a set of data.
However, the mean and the median do not always give us enough information
to draw meaningful conclusions about the data. For example, consider the following sets of data.
Ages of students on a
middle-school basketball team
Ages of students in a community
center tutoring program
11 12 12 13 13 13
13 13 14 14 14 14
6 8 9 9 10 13
13 15 17 18 19 19
The mean of both sets of data is 13 and the median of both sets of data is
13, but the two sets of data are quite different. We need a measure that indicates
how the individual data values are scattered or spread on either side of the
mean. A number that indicates the variation of the data values about the mean
is called a measure of dispersion.
Range
The simplest of the measures of dispersion is called the range. The range is the
difference between the highest value and the lowest value of a set of data. In the
sets of data given above, the range of ages of students on the middle-school basketball team is 14 2 11 or 3 and the range of the ages of the students in the community center tutoring program is 19 2 6 or 13. The difference in the ranges
indicates that the ages of the students on the basketball team are more closely
grouped about the mean than the ages of the students in the tutoring program.
14411C15.pgs
8/14/08
10:37 AM
Page 615
Measures of Dispersion
615
The range is dependent on only the largest data value and the smallest data
value. Therefore, the range can be very misleading. For example consider the
following sets of data:
Ages of members of the chess club: 11, 11, 11, 11, 15, 19, 19, 19, 19
Ages of the members of the math club: 11, 12, 13, 14, 15, 16, 17, 18, 19
For each of these sets of data, the mean is 15, the median is 15, and the range
is 8. But the sets of data are very different. The range often does not tell us critical information about a set of data.
Interquartile Range
Another measure of dispersion depends on the first and third quartiles of a distribution. The difference between the first and third quartile values is the
interquartile range. The interquartile range tells us the range of at least 50% of
the data. The largest and smallest values of a set of data are often not representative of the rest of the data. The interquartile range better represents the
spread of the data. It also gives us a measure for identifying extreme data values, that is, those that differ significantly from the rest of the data. For example,
consider the ages of the members of a book club.
21, 24, 25, 27, 28, 31, 35, 35, 37, 39, 40, 41, 69
↑
↑
↑
26
35
39.5
median
Q3
Q1
For these 13 data values, the median is the 7th value. For the six values
below the median, the first quartile is the average of the 3rd and 4th values from
27
the lower end: 25 1
5 26. The third quartile is the average of the 3rd and
2
40
4th values from the upper end: 39 1
5 39.5. The interquartile range of
2
the ages of the members of the book club is 39.5 2 26 or 13.5. The age of the
oldest member of the club differs significantly from the ages of the others. It is
more than 1.5 times the interquartile range above the upper quartile:
69 . 39.5 1 1.5(13.5)
We call this data value an outlier.
DEFINITION
An outlier is a data value that is greater than the upper quartile plus 1.5 times
the interquartile range or less than the lower quartile minus 1.5 times the
interquartile range.
When we draw the box-and-whisker plot for a set of data, the outlier is indicated by a ✸ and the whisker is drawn to the largest or smallest data value that
is not an outlier. The box-and-whisker plot for the ages of the members of the
book club is shown on page 616.
14411C15.pgs
8/14/08
616
10:37 AM
Page 616
Statistics
*
21
26
39.5 40
35
69
We can use the graphing calculator to graph a box-and-whisker plot with
outliers. From the STAT PLOT menu, choose the ı-▫-ı -- option. For example,
with the book club data entered into L1, the following keystrokes will graph a
box-and-whisker plot with outliers.
ENTER:
2nd
STAT PLOT
1
䉴
ENTER
䉲
䉴
ENTER
䉲
2nd
䉲
ALPHA
DISPLAY:
䉴
L1
1 ZOOM
9
EXAMPLE 1
The table shows the number of minutes, rounded to the nearest 5 minutes,
needed for each of 100 people to complete a survey.
a. Find the range and the interquartile range for this set of data.
b. Does this data set include outliers?
Minutes
30
35
40
45
50
55
60
65
70
85
Frequency
3
8
12
15
10
24
17
8
2
1
Cumulative
Frequency
3
11
23
38
48
72
89
97
99
100
Solution a. The range is the difference between the largest and smallest data value.
Range 5 85 2 30 5 55
To find the interquartile range, we must first find the median and the lower
and upper quartiles. Since there are 100 values, the median is the average of
the 50th and the 51st values. Both of these values lie in the interval 55.
Therefore, the median is 55. There are 50 values above the mean and 50 values below the mean. Of the lower 50 values, the lower quartile is the average of the 25th and 26th values. Both of these values lie in the interval 45.
The lower quartile is 45. The upper quartile is the average of the 25th and
the 26th from the upper end of the distribution (or the 75th and 76th from
the lower end). These values lie in the interval 60. The upper quartile is 60.
Interquartile range 5 60 2 45 5 15
14411C15.pgs
8/14/08
10:37 AM
Page 617
Measures of Dispersion
617
b. An outlier is a data value that is 1.5 times the interquartile range below the
first quartile or above the third quartile.
45 2 1.5(15) 5 22.5
60 1 1.5(15) 5 82.5
The data value 85 is an outlier.
Answers a. Range 5 55, interquartile range 5 15 b. The data value 85 is an outlier.
Exercises
Writing About Mathematics
1. In any set of data, is it always true that xi 5 i? For example, in a set of data with more than
three data values, does x4 5 4? Justify your answer.
2. In a set of data, Q1 5 12 and Q3 5 18. Is a data value equal to 2 an outlier? Explain why or
why not.
Developing Skills
In 3–6, find the range and the interquartile range for each set of data.
3. 3, 5, 7, 9, 11, 13, 15, 17, 19
4. 12, 12, 14, 14, 16, 18, 20, 22, 28, 34
5. 12, 17, 23, 31, 46, 54, 67, 76, 81, 93
6. 2, 14, 33, 34, 34, 34, 35, 36, 37, 37, 38, 40, 42
In 7–9, find the mean, median, range, and interquartile range for each set of data to the nearest
tenth.
7.
xi
fi
50
45
40
35
30
25
20
3
8
12
15
11
7
4
8.
xi
fi
10
9
8
7
6
5
4
3
2
2
4
6
9
3
3
2
0
1
9.
xi
fi
11
16
19
31
37
32
35
5
8
9
6
5
5
6
14411C15.pgs
8/14/08
618
10:37 AM
Page 618
Statistics
Applying Skills
10. The following data represents the yearly salaries, in thousands of dollars, of 10 basketball
players.
533 427 800 687 264 264 125 602 249 19,014
a. Find the mean and median salaries of the 10 players.
b. Which measure of central tendency is more representative of the data? Explain.
c. Find the outlier for the set of data.
d. Remove the outlier from the set of data and recalculate the mean and median salaries.
e. After removing the outlier from the set of data, is the mean more or less representative of the data?
11. The grades on a math test are shown in the stem-and-leaf
diagram to the right.
Stem
9
8
7
6
5
4
a. Find the mean grade.
b. Find the median grade.
c. Find the first and third quartiles.
d. Find the range.
Leaf
001269
0235557
88
46679
067
68
e. Find the interquartile range.
12. The ages of students in a Spanish class are shown in the table.
Find the range and the interquartile range.
Age
Frequency
19
18
17
16
15
1
8
8
6
2
13. The table shows the number of hours that 40 third graders reported studying a week. Find
the range and the interquartile range.
Hours
3
4
5
6
7
8
9
10
11
12
Frequency
2
1
3
3
5
8
8
5
4
1
14. The table shows the number of pounds lost during the first month by people enrolled in a
weight-loss program.
a. Find the range.
b. Find the interquartile range.
c. Which of the data values is an outlier?
Pounds Lost
1
2
3
4
5
6
7
8
9
11
15
Frequency
1
1
2
2
6
10
7
7
2
1
1
14411C15.pgs
8/14/08
10:37 AM
Page 619
Variance and Standard Deviation
619
15. The 14 students on the track team recorded the following number of seconds as their best
time for the 100-yard dash:
13.5 13.7 13.1 13.0 13.3 13.2 13.0
12.8 13.4 13.3 13.1 12.7 13.2 13.5
Find the range and the interquartile range.
16. The following data represent the waiting times, in minutes, at Post Office A and Post Office
B at noon for a period of several days.
A: 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 9, 10
B: 1, 2, 2, 3, 3, 5, 5, 6, 6, 7, 7, 8, 8, 9, 10
a. Find the range of each set of data. Are the ranges the same?
b. Graph the box-and-whisker plot of each set of data.
c. Find the interquartile range of each set of data.
d. If the data values are representative of the waiting times at each post office, which post
office should you go to at noon if you are in a hurry? Explain.
Hands-On Activity
Find the range and the interquartile range of the data from the survey in the Hands-On Activity of
Section 15-1 estimating the length of a minute. Does your data contain an outlier?
15-5 VARIANCE AND STANDARD DEVIATION
Variance
Let us consider a more significant measure of dispersion than either the
range or the interquartile range. Let xi represent a student’s grades on
eight tests.
– )2
x i 2 x–
(xi 2 x
Grade (xi)
95
92
88
87
86
82
80
78
8
9
6
2
1
0
24
26
28
8
81
36
4
1
0
16
36
64
8
a xi 5 688
– 2
a (xi 2 x ) 5 0
– 2
a (xi 2 x ) 5 238
i51
i51
i51
8
a xii
Mean 5
i51
8
5 688
8 5 86
The table shows the deviation,
or difference, of each grade from
the mean. Grades above the mean
are positive and grades below the
mean are negative. For any set of
data, the sum of these differences is
always 0.
14411C15.pgs
8/14/08
620
10:37 AM
Page 620
Statistics
In order to find a meaningful sum, we can use the squares of the differences
so that each value will be positive. The square of the deviation of each data value
from the mean is used to find another measure of dispersion called the variance.
To find the variance, find the sum of the squares of the deviations from the mean
and divide that sum by the number of data values. In symbols, the variance for a
set of data that represents the entire population is given by the formula:
n
Variance 5 n1 a (xi 2 x) 2
i51
For the set of data given above, the variance is 18 (238) or 29.75.
Note that since the square of the differences is involved, this method of finding a measure of dispersion gives greater weight to measures that are farther
from the mean.
EXAMPLE 1
A student received the following grades on five math tests: 84, 97, 92, 88, 79. Find
the variance for the set of grades of the five tests.
Solution The mean of this set of grades is:
84 1 97 1 92 1 88 1 79
5
5 440
5 5 88
xi
x i 2 x–
– )2
(xi 2 x
84
97
92
88
79
24
9
4
0
29
16
81
16
0
81
5
5
a xi 5 440
– 2
a (xi 2 x ) 5 194
i51
i51
– 2 5 A 1 B 194 5 384 5 38.8 Answer
Variance 5 n1 a (xi 2 x)
5
5
5
i51
When the data representing a population is listed in a frequency distribution table, we can use the following formula to find the variance:
n
2
a fi (xi 2 x)
Variance 5
i51
n
a fi
i51
14411C15.pgs
8/14/08
10:37 AM
Page 621
Variance and Standard Deviation
621
EXAMPLE 2
In a city, there are 50 math teachers who are under the age of 30. The table
below shows the number of years of experience of these teachers. Find the variance of this set of data. Let xi represent the number of years of experience and
fi represent the frequency for that number of years.
xi
0
1
2
3
4
5
6
7
8
fi
1
2
7
8
6
9
8
4
5
50
xifi
0
2
14
24
24
45
48
28
40
225
Solution
8
a xifi
Mean 5
i50
8
a fi
5 225
50 5 4.5
i50
For this set of data, the data value is equal to i for each xi.
The table below shows the deviation from the mean, the square of the deviation from the mean, and the square of the deviation from the mean multiplied
by the frequency.
xi
fi
(xi 2 x– )
(xi 2 x– ) 2
fi(xi 2 x– ) 2
8
7
6
5
4
3
2
1
0
5
4
8
9
6
8
7
2
1
3.5
2.5
1.5
0.5
20.5
21.5
22.5
23.5
24.5
12.25
6.25
2.25
0.25
0.25
2.25
6.25
12.25
20.25
61.25
25.00
18.00
2.25
1.50
18.00
43.75
24.50
20.25
8
8
a fi 5 50
– 2
a fi (xi 2 x ) 5 214.50
i51
i50
8
a
Variance 5
fi (xi 2 x–) 2
i50
8
a fi
5 214.50
50 5 4.29 Answer
i50
Note that the data given in this example represents a population, that is, data for
all of the teachers under consideration.
14411C15.pgs
8/14/08
622
10:37 AM
Page 622
Statistics
Standard Deviation Based on the Population
Although the variance is a useful measure of dispersion, it is in square units. For
example, if the data were a set of measures in centimeters, the variance would
be in square centimeters. In order to have a measure that is in the same unit of
measure as the given data, we find the square root of the variance. The square
root of the variance is called the standard deviation. When the data represents
a population, that is, all members of the group being studied:
n
Standard deviation based on a population 5
É
1
n a fi (xi
2 x) 2
i51
If the data is grouped in terms of the frequency of a given value:
n
2
a fi (xi 2 x)
i51
Standard deviation based on a population 5
é
n
a
fi
i51
The symbol for the standard deviation for a set of data that represents a
population is s (lowercase Greek sigma). Many calculators use the symbol sx.
EXAMPLE 3
Find the standard deviation for the number of years of experience for the 50
teachers given in Example 2.
Solution The standard deviation is the square root of the variance.
Therefore:
Standard deviation 5 !4.29 5 2.071231518
Calculator (1) Clear lists 1 and 2.
Solution (2) Enter the number of years of experience in L and the frequency in L .
1
2
(3) Locate the standard deviation for a population (sx) under 1-Var Stats.
ENTER:
STAT
䉴
2nd
L1
2nd
L2
ENTER
DISPLAY:
,
ENTER
>
1–Var Stats
–
x=4.5
x=225
x2=1227
Sx=2.092259788
x=2.071231518
n=50
(4) sx 5 2.071231518
Answer The standard deviation is approximately 2.07.
14411C15.pgs
8/14/08
10:37 AM
Page 623
Variance and Standard Deviation
623
Standard Deviation Based on a Sample
When the given data is information obtained from a sample of the population,
the formula for standard deviation is obtained by dividing the sum of the squares
of the deviation from the mean by 1 less than the number of data values.
If each data value is listed separately:
n
Standard deviation for a sample 5
É
1
n 2 1 a (xi
2 x–) 2
i51
If the data is grouped in terms of the frequency of a given value:
n
2
a fi (xi 2 x)
Standard deviation for a sample 5
i51
é
n
a a fi b 2 1
i51
The symbol for the standard deviation for a set of data that represents a
sample is s. Many calculators use the symbol sx.
EXAMPLE 4
From a high school, ten students are chosen at random to report their number
of online friends. The data is as follows: 15, 13, 12, 10, 9, 7, 5, 4, 3, and 2.
Solution The total number of online friends for these 10 students is 80 or a mean of 8
online friends (x– 5 8).
Online Friends (xi)
x i 2 x–
– )2
(xi 2 x
15
13
12
10
9
7
5
4
3
2
7
5
4
2
1
21
23
24
25
26
49
25
16
4
1
1
9
16
25
36
10
– 2
a (xi 2 x ) 5 182
i51
10
Standard deviation 5
É
1
10 2 1 a (xi
i51
2 x) 2 5 #19 (182) 4.5 Answer
14411C15.pgs
8/14/08
624
10:37 AM
Page 624
Statistics
EXAMPLE 5
In each of the following, tell whether the population or sample standard deviation should be used.
a. In a study of the land areas of the states of the United States, the area of
each of the 50 states is used.
b. In a study of the heights of high school students in a school of 1,200 students, the heights of 100 students chosen at random were recorded.
c. In a study of the heights of high school students in the United States, the
heights of 100 students from each of the 50 states were recorded.
Solution a. Use the population standard deviation since every state is included. Answer
b. Use the sample standard deviation since only a portion of the total school
population was included in the study. Answer
c. Use the sample standard deviation since only a portion of the total school
population was included in the study. Answer
EXAMPLE 6
A telephone survey conducted in Monroe County obtained information about
the size of the households. Telephone numbers were selected at random until a
sample of 130 responses were obtained. The frequency chart at the right shows
the result of the survey.
No. of People
per Household
Frequency
1
2
3
4
5
6
7
8
9
28
37
45
8
7
3
1
0
1
Solution The table below can be used to find the mean and the standard deviation.
xi
fi
fixi
(xi 2 x– )
(xi 2 x– ) 2
fi(xi 2 x– ) 2
1
2
3
4
5
6
7
8
9
28
37
45
8
7
3
1
0
1
28
74
135
32
35
18
7
0
9
21.6
20.6
0.4
1.4
2.4
3.4
4.4
5.4
6.4
2.56
0.36
0.16
1.96
5.76
11.56
19.36
29.16
40.96
71.68
13.32
7.20
15.68
40.32
34.68
19.36
0
40.96
9
9
9
a fi 5 130
a fi xi 5 338
– 2
a fi (xi 2 x ) 5 243.20
i51
i51
i51
14411C15.pgs
8/14/08
10:37 AM
Page 625
Variance and Standard Deviation
625
9
a fixi
Mean 5
i51
9
a fi
5 338
130 5 2.6
i51
9
a
Standard deviation 5
fi (xi 2 x) 2
i51
9
a fi b 2 1
è ia
51
243.20
243.20
5 #130
2 1 5 # 129 5 1.373051826
Calculator Enter the number of members of the households in L1 and the frequency for
Solution each data value in L2.
ENTER:
䉴
STAT
L1
,
ENTER
2nd
L2
2nd
ENTER
DISPLAY:
>
1–Var Stats
–
x=2.6
x=338
x2=1122
Sx=1.373051826
x=1.367760663
n=130
The standard deviation based on the data from a sample is sx 5 1.373051826
or approximately 1.37. Answer
Exercises
Writing About Mathematics
1. The sets of data for two different statistical studies are identical. The first set of data represents the data for all of the cases being studied and the second represents the data for a
sample of the cases being studied. Which set of data has the larger standard deviation?
Explain your answer.
2. Elaine said that the variance is the square of the standard deviation. Do you agree with
Elaine? Explain why or why not.
Developing Skills
In 3–9, the given values represent data for a population. Find the variance and the standard deviation for each set of data.
3. 9, 9, 10, 11, 5, 10, 12, 9, 10, 12, 6, 11, 11, 11
4. 11, 6, 7, 13, 5, 8, 7, 10, 9, 11, 13, 12, 9, 16, 10
5. 20, 19, 20, 17, 18, 19, 42, 41, 41, 39, 39, 40
6. 20, 101, 48, 25, 63, 31, 20, 50, 16, 14, 245, 9
14411C15.pgs
8/14/08
626
7.
10:37 AM
Page 626
Statistics
xi
fi
30
35
40
45
50
55
60
1
7
10
9
11
8
6
8.
xi
fi
2
4
6
8
10
12
14
16
0
1
6
10
13
21
7
1
9.
xi
fi
20
25
30
35
40
45
50
55
60
21
10
1
2
2
4
5
3
7
In 10–16, the given values represent data for a sample. Find the variance and the standard deviation
based on this sample.
10. 6, 4, 9, 11, 4, 3, 22, 3, 7, 10
11. 12.1, 33.3, 45.5, 60.1, 94.2, 22.2
12. 15, 10, 16, 19, 10, 19, 14, 17
13. 1, 3, 5, 22, 30, 45, 50, 55, 60, 70
14.
xi
fi
55
50
45
40
35
30
25
11
15
4
1
14
12
4
15.
xi
fi
33
34
35
36
37
38
39
3
1
4
6
5
11
6
16.
xi
fi
1
2
3
4
5
6
7
3
3
3
3
3
3
3
Applying Skills
17. To commute to the high school in which Mr. Fedora teaches, he can take either the Line A
or the Line B train. Both train stations are the same distance from his house and both stations report that, on average, they run 10 minutes late from the scheduled arrival time.
However, the standard deviation for Line A is 1 minute and the standard deviation for Line
B is 5 minutes. To arrive at approximately the same time on a regular basis, which train line
should Mr. Fedora use? Explain.
Stem
Leaf
18. A hospital conducts a study to determine if nurses need extra
staffing at night. A random sample of 25 nights was used. The
number of calls to the nurses’ station each night is shown in
the stem-and-leaf diagram to the right.
a. Find the variance.
b. Find the standard deviation.
9
8
7
6
5
4
0234466
68
01234
446789
037
19
14411C15.pgs
8/14/08
10:37 AM
Page 627
Variance and Standard Deviation
19. The ages of all of the students in a science class are
shown in the table. Find the variance and the
standard deviation.
Age
Frequency
18
17
16
15
1
2
9
9
627
20. The table shows the number of correct answers on a test consisting of 15 questions. The
table represents correct answers for a sample of the students who took the test. Find the
standard deviation based on this sample.
Correct
Answers
6
7
8
9
10
11
12
13
14
15
Frequency
2
1
3
3
5
8
8
5
4
1
21. The table shows the number of robberies during a given month in 40 different towns of a
state. Find the standard deviation based on this sample
Robberies
0
1
2
3
4
5
6
7
8
9
10
Frequency
1
1
1
2
2
6
10
7
7
2
1
22. Products often come with registration forms. One of the questions usually found on the registration form is household income. For a given product, the data below represents a random sample of the income (in thousands of dollars) reported on the registration form. Find
the standard deviation based on this sample.
38 40 26 42 39 25 40 40 39 36
46 41 43 47 49 43 39 35 43 37
Hands-On Activity
The people in your survey from the Hands-On Activity of Section 15-1 represent a random sample
of all people. Find the standard deviation based on your sample.
14411C15.pgs
8/14/08
628
10:37 AM
Page 628
Statistics
15-6 NORMAL DISTRIBUTION
The Normal Curve
....
......
Imagine that we were able to determine
......
........
........
..........
the height in centimeters of all 10-year..........
..........
............
............
old children in the United States. With a
.............
..............
..............
...............
scale along the horizontal axis that
................
................
..................
...................
includes all of these heights, we will place
....................
.....................
......................
........................
a dot above each height for each child of
.........................
.
...........................
.
.............................
. ................................
that height. For example, we will place
. ..................................
........................................
..........................................
above 140 on the horizontal scale a dot
for each child who is 140 centimeters tall. 123 128 133 138 143 148 153
Do this for 139, 138, 137, and so on for
Height in centimeters
each height in our data. The result would
be a type of frequency histogram. If we
draw a smooth curve joining the top dot for each height, we will draw a bellshaped curve called the normal curve. As the average height of 10-year-olds is
approximately 138 centimeters, the data values are concentrated at 138 centimeters and the normal curve has a peak at 138 centimeters. Since for each height
that is less than or greater than 138 centimeters there are fewer 10-year-olds, the
normal curve progressively gets shorter as you go farther from the mean.
Scientists have found that large sets of data that occur naturally such as
heights, weights, or shoe sizes have a bell-shaped or a normal curve. The highest
point of the normal curve is at the mean of the data. The normal curve is symmetric with respect to a vertical line through the mean of the distribution.
Standard Deviation and the Normal Curve
A normal distribution is a set of data that can be represented by a normal curve.
For a normal distribution, the following relationships exist.
1. The mean and the median of the
data values lie on the line of symmetry of the curve.
2. Approximately 68% of the data
values lie within one standard
deviation from the mean.
99.7% of data values
95% of data values
68% of
data values
3. Approximately 95% of the data
values lie within two standard
deviations from the mean.
4. Approximately 99.7% of the data
values lie within three standard
deviations from the mean.
13.5% 34% 34% 13.5%
x–23s x–22s x–2s x– x–1s x–12s x–13s
Normal distribution
14411C15.pgs
8/14/08
10:37 AM
Page 629
Normal Distribution
629
EXAMPLE 1
A set of data is normally distributed with a mean of 50 and a standard deviation
of 2.
a. What percent of the data values are less than 50?
b. What percent of the data values are between 48 and 52?
c. What percent of the data values are between 46 and 54?
d. What percent of the data values are less than or equal to 46?
Solution a. In a normal distribution, 50% of the
data values are to the left and 50% to
the right of the mean.
b. 48 and 52 are each 1 standard deviation away from the mean.
48 5 50 2 2
2.5%
13.5% 34% 34% 13.5%
52 5 50 1 2
46
48
50
50%
Therefore, 68% of the data values are
between 48 and 52.
52
54
50%
c. 46 and 53 are each 2 standard deviations away from the mean.
46 5 50 2 2(2)
54 5 50 1 2(2)
Therefore, 95% of the data values are between 48 and 52.
d. 50% of the data values are less than 50.
47.5% of the data values are more than 46 and less than 50.
Therefore, 50% 2 47.5% or 2.5% of the data values are less than or equal
to 46.
Answers a. 50% b. 68% c. 95% d. 2.5%
Z-Scores
The z-score for a data value is the deviation from the mean divided by the standard deviation. Let x be a data value of a normal distribution.
x2x
x2x
z-score 5 standard
deviation 5 s
The z-score of x, a value from a normal distribution, is positive when x is
above the mean and negative when x is below the mean. The z-score tells us how
many standard deviations x is above or below the mean.
14411C15.pgs
8/14/08
630
10:37 AM
Page 630
Statistics
1. The z-score of the mean is 0.
99.7%
95%
2. Of the data values, 34% have a
z-score between 0 and 1 and 34%
have a z-score between 21 and 0.
Therefore, 68% have a z-score
between 21 and 1.
68%
3. Of the data values, 13.5% have a
z-score between 1 and 2 and 13.5%
have a z-score between 22 and 21.
23
13.5% 34% 34% 13.5%
22 21
0
1
2
3
4. Of the data values, (34 1 13.5)% or
47.5% have a z-score between 0 and
2 and 47.5% have a z-score between 22 and 0. Therefore, 95% have a
z-score between 22 and 2.
5. Of the data values, 99.7% have a z-score between 23 and 3.
For example, the mean height of 10-year-old children is 138 centimeters
with a standard deviation of 5. Casey is 143 centimeters tall.
138
5 55 5 1
z-score for Casey 5 143 2
5
Casey’s height is 1 standard deviation above the mean. For a normal distribution, 34% of the data is between the mean and 1 standard deviation above the
mean and 50% of the data is below the mean. Therefore, Casey is as tall as or
taller than (34 1 50)% or 84% of 10-year-old children.
A calculator will give us this same answer.
The second entry of the DISTR menu is
normalcdf(. When we use this function, we must
supply a minimum value, a maximum value, the
mean, and the standard deviation separated by
commas:
>
DISTR DRAW
1: normalpdf(
2:normalcdf(
3:invNorm(
4:invT(
5:tpdf(
6:tcdf(
7 x2 p d f (
normalcdf(minimum, maximum, mean, standard deviation)
For the minimum value we can use 0. To find the proportion of 10-year-old
children whose height is 143 centimeters or less, use the following entries.
ENTER:
2nd
143
,
5
DISTR
)
,
ENTER
0
2
138
,
DISPLAY:
normalcdf(0,143,
138,5)
.8413447404
The calculator returns the number 0.8413447404, which can be rounded to 0.84
or 84%.
14411C15.pgs
8/14/08
10:37 AM
Page 631
Normal Distribution
631
If we wanted to find the proportion of the 10-year-old children who are
between 134.6 and 141.4 centimeters tall, we could make the following entry:
ENTER:
2nd
141.4
,
5
DISTR
)
,
2
138
134.6
DISPLAY:
,
ENTER
normalcdf(134.6,
141.4,138,5)
.5034956838
The calculator returns the number 0.5034956838, which can be rounded to 0.50
or 50%. Since 134.6 and 141.4 are equidistant from the mean, 25% of the data
is below the mean and 25% is above the mean. Therefore, for this distribution,
134.6 centimeters is the first quartile, 138 centimeters is the median or second
quartile, and 141.4 centimeters is the third quartile.
Note: 134.6 is 0.68 standard deviation
below the mean and 141.4 is 0.68 standard
deviation above the mean. For any normal
distribution, data values with z-scores of
20.68 are approximately equal to the first
quartile and data values with z-scores of
0.68 are approximately equal to the third
quartile.
25%
25%
20.68
25%
0
25%
0.68
EXAMPLE 1
On a standardized test, the test scores are normally distributed with a mean of
60 and a standard deviation of 6.
a. Of the data, 84% of the scores are at or below what score?
b. Of the data, 16% of the scores are at or below what score?
c. What is the z-score of a score of 48?
d. If 2,000 students took the test, how many would be expected to score at or
below 48?
Solution a. Since 50% scored at or below the mean and 34% scored within 1 standard
deviation above the mean, (50 1 34)% or 84% scored at or below 1 deviation above the mean:
x– 1 s 5 60 1 6 5 66
b. Since 50% scored at or below the mean and 34% scored within 1 standard
deviation below the mean, (50 2 34)% or 16% scored at or below 1 deviation below the mean:
x– 1 s 5 60 2 6 5 54
–
x
48 2 60
5 212
c. z-score 5 x 2
s 5
6
6 5 22
14411C15.pgs
8/14/08
632
10:37 AM
Page 632
Statistics
d. A test score of 48 has a z-score of 22. Since 47.5% of the scores are
between 22 and 0 and 50% of the scores are less than 0, 50% 2 47.5%
or 2.5% scored at or below 48.
2.5% 3 2,000 5 0.025 3 2,000 5 50 students
Answers a. 66 b. 54 c. 22 d. 50
EXAMPLE 2
For a normal distribution of weights, the mean weight is 160 pounds and a
weight of 186 pounds has a z-score of 2.
a. What is the standard deviation of the set of data?
b. What percent of the weights are between 155 and 165?
2 x–
s
186 2 160
s
Solution a. z-score 5 x
25
2s 5 26
s 5 13 Answer
b. ENTER: 2nd
155
13
DISTR
,
)
165
DISPLAY:
2
,
ENTER
160
,
normalcdf(155,16
5,160,13)
.2994775047
About 30% of the weights are between 155 and 165. Answer
Exercises
Writing About Mathematics
1. A student’s scores on five tests were 98, 97, 95, 93, and 67. Explain why this set of scores
does not represent a normal distribution.
2. If 34% of the data for a normal distribution lies between the mean and 1 standard deviation
above the mean, does 17% of the data lie between the mean and one-half standard deviation above the mean? Justify your answer.
Developing Skills
In 3–9, for a normal distribution, determine what percent of the data values are in each given range.
3. Between 1 standard deviation below the mean and 1 standard deviation above the mean
4. Between 1 standard deviation below the mean and 2 standard deviations above the mean
14411C15.pgs
8/14/08
10:37 AM
Page 633
Normal Distribution
633
5. Between 2 standard deviations below the mean and 1 standard deviation above the mean
6. Above 1 standard deviation below the mean
7. Below 1 standard deviation above the mean
8. Above the mean
9. Below the mean
10. A set of data is normally distributed with a mean of 40 and a standard deviation of 5. Find a
data value that is:
a. 1 standard deviation above the mean
b. 2.4 standard deviations above the mean
c. 1 standard deviation below the mean
d. 2.4 standard deviations below the mean
Applying Skills
In 11–14, select the numeral that precedes the choice that best completes the statement or answers
the question.
11. The playing life of a Euclid mp3 player is normally distributed with a mean of 30,000 hours
and a standard deviation of 500 hours. Matt’s mp3 player lasted for 31,500 hours. His mp3
player lasted longer than what percent of other Euclid mp3 players?
(1) 68%
(2) 95%
(3) 99.7%
(4) more than 99.8%
12. The scores of a test are normally distributed. If the mean is 50 and the standard deviation is
8, then a student who scored 38 had a z-score of
(1) 1.5
(2) 21.5
(3) 12
(4) 212
13. The heights of 10-year-old children are normally distributed with a mean of 138 centimeters
with a standard deviation of 5 centimeters. The height of a 10-year-old child who is as tall as
or taller than 95.6% of all 10-year-old children is
(1) between 138 and 140 cm.
(3) between 145 and 148 cm.
(2) between 140 and 145 cm.
(4) taller than 148 cm.
14. The heights of 200 women are normally distributed. The mean height is 170 centimeters
with a standard deviation of 10 centimeters. What is the best estimate of the number of
women in this group who are between 160 and 170 centimeters tall?
(1) 20
(2) 34
(3) 68
(4) 136
15. When coffee is packed by machine into 16-ounce cans, the amount can vary. The mean
weight is 16.1 ounces and the standard deviation is 0.04 ounce. The weight of the coffee
approximates a normal distribution.
a. What percent of the cans of coffee can be expected to contain less than 16 ounces of
coffee?
b. What percent of the cans of coffee can be expected to contain between 16.0 and 16.2
ounces of coffee?
8/14/08
634
10:37 AM
Page 634
Statistics
16. The length of time that it takes Ken to drive to work represents a normal distribution with a
mean of 25 minutes and a standard deviation of 4.5 minutes. If Ken allows 35 minutes to get
to work, what percent of the time can he expect to be late?
17. A librarian estimates that the average number of books checked out by a library patron is 4
with a standard deviation of 2 books. If the number of books checked out each day approximates a normal distribution, what percent of the library patrons checked out more than 7
books yesterday?
18. The heights of a group of women are normally distributed with a mean of 170 centimeters
and a standard deviation of 10 centimeters. What is the z-score of a member of the group
who is 165 centimeters tall?
19. The test grades for a standardized test are normally distributed with a mean of 50. A grade
of 60 represents a z-score of 1.25. What is the standard deviation of the data?
20. Nora scored 88 on a math test that had a mean of 80 and a standard deviation of 5. She also
scored 80 on a science test that had a mean of 70 and a standard deviation of 3. On which
test did Nora perform better compared with other students who took the tests?
15-7 BIVARIATE STATISTICS
Statistics are often used to compare two sets of data. For example, a pediatrician
may compare the height and weight of a child in order to monitor growth. Or
the owner of a gift shop may record the number of people who enter the store
with the revenue each day. Each of these sets of data is a pair of numbers and is
an example of bivariate statistics.
Representing bivariate sta550
tistics on a two-dimensional
graph or scatter plot can help
540
us to observe the relationship
530
between the variables. For example the mean value for the critical
520
reading and for the math sections
of the SAT examination for nine
510
schools in Ontario County are
500
listed in the table and shown on
the graph on the right.
510 520 530 540 550 560 570
Critical Reading
14411C15.pgs
Math
Math
530
551
521
522
537
511
516
537
566
Critical Reading
530
529
512
518
526
500
504
515
543
14411C15.pgs
8/14/08
10:37 AM
Page 635
Bivariate Statistics
635
The graph shows that there appears to be a linear relationship between the
critical reading scores and the math scores. As the math scores increase, the critical reading scores also increase. We say that there is a correlation between the
two scores. The points of the graph approximate a line.
These data can also be shown on a calculator. Enter the math scores as L1
and the corresponding critical reading scores as L2. Then turn on Plot 1 and use
ZoomStat from the ZOOM menu to construct a window that will include all
values of x and y:
ENTER:
2nd
STAT PLOT
䉲
ENTER
䉲
2nd
DISPLAY:
2nd
ENTER:
1
ZOOM
9
ENTER
L1
䉲
L2
DISPLAY:
Plot1 Plot2 Plot3
On
Off
.
Ty p e : ......
Xlist:L1
Xlist:L2
Mark:
+ .
The calculator will display a graph similar to
that shown above. To draw a line that approximates the data, use the regression line on the calculator. A regression line is a special line of best
fit that minimizes the square of the vertical distances to each data point. In this course, you do
not have to know the formula to find the regression line. The calculator can be used to determine
the regression line:
ENTER:
STAT
䉴
4
VARS
䉴
1
1
The calculator displays values for a and b for
the linear equation y = ax + b and stores the
regression equation into Y1 in the Y menu.
9
Press ZOOM
to display the scatter plot
and the line that best approximates the data. If
we round the given values of a and b to three decimal places, the linear regression equation is:
y 5 0.693x 1 151.013
LinReg
y=ax+b
a=.6925241158
b=151.0129957
r2=.7986740639
r=.8936856628
ENTER
14411C15.pgs
8/14/08
636
10:37 AM
Page 636
Statistics
We will study other data that can be approximated by a curve rather than a line
in later sections.
The scatter plots below show possible linear correlation between elements
of the pairs of bivariate data. The correlation is positive when the values of the
second element of the pairs tend to increase when the values of the first elements of the pairs increase. The correlation is negative when the values of the
second element of the pairs tend to decrease when the values of the first elements of the pairs increase.
Strong positive correlation
Moderate positive correlation
No linear correlation
Moderate negative correlation
Strong negative correlation
EXAMPLE 1
Jacob joined an exercise program to try to lose weight. Each month he records
the number of months in the program and his weight at the end of that month.
His record for the first twelve months is shown below:
Month
Weight
1
2
3
4
5
6
7
248 242 237 228 222 216 213
8
9
206
197
10
11
12
193 185 178
8/14/08
10:37 AM
Page 637
Bivariate Statistics
637
a. Draw a scatter plot and describe the correlation between the data (if any).
b. Draw a line that appears to represent the data.
c. Write an equation of a line that best represents the data.
Solution a. Use a scale of 0 to 13 along a horizontal line for the number of months in
the program and a scale from 170 to 250 along the vertical axis for his
weight at the end of that month. The scatter plot
250
is shown here. There appears to be a strong neg240
ative correlation between the number of months
230
and Jacob’s weight.
Weight
14411C15.pgs
220
210
200
190
180
170
b. The line on the graph that appears to approximate the data intersects the points (1, 248)
and (11, 185). We can use these two points to
write an equation of the line.
2 185
y 2 248 5 248
1 2 11 (x 2 1)
y 2 248 5 26.3x 1 6.3
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Month
y 5 26.3x 1 254.3
c. On a calculator, enter the number of the
month in L1 and the weight in L2. To use the
calculator to determine a line that approximates the data, use the following sequence:
䉴
ENTER: STAT
1
䉴
VARS
4
1
LinReg
y=ax+b
a=-6.283216783
b=254.5909091
r2=.9966845178
r=-.9983408825
ENTER
The calculator displays values for a and b for the linear equation y = ax 1 b
and stores the equation as Y1 in the Y menu. If we round the given values of a and b to three decimal places, the linear regression equation is:
y 5 26.283x 1 254.591
Turn on Plot 1, then use ZoomStat to graph
the data:
ENTER:
2nd
䉲
STAT PLOT
ENTER
ENTER
2nd
ZOOM
9
䉲
ENTER
2nd
L2
ENTER
L1
ENTER
The calculator will display the scatter plot of the data along with the
regression equation.
14411C15.pgs
8/14/08
638
10:37 AM
Page 638
Statistics
EXAMPLE 2
In order to assist travelers in planning a trip, a travel guide lists the average high
temperature of most major cities. The listing for Albany, New York is given in
the following table.
Month
1
2
3
4
5
6
7
8
9
10
11
12
Temperature
31
33
42
51
70
79
84
81
73
62
48
35
Can these data be represented by a regression line?
Solution Draw a scatter plot the data. The
graph shows that the data can be
characterized by a curve rather
than a line. Finding the regression
line for this data would not be
appropriate.
SUMMARY
• The slope of the regression line gives the direction of the correlation:
A positive slope shows a positive correlation.
A negative slope shows a negative correlation.
• The regression line is appropriate only for data that appears to be linearly
related. Do not calculate a regression line for data with a scatter plot
showing a non-linear relationship.
• The regression equation is sensitive to rounding. Round the coefficients to
at least three decimal places.
Exercises
Writing About Mathematics
1. Explain the difference between univariate and bivariate data and give an example of each.
2. What is the relationship between slope and correlation? Can slope be used to measure the
strength of a correlation? Explain.
Developing Skills
In 3–6, is the set of data to be collected univariate or bivariate?
3. The science and math grades of all students in a school
4. The weights of the 56 first-grade students in a school
14411C15.pgs
8/14/08
10:37 AM
Page 639
Bivariate Statistics
639
5. The weights and heights of the 56 first-grade students in a school
6. The number of siblings for each student in the first grade
In 7–10, look at the scatter plots and determine if the data sets have high linear correlation, moderate linear correlation, or no linear correlation.
7.
8.
9.
10.
Applying Skills
In 11–17: a. Draw a scatter plot. b. Does the data set show strong positive linear correlation, moderate positive linear correlation, no linear correlation, moderate negative linear correlation, or
strong negative linear correlation? c. If there is strong or moderate correlation, write the equation
of the regression line that approximates the data.
11. The following table shows the number of gallons of gasoline needed to fill the tank of a car
and the number of miles driven since the previous time the tank was filled.
Gallons
8.5
7.6
9.4
8.3 10.5 8.7
9.6
4.3
6.1
7.8
Miles
255 230 295 250 315 260 290 130
180
235
12. A business manager conducted a study to examine the relationship between number of ads
placed for each month and sales. The results are shown below where sales are in the thousands.
Number of Ads
10
12
14
Sales
20 26.5 32
16
18
34.8
40
20
22
24
26
28
30
47.2 49.1 56.9 57.9 65.8 66.4
13. Jack Sheehan looked through some of his favorite recipes to compare the number of calories per serving to the number of grams of fat. The table below shows the results.
Calories
310 210 260 330 290 320 245 293
Fat
11
5
11
12
14
16
7
10
220 260 350
8
8
15
14411C15.pgs
8/14/08
640
10:37 AM
Page 640
Statistics
14. Greg did a survey to support his theory that the size of a family is related to the size of the
family in which the mother of the family grew up. He asked 20 randomly selected people to
list the number of their siblings and the number of their mother’s siblings. Greg made the
following table.
Family
2
2
3
1
0
3
5
2
1
2
Mother’s Family
4
0
3
7
4
2
2
6
4
7
Family
3
6
0
2
1
4
1
0
4
1
Mother’s Family
5
4
6
1
0
2
1
3
3
2
15. When Marie bakes, it takes about five and a half minutes for the temperature of the oven to
reach 350°. One day, while waiting for the oven to heat, Marie recorded the temperature
every 20 seconds. Her record is shown below.
Seconds
0
20
40
60
80
100 120
140
160
Temperature
100 114 126 145 160 174 193
207
222
Seconds
180 200 220 240 260 280 300
320
340
Temperature
240 255 268 287 301 318 331
342
350
16. An insurance agent is studying the records of his insurance company looking for a relationship between age of a driver and the percentage of accidents due to speeding. The table
shown below summarizes the findings of the insurance agent.
Age
17
18
21
25
30
35
40
45
50
55
60
65
% of Speeding
Accidents
49
49
48
39
31
33
24
25
16
10
5
6
17. A sociologist is interested in the relationship between body weight and performance on the
SAT. A random sample of 10 high school students from across the country provided the following information:
Weight
Score
197
193
194
157
159
170
149
169
157
185
1,485 1,061 1,564 1,729 1,668 1,405 1,544 1,752 1,395 1,214
14411C15.pgs
8/14/08
10:37 AM
Page 641
Correlation Coefficient
641
15-8 CORRELATION COEFFICIENT
We would like to measure the strength of the linear relationship between the
variables in a set of bivariate data. The slope of the regression equation tells us
the direction of the relationship but it does not tell us the strength of the relationship. The number that we use to measure both the strength and direction of
the linear relationship is called the correlation coefficient, r. The value of the
correlation coefficient does not depend on the units of measurement. In more
advanced statistics courses, you will learn a formula to derive the correlation
coefficient. In this course, we can use the graphing calculator to calculate the
value of r.
EXAMPLE 1
The coach of the basketball team made the following table of attempted and
successful baskets for eight players.
Attempted
Baskets (xi)
10
12
12
13
14
15
17
19
Successful
Baskets (yi)
6
7
9
8
10
11
14
15
Find the value of the correlation coefficient.
Solution Enter the given xi in L1 and yi in L2.
Then choose LinReg(ax+b) from the
CALC STAT menu:
ENTER:
STAT
䉴
4
ENTER
LinReg
y=ax+b
a=1.066666667
b=-4.933333333
r2=.9481481481
r=.9737289911
The calculator will list both the regression equation and r, the correlation coefficient.
Answer r 5 0.97
Note: If the correlation coefficient does not appear on your calculator, enter
2nd
CATALOG
D , scroll down to DiagnosticOn, press ENTER , and
press ENTER again.
When the absolute value of the correlation coefficient is close to 1, the data
have a strong linear correlation. When the absolute value of the correlation
coefficient is close to 0, there is little or no linear correlation. Values between 0
and 1 indicate various degrees of positive moderate correlation and values
between 0 and 21 indicate various degrees of negative moderate correlation.
14411C15.pgs
8/14/08
642
10:37 AM
Page 642
Statistics
Properties of the Correlation Coefficient, r
21 r 1.The correlation coefficient is a number between 21 and 1.
When r 5 1 or 21, there is a perfect linear
relationship between the data values.
r 5 1
r 5 21
When r 5 0, no linear relationship exists
between the data values.
When |r| is close to 1, the data have a
strong linear relationship.Values between
0 and 1 indicate various degrees of
moderate correlation.
The sign of r matches the sign of the slope of
the regression line.
EXAMPLE 2
An automotive engineer is studying the fuel efficiency of a new prototype. From
a fleet of eight prototypes, he records the number of miles driven and the number of gallons of gasoline used for each trip.
Miles Driven
310
270
350
275
380
320
290
405
Gallons of
Gasoline Used
10.0
9.0
11.2
8.7
12.3
10.2
9.5
12.7
14411C15.pgs
8/14/08
10:37 AM
Page 643
Correlation Coefficient
643
a. Based on the context of the problem, do you think the correlation coefficient will be positive, negative, or close to 0?
b. Based on the scatter plot of the data, do you expect the correlation coefficient to be close to 21, 0, or 1?
c. Use a calculator to find the equation of the regression line and determine
the correlation coefficient.
Solution a. As gallons of gasoline used tend to increase with miles driven, we expect
the correlation coefficient to be positive.
Enter the given data as L1 and L2 on a calculator.
b.
c.
There appears to be a strong positive
correlation, so r will be close to 1.
y 5 0.030x 1 0.748, r 5 0.99
LinReg
y=ax+b
a=.0298533724
b=.7476539589
r2=.9879952255
r=.9939794895
EXAMPLE 3
The produce manager of a food store noted the relationship between the
amount the store charged for a pound of fresh broccoli and the number of
pounds sold in one week. His record for 11 weeks is shown in the following
table.
Cost per
Pound
Pounds
Purchased
$0.65 $0.85 $0.90 $1.00 $1.25 $1.50 $1.75 $1.99 $2.25 $2.50 $2.65
58
43
49
23
39
16
56
32
12
35
What conclusion could the product manager draw from this information?
11
14411C15.pgs
8/14/08
644
10:37 AM
Page 644
Statistics
Solution Enter the given data as L1 and L2 on the calculator, graph the scatter plot, and
find the value of the correlation coefficient.
LinReg
y=ax+b
a=-13.82881393
b=55.73638117
r2=.331571737
r=-.5758226611
From the scatter plot, we can see that there is a moderate negative correlation.
Since r 5 20.58, the product manager might conclude that a lower price does
explain some of the increase in sales but other factors also influence the number of sales.
A Warning About Cause-and-Effect
The correlation coefficient is a number that measures the strength of the linear
relationship between two data sets. However, simply because there appears to
be a strong linear correlation between two variables does not mean that one
causes the other. There may be other variables that are the cause of the
observed pattern. For example, consider a study on the population growth of a
city. Although a statistician may find a linear pattern over time, this does not
mean that time causes the population to grow. Other factors cause the city grow,
for example, a booming economy.
SUMMARY
• 21 r 1. The correlation coefficient is a number between 21 and 1.
• When r 5 1 or r 5 21, there is a perfect linear relationship between the
data values.
• When r 5 0, no linear relationship exists between the data values.
• When r is close to 1, the data have a strong linear relationship. Values
between 0 and 1 indicate various degrees of moderate correlation.
• The sign of r matches the sign of the slope of the regression line.
• A high correlation coefficient does not necessarily mean that one variable
causes the other.
14411C15.pgs
8/14/08
10:37 AM
Page 645
Correlation Coefficient
645
Exercises
Writing About Mathematics
1. Does a correlation coefficient of 21 indicate a lower degree of correlation than a correlation coefficient of 0? Explain why or why not.
2. If you keep a record of the temperature in degrees Fahrenheit and in degrees Celsius for a
month, what would you expect the correlation coefficient to be? Justify your answer.
Developing Skills
In 3–6, for each of the given scatter plots, determine whether the correlation coefficient would be
close to 21, 0, or 1.
3.
4.
5.
6.
In 7–14, for each of the given correlation coefficients, describe the linear correlation as strong positive, moderate positive, none, moderate negative, or strong negative.
7. r 5 0.9
11. r 5 1
8. r 5 21
9. r 5 20.1
12. r 5 20.5
10. r 5 0.3
13. r 5 0
14. r 5 20.95
Applying Skills
In 15–19: a. Draw a scatter plot for each data set. b. Based on the scatter plot, would the correlation
coefficient be close to 21, 0, or 1? Explain. c. Use a calculator to find the correlation coefficient for
each set of data.
15. The following table shows the number of gallons of gasoline needed to fill the tank of a car
and the number of miles driven since the previous time the tank was filled.
Gallons 12.5 3.4
Miles
7.9
9.0 15.7 7.0
5.1 11.9 13.0 10.7
392 137 249 308 504 204 182 377
407
304
16. A man on a weight-loss program tracks the number of pounds that he lost over the course
of 10 months. A negative number indicates that he actually gained weight for that month.
Month
No. of
Pounds Lost
1
2
3
4
5
6
7
8
9.2
9.1
4.8
4.5
2.8
1.8
1.2
0
9
10
0.8 22.6
14411C15.pgs
8/14/08
646
10:37 AM
Page 646
Statistics
17. An economist is studying the job market in a large city conducts of survey on the number of
jobs in a given neighborhood and the number of jobs paying $100,000 or more a year. A
sample of 10 randomly selected neighborhood yields the following data:
Total Number
of Jobs
24
28
17
39
32
21
39
39
24
29
No. of HighPaying Jobs
3
3
4
5
7
3
4
7
7
4
18. The table below shows the same-day forecast and the actual high temperature for the day
over the course of 18 days. The temperature is given in degrees Fahrenheit.
Same-Day Forecast
56
52
67
55
58
56
59
57
53
Actual Temperature
53
54
63
49
66
54
54
56
59
Same-Day Forecast
45
55
45
58
59
55
48
53
54
Actual Temperature
48
60
36
59
59
47
46
52
48
19. The table below shows the five-day forecast and the actual high temperature for the fifth
day over the course of 18 days. The temperature is given in degrees Fahrenheit.
Five-Day Forecast
56
52
67
55
58
56
59
57
53
Actual Temperature
50
50
84
54
57
40
70
79
48
Five-Day Forecast
45
55
45
58
59
55
48
53
54
Actual Temperature
40
61
40
70
46
75
49
46
88
20. a. In Exercises 18 and 19, if the forecasts were 100% accurate, what should the value of r
be?
b. Is the value of r for Exercise 18 greater than, equal to, or less than the value of r for
Exercise 19? Is this what you would expect? Explain.
8/14/08
10:37 AM
Page 647
Non-Linear Regression
647
15-9 NON-LINEAR REGRESSION
Not all bivariate data can be represented
by a linear function. Some data can be
better approximated by a curve. For
example, on the right is a scatter plot of
the file size of a computer program called
Super Type over the course of 6 different
versions. The relationship does not appear
to be linear. For this set of data, a linear
regression would not be appropriate.
There are a variety of non-linear
functions that can be applied to non-linear data. In a statistics course, you will
learn more rigorous methods of determining the regression model. In this
course, we will use the scatter plot of the
data to choose the regression model:
Regression to Use
Description of Scatter Plot
Exponential
• An exponential curve that
does not pass through (0, 0)
• y-intercept is positive
• Data constraint: y . 0
Logarithmic
• A logarithmic curve that
does not pass through (0, 0)
• y-intercept is positive or
negative
• Data constraint: x . 0
Power
• Positive half of power curve
passing through (0, 0)
• Data constraints: x . 0, y . 0
1,800
Size (megabytes)
14411C15.pgs
1,600
1,400
1,200
1,000
800
600
400
200
0
0
1
2
3 4
5
6
7
Version
Examples
(continued on next page)
14411C15.pgs
8/14/08
648
10:37 AM
Page 648
Statistics
Regression to Use
Description of Scatter Plot
Examples
Specific types of power regression:
Quadratic
• A quadratic curve
Cubic
• A cubic curve
The different non-linear regression models can be found in the STAT
CALC menu of the graphing calculator.
• 5:QuadReg is quadratic regression.
• 6:CubicReg is cubic regression.
• 9:LnReg is logarithmic regression.
• 0:ExpReg is exponential regression.
• A:PwrReg is power regression.
In the example of the file size of Super Type, the scatter plot appears to be
exponential or power. The table below shows the data of the scatter plot:
Version
Size (megabytes)
1
2
3
4
5
6
155
240
387
630
960
1,612
To find the exponential regression model, enter the data into L1 and L2.
Choose ExpReg from the STAT CALC menu:
ENTER:
STAT
䉴
0
VARS
䉴
1
1
ENTER
The calculator will display the regression equation and store the equation into
Y1 of the Y menu. To the nearest thousandth, the regression equation is
y 5 95.699(1.596x). Press ZOOM
sion equation.
9
to graph the scatter plot and the regres-
14411C15.pgs
8/14/08
10:37 AM
Page 649
Non-Linear Regression
649
ExpReg
y=a* b^x
a=95.69902108
b=1.595666689
r2=.9994598905
r=.9997299088
To find the power regression model, with the data in L1 and L2, choose
PwrReg from the STAT CALC menu:
ENTER:
STAT
䉴
ALPHA
䉴
VARS
A
1
ENTER
1
The calculator will display the regression equation and store the equation into
Y1 of the Y menu. To the nearest thousandth, the regression equation is
y 5 121.591x1.273. Press ZOOM
to graph the scatter plot and the regres-
9
sion equation.
PwrReg
y=a* x^b
a=121.5914377
b=1.273149963
r2=.9307655743
r=.9647619263
From the scatter plots, we see that the exponential regression equation is a
better fit for the data.
EXAMPLE 1
A stone is dropped from a height of 1,000 feet. The trajectory of the stone is
recorded by a high-speed video camera in intervals of half a second. The
recorded distance that the stone has fallen in the first 5 seconds in given below:
Seconds
1
1.5
2
2.5
3
3.5
4
4.5
Distance
16
23
63
105 149 191 260 321
a. Determine which regression model is most appropriate.
b. Find the regression equation. Round all values to the nearest thousandth.
14411C15.pgs
8/14/08
650
10:37 AM
Page 650
Statistics
Solution a. Draw a scatter plot of the data. The data
appears to approximate an exponential
function or a power function. Enter the
data in L1 and L2. Find and graph the exponential and power models. From the
displays, it appears that the power model is
the better fit.
exponential
power
b. To the nearest thousandth, the calculator will display the power equation
y = axb for a 5 13.619, b 5 2.122.
Answers a. Power regression
b. y 5 13.619(x2.122)
EXAMPLE 2
A pediatrician has the following table that lists the head circumferences for a
group of 12 baby girls from the same extended family. The circumference is
given in centimeters.
Age in Months
Circumference
2
2
5
4
1
17
11
14
7
11
10
19
36.8 37.2 38.6 38.2 35.9 40.4 39.7 39.9 39.2 41.1 39.3 40.5
a. Make a scatter plot of the data.
b. Choose what appears to be the curve that best fits the data.
c. Find the regression equation for this model.
14411C15.pgs
8/14/08
10:37 AM
Page 651
Non-Linear Regression
651
Solution a.
b. The data appears to approximate a log function. Answer
c. With the ages in L1 and the circumferences in
L2, choose LnReg from the STAT CALC
menu:
ENTER:
STAT
䉴
9
ENTER
` The equation of the regression equation, to
the nearest thousandth, is
LnReg
y=a+blnx
a=35.9381563
b=1.627161495
r2=.9281619959
r=.9634116441
y 5 35.938 1 1.627 ln x Answer
Exercises
Writing About Mathematics
1. At birth, the average circumference of a child’s head is 35 centimeters. If the pair (0, 35) is
added to the data in Example 2, the calculator returns an error message. Explain why.
2. Explain when the power function, y = axb, has only positive or only negative y-values and
when it has both positive and negative y-values.
Developing Skills
In 3–8, determine the regression model that appears to be appropriate for the data.
3.
4.
5.
14411C15.pgs
8/14/08
652
10:37 AM
Page 652
Statistics
6.
7.
8.
In 9–13: a. Create a scatter plot for the data. b. Determine which regression model is the most appropriate for the data. Justify your answer. c. Find the regression equation. Round the coefficient of the
regression equation to three decimal places.
9.
10.
11.
12.
x
4
7
3
8
6
5
6
3
9
4.5
y
10
7
15
9
5
6
6
14
14
8
x
–2.3 1.8 21.0 4.6
1.4
3.7
5.3
1.9
0.7
4.2
y
2.2 18.3 8.2
x
3.3
23.8 22.1
0.4
3.5
23.8 21.8 20.4
2.4
1.2
y
12.5 217.1 23.6
0.4
15.0 218.9 22.1 20.4
4.2
3.4
x
y
13.
1
2
63.7 15.3 43.0 89.5 22.7 12.1 54.7
3
4
5
6
7
8
9
10
27 25.6 24.8 24.2 23.8 23.4 23.1 22.8 22.6 22.4
x
1
2
3
4
5
6
7
8
9
10
y
2.7
2.3
2.0
1.7
1.5
1.2
1.1
0.9
0.8
0.7
Applying Skills
14. Mrs. Vroman bought $1,000 worth of shares in the Acme Growth Company. The table below
shows the value of the investment over 10 years.
Year
Value ($)
1
2
3
4
1,045 1,092 1,141 1,192
5
6
7
8
9
10
1,246
1,302
1,361
1,422
1,486
1,553
a. Find the exponential regression equation for the data with the coefficient and base
rounded to three decimal places.
b. Predict, to the nearest dollar, the value of the Vromans’ investment after 11 years.
14411C15.pgs
8/14/08
10:37 AM
Page 653
Non-Linear Regression
653
15. The growth chart below shows the average height in inches of a group of 100 children from
2 months to 36 months.
Month
2
Height in Inches
4
8
10
12
14
16
18
22.7 26.1 27.5 28.9 32.1 31.7 33.1 32.7 34.0
Month
20
Height in inches
6
22
24
26
28
30
32
34
36
34.4 34.6 36.0 34.6 35.2 36.6 35.6 37.2 37.6
a. Find the logarithmic regression equation for the data with the coefficients rounded to
three decimal places.
b. Predict, to the nearest tenth of an inch, the average height of a child at 38 months.
16. The orbital speed in kilometers per second and the distance from the sun in millions of kilometers of each of six planets is given in the table.
Planet
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Orbital Speed
34.8
29.6
23.9
12.9
9.6
6.6
Distance from the Sun
108.2
149.6
227.9
778.0
1,427
2,871
a. Find the regression equation that appears to be the best fit for the data with the coefficient rounded to three decimal places.
b. Neptune has an orbital speed of 5.45 km/sec and is 4,504 million kilometers from the
sun. Does the equation found for the six planets given in the table fit the data for
Neptune?
17. A mail order company has shipping boxes that have square bases and varying heights
from 1 to 5 feet. The relationship between the height of the box and the volume is shown
in the table.
Height (ft)
3
Volume (ft )
1
1.5
2
2.5
3
3.5
2
7
16
31
54
86
4
4.5
128 182
5
250
a. Create a scatter plot for the data. Let the horizontal axis represent the height of the
box and the vertical axis represent the volume.
b. Determine which regression model is most appropriate for the data. Justify your
answer.
c. Find the regression equation. Round the coefficient of the regression equation to three
decimal places.
14411C15.pgs
8/14/08
654
10:37 AM
Page 654
Statistics
18. In an office building the thermostats have six settings. The table below shows the average
temperature in degrees Fahrenheit for a month that each setting produced.
Setting
1
2
3
4
5
6
Temperature (°F)
61
64
66
67
69
70
a. Create a scatter plot for the data. Let the horizontal axis represent the setting and the
vertical axis represent temperature.
b. Find the equation of best fit using a power regression. Round the coefficient of the
regression equation to three decimal places.
19. The following table shows the speed in megahertz of Intel computer chips over the course
of 36 years. The time is given as the number of years since 1971.
Year
0
1
3
7
11
14
18
22
Speed 0.108
0.8
2
5
6
16
25
66
Year
24
26
28
29
31
34
35
36
Speed
200
300
500
1,500 1,700 3,200 2,900 3,000
One application of Moore’s Law is that the speed of a computer processor should double
approximately every two years. Use this information to determine the regression model.
Does Moore’s Law hold for Intel computer chips? Explain.
Hands-On Activity: Sine Regression
If we make a scatter plot of the following set of data on a graphing calculator, we may observe that
the data points appear to form a sine curve.
x
13
14
15
16
17
y 211.6 217.2 218.0 215.0 28.4
18
19
20
21
22
23
24
6.2
12.0
20.1
18.4
11.9
3.4
26.9
A sine function should be used to model the data. We
can use a graphing calculator to find the sinusoidal regression equation. Enter the x-values into L1 and the y-values
into L2. Then with the calculator in radian mode, choose
SinReg from the STAT CALC menu:
ENTER:
䉴
STAT
1
1
ALPHA
ENTER
C
VARS
䉴
25
20
15
10
5
0
25
210
215
220
10 12 14 16 18 20 22 24 26
14411C15.pgs
8/14/08
10:37 AM
Page 655
Interpolation and Extrapolation
655
To the nearest thousandth, the sinusoidal regression equation is:
y 5 19.251 sin (0.551x 1 2.871) 2 0.029
Press ZOOM
9
to graph the scatter plot and the regression equation.
Note: The sinusoidal regression model on the graphing calculator assumes that the x-values are
equally spaced and in increasing order. For arbitrary data, you need to give the calculator an estimate of the period. See your calculator manual for details.
The average high temperature of a city is recorded for 14 months. The table below shows this
data.
Month
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Temp (°F)
40
48
61 71
81
85
83
77
67
54
41
39
42
49
a. Create a scatter plot for the data.
b. Find the sinusoidal regression equation for the data with the coefficient and base rounded to
three decimal places.
c. Predict the average temperature of the city at 15 months. Round to the nearest degree.
d. Predict the average temperature of the city at 16 months. Round to the nearest degree.
15-10 INTERPOLATION AND EXTRAPOLATION
Data are usually found for specific values of one of the variables. Often we wish
to approximate values not included in the data.
Interpolation
The process of finding a function value between given values is called
interpolation.
EXAMPLE 1
Each time Jen fills the tank of her car, she records the number of gallons of gas
needed to fill the tank and the number of miles driven since the last time that
she filled the tank. Her record is shown in the table.
Gallons of Gas
7.5
8.8
5.3
9.0
8.1
4.7
6.9
8.3
Miles
240 280 170 290 260 150 220
270
a. If Jen needs 8.0 gallons the next time she fills the tank, to the nearest mile,
how many miles will she have driven?
b. If Jen has driven 200 miles, to the nearest tenth, how many gallons of gasoline can she expect to need?
14411C15.pgs
8/14/08
656
10:37 AM
Page 656
Statistics
Solution Graph the scatter plot of the data.
There appears to be positive linear
correlation.
With the data in L1 and L2, use the
calculator to find the regression
equation:
ENTER:
STAT
䉴
4
ENTER
When rounded to the nearest thousandth, the linear equation returned
is
LinReg
y=ax+b
a=32.36537919
b=-2.076402594
r2=.9988025622
r=.9994011018
y 5 32.365x 2 2.076
a. Substitute 8.0 for x in the equation given by the calculator.
y 5 32.365x 2 2.076
5 32.365(8.0) 2 2.076
5 256.844
Jen will have driven approximately 257 miles. Answer
b. Substitute 200 for y in the equation given by the calculator.
y 5 32.365x 2 2.076
200 5 32.365x 2 2.076
202.076 5 32.365x
6.244 x
Jen will need approximately 6.2 gallons of gasoline. Answer
Extrapolation
Often we want to use data collected about past events to predict the future. The
process of using pairs of values within a given range to approximate values outside of the given range of values is called extrapolation.
14411C15.pgs
8/14/08
10:37 AM
Page 657
Interpolation and Extrapolation
657
EXAMPLE 2
The following table shows the number of high school graduates in the U.S. in the
thousands from 1992 to 2004.
Year
1992 1993 1994
1995 1996 1997 1998
No. of Graduates 2,478 2,481 2,464 2,519 2,518 2,612 2,704
Year
1999 2000 2001
2002 2003 2004
No. of Graduates 2,759 2,833 2,848 2,906 3,016 3,081
a. Write a linear regression equation for this data.
b. If the number of high school graduates continued to grow at this rate, how
many graduates would there have been in 2006?
c. If the number of high school graduates continues to grow at this rate,
when is the number of high school graduates expected to exceed
3.5 million?
Solution a. Enter the year using the number of years
since 1990, that is, the difference between
the year and 1990, in L1. Enter the corresponding number of high school graduates
in L2. The regression equation is:
y 5 53.984x 1 2,277.286 Answer
b. Use the equation y 5 53.984x 1 2,277.286
and let x 5 16:
y 5 53.984(16) 1 2,277.286 3,141
If the increase continued at the same rate, the expected number of graduates in 2006 would have been approximately 3,141,000.
c. Use the equation y 5 53.984x 1 2,277.286 and let y 5 3,500:
3,500 5 53.984x 1 2,277.286
1,222.714 5 53.984x
22.650 x
If the rate of increase continues, the number of high school graduates can
be expected to exceed 3.5 million in the 23rd year after 1990 or in the year
2013.
14411C15.pgs
8/14/08
658
10:37 AM
Page 658
Statistics
Unlike interpolation, extrapolation is not usually accurate. Extrapolation is
valid provided we are sure that the regression model continues to hold outside
of the given range of values. Unfortunately, this is not usually the case. For
instance, consider the data given in Exercise 15 of Section 15-9.
The growth chart below shows the average height in inches of a group of 100
children from 2 months to 36 months.
Month
2
Height in Inches
4
6
8
10
12
14
16
18
22.7 26.1 27.5 28.9 32.1 31.7 33.1 32.7 34.0
Month
20
Height in inches
22
24
26
28
30
32
34
36
34.4 34.6 36.0 34.6 35.2 36.6 35.6 37.2 37.6
The data appears logarithmic. When the
coefficient and the exponent are rounded to
three decimal places, an equation that best fits
the data is y 5 19.165 1 5.026 ln x. If we use this
equation to find the height of child who is 16
years old (192 months), the result is approximately 45.6 inches or less than 4 feet. The average 16-year-old is taller than this. The chart is
intended to give average growth for very young
children and extrapolation beyond the given
range of ages leads to errors.
Exercises
Writing About Mathematics
1. Explain the difference between interpolation and extrapolation.
2. What are the possible sources of error when using extrapolation based on the line of best fit?
Developing Skills
In 3–5: a. Determine the appropriate linear regression model to use based on the scatter plot of the
given data. b. Find an approximate value for y for the given value of x. c. Find an approximate value
for x for the given value of y.
3. b. x 5 5.7 c. y 5 1.25
x
y
1
2
3
4
5
6
7
8
9
10
1.05 1.10 1.16 1.22 1.28 1.34 1.41 1.48 1.55 1.62
14411C15.pgs
8/14/08
10:37 AM
Page 659
Interpolation and Extrapolation
659
4. b. x 5 12 c. y 5 140
x
1
2
3
4
5
6
7
8
9
10
y
3.1
3.6
4.0
4.5
5.1
5.6
6.0
6.5
6.9
7.5
6
7
8
9
10
5. b. x 5 0.5 c. y 5 0.5
x
1
2
3
4
5
y
1
2.3
3.7
5.3
6.9
8.6 10.3 12.1 14.0 15.9
In 6–9: a. Determine the appropriate non-linear regression model to use based on the scatter plot
of the given data. b. Find an approximate value for y for the given value of x. c. Find an approximate
value for x for the given value of y.
6. b. x 5 1.4 c. y 5 1.50
x
y
1
2
3
4
5
6
7
8
9
10
0.80 1.09 1.25 1.37 1.46 1.54 1.60 1.66 1.71 1.75
7. a. x 5 12 b. y 5 80
x
y
23
26
13
14
11.6 33.3 52.5 43.5
20
17
29
18
18
17
4.0 18.7 84.0 8.0 12.4 11.5
8. a. x 5 10.5 b. y 5 100.0
x
–2.0 21.0 20.5 0.1
0.5
0.8
y
1.0
7.7
9.3 12.0 16.5 21.6 28.0
2.3
3.3
5.3
1.1
1.5
1.8
2.1
9. a. x 5 12 b. y 5 215
x
0.5
1.7
2.7
3.9
4.9
5.7
7.0
8.2
9.2
10.0
y
0.1
2.1
7.8
23.1
43.4
65.1 114.9 183.8 248.3 311.3
14411C15.pgs
8/14/08
660
10:37 AM
Page 660
Statistics
Applying Skills
In 10–12, determine the appropriate linear regression model to use based on the scatter plot of the
given data.
10. The following table represents the percentage of the Gross Domestic Product (GDP) that a
country spent on education.
Year
1960
1965
1970
1975
1980
1985
1990
1995
2000
2005
Percent
2.71
3.17
5.20
5.61
7.20
8.14
8.79
10.21
10.72
11.77
a. Estimate the percentage of the GDP spent on education in 1998.
b. Assuming this model continues to hold into the future, predict the percentage of the
GDP that will be spent on education in 2015.
11. The following chart gives the average time in seconds that a group of 10 F1 racing cars went
from zero to the given miles per hour.
Speed
75
100 125 150 175 200 225 250 275 300
Time
1.2
2.2
2.9
4.0
5.6
6.8
7.3
8.7
9.3
9.9
a. What was the average time it took the 10 racing cars to reach 180 miles per hour?
b. Estimate the average time it will take the 10 racing cars to reach 325 miles per hour.
12. The relationship between degrees Celsius and degrees Fahrenheit is shown in the table at
intervals of 10° Fahrenheit.
Celsius
0
10
20
30
40
50
60
70
80
90 100
Fahrenheit
32
50
68
86 104 122 140 158 176 194 212
a. Find the Fahrenheit temperature when the Celsius temperature is 25°.
b. Find the Celsius temperature when the Fahrenheit temperature is 24°.
13. The following table gives the number of compact cars produced in a country over the
course of several years.
Year
1981
1984
1987
1990
1993
1996
1999
2002
2005
2008
No. of Cars
100
168
471
603
124
1,780 1,768
4,195
6,680
10,910
a. Estimate the number of cars produced by the country in 2000 using an exponential
model.
b. Estimate the number of cars produced by the country in 1978 using the model from
part a.
14411C15.pgs
8/14/08
10:37 AM
Page 661
Interpolation and Extrapolation
661
14. In an office building the thermostats have six settings. The table below shows the average
temperature in degrees Fahrenheit for a month that each setting produced.
Setting
1
2
3
4
5
6
Temperature (°F)
61
64
66
67
69
70
Using a power model and assuming that it is possible to choose a setting between the given
settings:
a. what temperature would result from a setting halfway between 2 and 3?
b. where should the setting be placed to produce a temperature of 68 degrees?
In 15 and 16, determine the appropriate non-linear regression model to use based on the scatter plot
of the given data.
15. A mail order company has shipping boxes that have square bases and varying height from
1 to 5 feet. The relationship between the height of the box and the volume in cubic feet is
shown in the table.
Height
1
1.5
2
2.5
3
3.5
4
4.5
5
Volume
2
6.75
16
31.25
54
85.75
128
182.25
250
a. If the company introduces a box with a height of 1.25 feet, what would be the volume
to the nearest hundredth cubic foot?
b. If the company needs a box with a volume of at least 100 cubic feet, what would be the
smallest height to the nearest tenth of a foot?
c. If the company needs a box with a volume of 800 square feet, what would be the
height to the nearest foot?
16. Steve kept a record of the height of a tree that he planted. The heights are shown in the
table.
Age of Tree in Years
1
3
5
Height in Inches
7
12
15
7
9
11
13
16.5 17.8 19
20
a. Write an equation that best fits the data.
b. What was the height of the tree after 2 years?
c. If the height of the tree continues in this same pattern, how tall will the tree be after
20 years?