Download Their Measure

Their Measure
E Introduction We begin our study of trigonometry by discussing angles and two
methods of measuring them: degrees and radians. As we will see in Section 3.1, it is the
radian measure of an angle that enables us to define trigonometric functions on sets of
real numbers.
Terminal
side
Vertex
ial
side
(a) Two half-rays
V
Terminal
side
Initial
side
(b) Standard position
FIGURE 2.1.1 Initia and
termina sides of an ange
D AngLes An angle is formed by two half-rays, or half-lines, which have a common
endpoint, called the vertex. We designate one ray the initial side of the angle and the
other the terminal side. It is useful to consider the angle as having been formed by a rota
tion from the initial side to the terminal side as shown in FIGURE 2.1.1(a). An angle is said to
be in standard position if its vertex is placed at the origin of a rectangular coordinate
system with its initial side coinciding with the positive x-axis, as shown in Figure 2.1.1(b).
D Degree Measure The degree measure of an angle is based on the assignment of
360 degrees (written 3600) to the angle formed by one complete counterclockwise rota
tion, as shown in FIGURE 2.1.2. Other angles are then measured in terms of a 360° angle,
of a complete rotation. If the rotation is counter
with a 1° angle being formed by
clockwise, the measure will be positive; if clockwise, the measure is negative. For exam
ple, the angle in FIGURE 2.1.3(a) obtained by one-fourth of a complete counterclockwise
rotation will be
(360°)
90°.
=
Shown in Figure 2.1.3(b) is the angle formed by three-fourths of a complete clockwise
rotation. This angle has measure
(—360°)
=
—270°.
V
4
360°
(a) 900 angle
FIGURE 2.1.2
FIGURE 2.1.3
360 degrees
+ 3600
O—360,
Three
coterminal anges
FIGURE 2.1.4
90
(b) —270° angle
Positive measure in (a);
negative measure in (b)
Angle of
O Coterminat AngLes Comparison of Figure 2,1.3(a) with Figure 2.1.3(b) shows
that the terminal side of a 90° angle coincides with the terminal side of a —270° angle.
When two angles in standard position have the same terminal sides we say they are
coterminal. For example, the angles 0, 0 + 360°, and 0
360° shown in FIGURE 2.1.4
are coterminal. In fact, the addition of any integer multiple of 360° to a given angle
results in a coterminal angle. Conversely, any two coterminal angles have degree meas
ures that differ by an integer multiple of 360°.
—
Ancjtes and Coterminat Anqtes
For a 960° angle:
(a) Locate the terminal side and sketch the angle.
(b) Find a coterminal angle between 0° and 3 60°.
(c) Find a coterminal angle between —360° and 0°.
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
olution (a) We first determine how many full rotations are made in forming this
gle. Dividing 960 by 360 we obtain a quotient of 2 and a remainder of 240. Equivalently
e can write
960 = 2(360) + 240.
hus, this angle is formed by making two counterclockwise rotations before complet
of another rotation. As illustrated in FIGURE 2.1.5(a), the terminal side of 960°
ig
=
es in the third quadrant.
) Figure 2.1.5(b) shows that the angle 240° is coterminal with a 960° angle.
) Figure 2.1.5(c) shows that the angle 120° is coterminal with a 960° angle.
—
y
y
9600
240°
/
(a)
FIGURE 2.1.5
220:
(c)
(b)
Ang[es in (b) and (c) are coterminal with the angLe in (a)
I Minutes and Seconds With calculators it is convenient to represent fractions of
grees by decimals, such as 42.23°. Traditionally, however, fractions of degrees were
pressed in minutes and seconds, where
1°
=
60 minutes (written 60’)*
(1)
1’
=
60 seconds (written 60”).
(2)
rexample, an angle of 7 degrees, 30 minutes, and 5 seconds is expressed as 7°30’5”.
me calculators have a special DMS key for converting an angle given in decimal
grees to Degrees, Minutes, and Seconds (DMS notation), and vice versa. The fol
wing examples show how to perform these conversions by hand.
EXAMPLE 2
Usinq (1) and (2
onvert:
86.23° to degrees, minutes, and seconds,
) 17°47’13” to decimal notation.
i)
iilution In each case we will use (1) and (2).
i)
Since 0.23° represents
j
of 1° and 1°
86.23°
=
=
=
60 we have
86° + 0.23°
86° + (0.23)(60’)
86° + 13.8
ow 13.8’ = 13’ + 0.8’, so we must convert 0.8’ to seconds. Since 0.8’ represents
and l’= 60”, we have
of
86° + l3’+ 0.8’= 86° + 13’+ (0.8)(60”)
= 86° + 13’+ 48”.
ence, 86.23°
86°13’48”.
lie use of the number 60 as a base dates back to the Babylonians. Another example of the
e of this base in our culture in the measurement of time (1 hour = 60 minutes and
minute = 60 seconds).
2.1 Angles and Their Measure
91
(b) Since 1°
have
= 60’,itfollowsthat 1’= ()°.Simi1arly, I” = ()‘=
17°47’13”
Thuswe
= 17° + 47’ + 13”
= 17° + 47()° + l3()°
17° + 07833° + 00036°
= 17.7869°.
D Radian Measure Another measure for angles is radian measure, which is gen
erally used in almost all applications of trigonometry that involve calculus. The radian
measure of an angle 0 is based on the length of an arc on a circle. If we place the vertex
of the angle U at the center of a circle of radius r, then 0 is called a central angle. As we
know, an angle 0 in standard position can be viewed as having been formed by the ini
tial side rotating from the positive x-axis to the terminal side. The region formed by the
initial and terminal sides with a central angle 0 is called a sector of a circle. As shown
ifl FIGURE 2.1.6(a), if the initial side of 0 traverses a distances along the circumference of
the circle, then the radian measure of 0 is defined by
5
In the case when the terminal side of 0 traverses an arc of length s along the cir
cumference of the circle equal to the radius r of the circle, then we see from (3) that the
measure of the angle 0 is 1 radian. See Figure 2.1.6(b).
Terminal
side
Terminal
r/
side
/o
S
r
/ Initial
_____//
\
r
side
r
/ Initial
__,,i/ side
(a)
(b)
FIGURE 2.1.6 CentraL angLe in (a); angLe of 1 radian in (b)
The definition given in (3) does not depend on the size of the circle. To see this, all
we need do is to draw another circle centered at the vertex of 0 of radius r’and sub
tended arc length s See FIGURE 2.1.7. Because the two circular sectors are similar the
ratios s/r and s7r’ are equal. Therefore, regardless of which circle we use, we obtain the
same radian measure for 0.
In equation (3) any convenient unit of length maybe used for s and r, but the same
unit must be used for both s and r. Thus,
FIGURE 2.1.7
circLes
Concentric
0(in radians) =
s(units of length)
r(units of length)
appears to be a “dimensionless” quantity. For example, ifs = 6 in. and r = 2 in., then
the radian measure of the angle is
0 =
4 in.
—----
2 In.
2,
where 2 is simply a real number. This is the reason why sometimes the word radians is omit
ted when an angle is measured in radians. We will come back to this idea in Section 3.1.
92
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
One complete rotation of the initial side of 0 will traverse an arc equal in length to
the circumference of the circle 2irr. It follows from (3) that
one rotation
s
=
—
r
27rr
=
—
r
2 radians.
We have the same convention as before: An angle formed by a counterclockwise rota
tion is considered positive, whereas an angle formed by a clockwise rotation is nega
and 3
tive. In FIGURE 2.1.8 we illustrate angles in standard position ofr/2, —r/2,
radians, respectively. Note that the angle of ir/2 radians shown in (a) is obtained by
one-fourth of a complete counterclockwise rotation; that is
,
radians)
=
—
radians.
The angle shown in Figure 2.1.8(b), obtained by one-fourth of a complete clockwise rota
tion, is —ir/2 radians. The angle shown in Figure 2.1.8(c) is coterminal with the angle
shown in Figure 2.1.8(d). In general, the addition of any integer multiple of 2ir radians
to an angle measured in radians results in a coterminal angle. Conversely, any two coter
minal angles measured in radians will differ by an integer multiple of 2r.
y
y+
(a)
FIGURE 2.1.8
(b)
AngLes measured in radians
EXAMPLE 3
y
yt
(d)
(c)
A Coterminat AngLe
Find an angle between 0 and 2- radians that is coterminal with 0
Sketch the angle.
=
1 17r14 radians.
Solution Since 2r < 11 ‘w/4 < 3ir, we subtract the equivalent of one rotation, or 2
radians, to obtain
llr
1lr
8w
32r
3r
Alternatively, we can proceed as in part (a) of Example 1 and divide: llir!4 =
2r + 3ir/4. Thus, an angle of 31T/4 radians is coterminal with 0, as illustrated in
FIGURE 2.1.9.
CoterminaL
ang’es in ExampLe 3
FIGURE 2.1.9
0 Conversion FormuLas Although many scientific calculators have keys that con
vert between degree and radian measure, there is an easy way to remember the rela
tionship between the two measures. Since the circumference of a unit circle is 27r, one
complete rotation has measure 2 radians as well as 360°. It follows that 360° = 2r
radians or
180°
rradians.
(4)
If we interpret (4) as 180(10) = ir (1 radian), then we obtain the following two formulas
for converting between degree and radian measure.
2.1 AngLes and Their Measure
93
CONVERSION BETWEEN DEGREES AND RADIANS
10
(5)
—fl—radian
180
=
1 radian
(\0
(6)
\ITJ
Using a calculator to carry out the divisions in (5) and (6), we find that
10
0.0174533 radian
:EXAMPLE4
and
57.29578°.
1 radian
Conversion Between Degree and Radians
Convert:
(a) 20° to radians, (b) 7-/6 radians to degrees, (c) 2 radians to degrees.
Solution (a) To convert from degrees to radians we use (5):
20°
=
20(1°)
=
F’
20( —radian 1=—radian.
\\180
9
J
(b) To convert from radians to degrees we use (6):
1T
7
—radians
6
71T
=
—
6
.
(1 radian)
=
7irul8ON°
—I
j
—
6\rJ
(c) We again use (6):
2radians
TABLE 2.1.1
Degrees 0
Radians 0
=
2(lradian)
ul8ON°
2(
1
1360N°
=1—)
1
210°.
approximate answer
rounded to two
decimai places
114.59°.
Table 2.1.1 provides the radian and degree measure of the most commonly used
angles.
30 45 60 90 180
D TerminoLogy You may recall from geometry that a 90° angle is called a right
angle and a 180° angle is called a straight angle. In radian measure, r/2 is a right
angle and IT is a straight angle. An acute angle has measure between 0° and 90° (or
between 0 and ir/2 radians); and an obtuse angle has measure between 90° and 180°
(or between ir/2 and IT radians). Two acute angles are said to be complementary if
their sum is 90° (or ii-/2 radians). Two positive angles are supplementary if their sum
is 180° (or IT radians). The angle 180° (or radians) is a straight angle. An angle
whose terminal side coincides with a coordinate axis is called a quadrantal angle. For
example, 90° (or IT/2 radians) is a quadrantal angle. A triangle that contains a right
angle is called a right triangle. The lengths a, b, and c of the sides of a right triangle
satisfy the Pythagorean relationship a
2 + b
2 = c
, where c is the length of the side
2
opposite the right angle (the hypotenuse).
94
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
EXAMPLE 5
Complementary and Supplementary Angles
Find the angle that is complementary to 0
) Find the angle that is supplementary to q
i)
=
74.23°.
rr/3 radians.
olution (a) Since two angles are complementary if their sum is 90°, we find the angle
tat is complementary to 0
74.23° is
90°
—
0
=
90°
—
74.23°
15.77°.
=
) Since two angles are supplementary if their sum is
tat is supplementary to = /3 radians is
i.
3
7T
2r
ir
radians, we find the angle
radians.
I Arc Length In many applications it is necessary to find the lengths of the arc sub
nded by a central angle 0 in a circle of radius r See FIGURE 2.1.10. From the definition
f radian measure given in (3),
0 (in radians)
5
=
—.
r
y multiplying both sides of the last equation by r we obtain the arc length formula
rO. We summarize the result.
HEOREM 2.1.1
FIGURE 2.1.10
Length
of arcs determined
by a centra ange 0
Arc Length Formula
or a circle of radius r, a central angle of 0 radians subtends an arc of length
(7)
srO.
EXAMPLE 6
Finding Arc Length
md the arc length subtended by a central angle of(a) 2 radians in a circle of radius 6 inches,
) 30° in a circle of radius 12 feet.
6 inches, we
olution (a) From the arc length formula (7) with 0 = 2 radians and r
ave s = rO
2 6 = 12. So the arc length is 12 inches.
rI6 radians. Then from the
) We must first express 30° in radians. Recall that 30°
rc length formula (7) we have s = r0 = (12)(/6) = 2r. So the arc length is
6.28 feet.
Exercises
I Studenis ofien apply ihe arc length formula
incorrectly by using degree measure.
Remember .c = rO is valid only if 0 is
measured in radians.
Answers to selected odd-numbered problems
begin on page ANS-7.
Problems 1—16, draw the given angle in standard position. Bear in mind that the
ick of a degree symbol (°) in an angular measurement indicates that the angle is
teasured in radians.
i
1 60°
5. 1140°
2 —120°
6. —315°
3 135°
7. —240°
4 150°
8. —210°
2.1 Angtes and Their Measure
95
9.
13.
—
10.
11.
12.
14. —3-
15. 3
16. 4
—
In Problems 17—20, express the given angle in decimal notation.
17. 10°39’l7”
19. 5°10’
18. l43°7’2”
20. 10°25’
In Problems 2 1—24, express the given angle in terms of degrees, minutes, and seconds.
21. 210.78°
23. 30.810
22. 15.45°
24. 110.5°
In Problems 25—32, convert from degrees to radians.
25 10°
29. 270°
27 45°
31. —230°
26 15°
30. —120°
28 215°
32. 540°
In Problems 33-40, convert from radians to degrees.
33.
2r
9
—
34.
117r
35.
6
39. 3.1
38. 7r
37.
27r
3
—
36.
512
40. 12
In Problems 41—44, find the angle between 0° and 360° that is coterminal with the given
angle.
41. 875°
43. —610°
42. 400°
44. —150°
45. Find the angle between —360° and 0° that is coterminal with the angle in
Problem 41.
46. Find the angle between —360° and 0° that is coterminal with the angle in
Problem 43.
In Problems 47—52, find the angle between 0 and 2ir that is coterminal with the given
angle.
_2
50.
48.
17r
49. 5.3’w
52. 7.5
51. —4
—
53. Find the angle between —2ir and 0 radians that is coterminal with the angle in
Problem 47.
54. Find the angle between 2r and 0 radians that is coterrninal with the angle in
Problem 49.
—
In Problems 55—62, find an angle that is (a) complementary and (b) supplementary to
the given angle, or state why no such angle can be found.
96
55. 48.25°
56. 93°
57. 98.4°
58. 63.08°
59.
60.
61.
62.
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
63. Find both the degree and the radian measures of the angle formed by (a) threefifths of a counterclockwise rotation and (b) five and one-eighth clockwise
rotations.
64. Find both the degree and the radian measures of the obtuse angle formed by the
hands of a clock (a) at 8:00, (b) at 1:00, and (c) at 7:30.
65. Find both the degree and the radian measures of the angle through which the
hour hand on a clock rotates in 2 hours.
66. Answer the question in Problem 65 for the minute hand.
67. The Earth rotates on its axis once every 24 hours. How long does it take the
Earth to rotate through an angle of (a) 240°and (b) 7r16 radians?
68. The planet Mercury completes one rotation on its axis every 59 days. Through
what angle (measured in degrees) does it rotate in (a) 1 day, (b) 1 hour, and
(c) 1 minute?
69. Find the arc length subtended by a central angle of 3 radians in a circle of
(a) radius 3 and (b) radius 5.
70. Find the arc length subtended by a central angle of 30° in a circle of (a) radius 2
and (b) radius 4.
71. Find the measure of a central angle 0 in a circle of radius 5 if 0 subtends an arc of
length 7.5. Give 0 in (a) radians and (b) degrees.
72. Find the measure of a central angle 0 in a circle of radius 1 if 0 subtends an arc of
length ir/3. Give 0 in (a) radians and (b) degrees.
73. Show that the area A of a sector formed by a central angle of 0 radians in a circle
0. See Figure 2.1.10. [Hint: Use the proportional
2
of radius ris given byA = r
ity property from geometry that the ratio of the area A of a circular sector to the
2 of the circle equals the ratio of the central angle 0 to one complete
total area rr
revolution 2’ir.]
74. What is the area of the red shaded circular band shown in FIGURE 2.1.11 if 0 is measured
(a) in radians and (b) in degrees? [Hint: Use the result of Problem 73.]
Planet Mercury in
ProbLem 68
FIGURE 2.1.11
Circular band in
Problem 74
MisceLLaneous AppLications
75. Angular and Linear Speed If we divide (7) by time t we get the relationship
v = rw, where v = sit is called the linear speed of a point on the circumference
of a circle and w = 0,/t is called the angular speed of the point. A comrnunica
tions satellite is placed in a circular geosynchronous orbit 35,786 km above the
surface of the Earth. The time it takes the satellite to make one full revolution
around the Earth is 23 hours, 56 minutes, 4 seconds and the radius of the Earth is
Satellite
6378 km. See FIGURE 2.1.12.
(a) What is the angular speed of the satellite in rad/s?
(b) What is the linear speed of the satellite in kmis?
76. Pendulum Clock A clock pendulum is 1.3 m long and swings back and forth
along a 15-cm arc. Find (a) the central angle and (b) the area of the sector
through which the pendulum sweeps in one swing. [Hint: To answer part (b), use
the result of Problem 73.)
77. Sailing at Sea A nautical mile is defined as the arc length subtended on the sur
face of the Earth by an angle of measure 1 minute. If the diameter of the Earth is
7927 miles, find how many statute (land) miles there are in a nautical mile.
78. Circumference of the Earth Around 230 BCE Eratosthenes calculated the cir
cumference of the Earth from the following observations. At noon on the longest
day of the year, the Sun was directly overhead in Syene, while it was inclined
7.2° from the vertical in Alexandria. He believed the two cities to be on the same
longitudinal line and assumed that the rays of the Sun are parallel. Thus he con2.1
AngLes and Their Measure
FIGURE 2.1.12
Satellite in
Problem 75
97
Rays
.-
-
from
00 stades
-(
-
2
Sun
Syene
‘exandria
FIGURE 2.1.13
— -
Earth in
79.
Problem 78
80.
81.
Yo-yo in
Problems 79 and 80
82.
cluded that the arc from Syene to Alexandria was subtended by a central angle of
7.2° at the center of the Earth. See FIGURE 2.1.13. At that time the distance from
559 feet, find
Syene to Alexandria was measured as 5000 stades. If one stade
that
Show
miles.
(b)
the circumference of the Earth in (a) stades and
Eratosthenes’ data gives a result that is within 7% of the correct value if the polar
diameter of the Earth is 7900 miles (to the nearest mile).
Circular Motion of a Yo-Yo A yo-yo is whirled around in a circle at the end of
its 100-cm string.
(a) If it makes six revolutions in 4 seconds, find its rate of turning, or angular
speed, in radians per second.
(b) Find the speed at which the yo-yo travels in centimeters per second; that is
its linear speed.
More Yo-Yos If there is a knot in the yo-yo string described in Problem 79 at a
point 40 cm from the yo-yo, find (a) the angular speed of the knot and (b) the
linear speed.
Circular Motion of a Tire An automobile with 26-in, diameter tires is traveling
atarateof55 mi/h.
(a) Find the number of revolutions per minute that its tires are making.
(b) Find the angular speed of its tires in radians per minute.
Diameter of the Moon The average distance from the Earth to the Moon as
given by NASA is 238,855 miles. If the angle subtended by the Moon at the eye
of an observer on Earth is 0.52°, then what is the approximate diameter of the
Moon? FIGURE 2.1.14 is not to scale.
The curved red arc represents the
approximate diameter of the Moon
FIGURE 2.1.14
Right TriangLe Trigonometry
Introduction As we said in the introduction to this chapter, the word trigonotn
etry refers to the measurement of triangles. In this section we define the six trigonometric
functions—sine, cosine, tangent, cotangent, secant, and cosecant—as ratios of the
lengths of the sides of a right triangle. The names of these functions are abbreviated as
sin, cos, tan, cot. sec, and csc, respectively.
We begin with some terminology and a definition.
Hypotenuse
C
b
Side
opposite
Defining the trigono
function
of the angLe U
metric
O TerminoLogy In FIGURE 2.2.1 we have drawn a right triangle with sides labeled
a, b, and c (indicating their respective lengths) and one of the two acute angles denoted by
2 + b
. The side opposite the right
2
2 = c
0. From the Pythagorean theorem we know that a
angle is called the hypotenuse; the remaining sides are referred to as the legs of the tri
angle. The legs labeled a and b are, in turn, said to be the side adjacent to the angle 0
98
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
a
Side adjacent
FIGURE 2.2.1
I-
id the side opposite the angle 0. We will also use the abbreviations hyp, adj, and opp
denote the lengths of these sides.
We next define the six trigonometric functions of an acute angle.
)EFINITION 2.2.1
Trigonometric Functions
‘he trigonometric functions of an acute angle 0 in a right triangle are
opp
b
sinO=—=—
C
hyp
opp
b
tan0=—-=—
a
adj
hyp
c
sec0=——
a
adj
.
adj
a
cosO=—=—
C
hyp
adj
a
cot0=—=—
opp
b
hyp
c
csc0=—=—
opp
b
(1)
Domains The domain of each of these trigonometric functions is the set of all
•ute angles. In Section 2.4 we will extend these domains to include angles other than
ute angles. Then in Chapter 3 we will see how the trigonometric functions can be
fined with domains consisting of real numbers rather than angles.
The values of the six trigonometric functions depend only on the size of the angle
and not on the size of the right triangle. To see this, consider the two right triangles
own in FIGURE 2.2.2. Because the right triangles have the same acute angle 0 they are
nilar and thus the ratios of the corresponding sides are equal. For example, from the
d triangle in Figure 2.2.2(a) we have
•
sin 0
opp
=
hyp
C
b
a
(a)
lb
b
=
—,
C
[iereas in the smaller blue triangle in Figure 2.2.2(b) we find that
•
sin 0
opp
hyp
(b)
FIGURE 2.2.2 Similar triangles
b’
c
—,.
it since the red triangle is similar to the blue triangle we must have
b’
b
C’
C•
other words, we get the same value for sin 0 regardless of which right triangle we use
compute it. A similar argument can be made for the remaining five trigonometric
nctions.
EXAMPLE I
Values of the Trigonometric Functions
nd the exact values of the six trigonometric functions of the acute angle 0 in the right
angle shown in FIGURE 2.2.3.
ilution Matching Figure 2.2.3 with Figure 2.2.1 we see that the side opposite the
gleO has length 8 and the side adjacent toO has length 15; that is, b = 8 and a = 15.
om the Pythagorean theorem the hypotenuse c
hyp is
32
+ 152
=
289
and so
c
=
=
17.
2.2 Right Triangle Trigonometry
hyp
8
15
FIGURE 2.2.3
Right triangle in
Example 1
99
Thus from (1) the values of the six trigonometric functions are
15
adj
a
cosO=—=—=—,
C
17
hyp
adj
15
b
cotO=—=—=—,
opp
a
8
17
hyp
c
cscQ=—=—=—.
opp
8
b
b
opp
8
sinO=—=—=—,
C
17
hyp
a
8
opp
tanO__=—=—-—
15
b
adj
17
c
hyp
secO=—=—=—
a
15
adj
D Quotient and ReciprocaL Identities There are many important relationships
among the trigonometric functions. The basic ones listed next are referred to as the
fundamental identities and should be memorized.
Quotient Identities:
tanO
sinQ
cosO
=
cotO
,
=
cosQ
sinO
Reciprocal Identities:
secO
=
1
cosO
1
sinO
cscO
,
1
tanO
cotO
—--—,
Identities (2) and (3) can be obtained from Definition 2.2.1. For example, the first of the
quotient identities is verified as follows:
sinQ
cosO
=
opp/hyp
adj/hyp
=
opp
adj
=
tan ü
The others can be verified in a similar manner. Using these identities, we can find the
values of all six trigonometric functions once we know the values of sinO and cos 0.
EXAMPLE
Using (2) and (3)
2
and cos 0
Given sin 0
,
find the values of the remaining four trigonometric functions.
Solution From the fundamental identities, we have
tan0
=
secO
=
cscO
=
cot0
=
sinO
cosO
1
cos0
1
sinO
1
tan0
—
-
4
=
=
=
=
4
3
15
3
15
=
4
13
=
4
=
—
from (2)
—
—
4—from(3)
—
—
—.
Although we computed cot 0 using the reciprocal identity in (3), we could have also com
puted cot 0 using the quotient identity in (2).
EXAMPLE 3
Given cosO
=
Using a Right TriangLe
andtan0
=
2V’. Find sin 0.
Solution We can obtain sin 0 by multiplying the first identity in (2) by cos 0:
sin0
100
=
cos0tan0
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
1
=
—.
2V’
2V
=
The next example illustrates that ifjust one trigonometric function value of an acute
angle is known, it is possible to find the other five function values by drawing an
appropriate triangle.
EXAMPLE 4
Using a Right Triangle
[f 0 is an acute angle and sin 0
=
,
find the values of the other trigonometric functions
Solution We sketch a right triangle with an acute angle 0 satisfying sin 0 = by making
pp = 2 and hyp = 7 as shown in FIGURE 2.2.4. From the Pythagorean theorem we have
,
22 + (adj)
2
=
72
so that
2
(adj)
72
=
—
22
=
45.
adj
= 3\/5
fhus, adj =
The values of the remaining five trigonometric functions are obtained from the
iefinitions in (1):
cosO
adj
=
hyp
opp
3\/
7
2
tan0=—=
adj
3\/
hyp
7
secO=—=
3\/
adj
3\/
adj
cotO =
=
opp
2
hyp
7
=
opp
2
2V’
=
FIGURE 2.2.4
Example 3
7v
=
15
—
15
csc0
=
—.
] Cofunctions The use of the terminology sine and cosine, tangent and cotangent,
;ecant and cosecant is a result of the following observation. As shown in FIGURE 2.2.5, if
he two acute angles of a right triangle ABC are labeled a and /3 and a is the length of
he side opposite a, b is the length of the side opposite ,13, and c is the length of the side
)pposite the right angle, then by Definition 2.2.1,
a
sin a
Right triangle in
cos [3,
—
cos a
b
=
—
=
B
a
A
b
C
Acute angles
ci’and/3in a righttriangle
sin [3,
FIGURE 2.2.5
a
tana
=
seca
=
-
I,
=
cot/3,
cota
=
=
csc/3,
csca
=
—
=
tan/’3
=
secl3.
3ecause the sum of the angles in any triangle is 1800 (or ir radians), the acute angles a
md /3 in a right triangle are complementary. Thus the cosine of an acute angle equals
he sine of the complementary angle, the cotangent of an acute angle equals the tangent
)f the complementary angle, the cosecant of an acute angle equals the secant of the
:omplementary angle, and conversely. For this reason we say that sine and cosine,
angent and cotangent, and secant and cosecant are cofunctions of one another. We can
;ummarize the discussion in one simple sentence:
• Cofunctions of co,nplementaiy angles are equal.
] Cofunction
hen
(4)
Identities If a and /3 are the acute angles in the triangle in Figure 2.2.5,
or
2.2 Right Triangte Trigonometry
101
Because cos f3
=
sin c we obtain
sin(
13).
—
The last expression is one of six cofunction identities.
Cofunction Identities:
cosO
sin(
sinO
cosQ-
—
o)
tan(
cotO
tanO
—
=
cot(
—
—
cscO
=
sec(
secO
=
csc(
—
—
e)
o)
or equivalently
cosO
sinO
=
sin(90°
cos(90°
—
—
cotO
tanO
0)
0)
tan(90°
cot (90°
=
—
—
cscO
secO
0)
0)
=
sec(90°
csc(90°
—
—
0)
0).
(6)
In (5) and (6) it is understood that 0 is measured in radians and degrees, respectively.
Using (5) and (6)
EXAMPLE 5
From (5):
complementary angles
,,
(a) cot
=
tan(
—
(b) cos
sinQ
—
)
)
,
=
tan
=
sin.
From (6):
(c) csc27°
(d) cot 15°
=
sec(90°
tan(90°
—
—
27°)
15°)
=
=
sec63°
tan75°.
D Pythagorean Identities If just one trigonometric function value of an acute
angle is known, it is possible to find the values of the other five functions without using
the relationships in (1). Because the triangle in Figure 2.2.1 is a right triangle, the
Pythagorean theorem relates the lengths of the sides of the triangle by
2 + b
a
2
=
.
2
C
/c
b
/c + 2
a
, we obtain 2
2
If we divide this last result by c
=
I or
2
(bN
(aN
2
+1—I =1
\\cJ
\\cJ
(7)
I—I
2
2 + b
Similarly, if we divide a
=
2 we obtain, in turn,
2 and b
2 by a
c
(bV
1+1—
/
(C
\7
=
-
(8)
and
2
(aN
(CN
+1=1—
\b
\\b)
102
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
(9)
ig the appropriate definition in (1) in the results (7), (8), and (9) yield another set
nportant identities.
aagorean Identities:
O = 1
2
sin + cos
O
2
O = sec
2
1 + tan
O
2
O + 1 = csc
2
cot
O.
2
(10)
(11)
(12)
the square of the trigonometric functions are written (sin 0)2 =
), (10), and (11),
7
=
cos0, (tanO)
=
IXAMPLE 6
sin
O
2
,
tan 2 0, and so on.
Using (11)
is an acute angle and tan 0
=
\/3 find the value cos 0.
There are several ways of solving this problem. One way is to use the
agorean identity (11):
[tion
0
sec
0
2
+
1=5+1=6
+l=(V)
=tan
so secO = \/. Because secO
= 1/V = V/6.
1/cosO we have cosO
=
1/sec 0. Therefore
NOTES FROM THE CLASSROOM
we will see in Section 3.4, all the identities given in this section hold for any angle
not just acute angles).
Exercises
Answers to selected odd-numbered problems
begin on page ANS-7.
oblems 1—10, find the values of the six trigonometric functions of the angle Gin
iven triangle.
5
12
FIGURE 2.2.7
3
FIGURE 2.2.6
TriangLe for
TriangLe for
ProbLem 2
ProbLem 1
FIGURE 2.2.8
ProbLem 3
TnangLe for
FIGURE 2.2.9
TnangLe for
ProbLem 4
2.2 Right Triangle Trigonometry
103
>12
Triangle for
FIGURE 2.2.10
FIGURE 2.2.11
Problem 5
Triangle for
Problem 6
7.
1.2
FIGURE 2.2.12
TriangLe for
FIGURE 2.2.13
ProbLem 7
Triangle for
ProbLem 8
10.
9.
y
x
FIGURE 2.2.14
S
TriangLe for
FIGURE 2.2.15
Problem 9
Triangle for
Problem 10
In Problems 11—20, use the identities given in this section to find the values of the four
remaining trigonometric functions at the acute angle 0.
2
11. sinO
=
13. sin 0
=
15. sinO
=
17. csc0
=
19. cos0
=
,
2
cos0
cos 0
—,
1
,
5
—,
1
—,
3
3\/
=
tanO
sec0
cscO
3
=
1
8
=
—
5
=
—
3
2V
12. sinO
=
cos 0
=
14. sin0
=
cos 0
=
cotO
=
,
cot 0
=
7
,
cot 0
=
7
16. cos0
=
18. sin 0
=
20. sin 0
=
v
2
In Problems 2 1—28, find the value of the remaining trigonometric functions by drawing
an appropriate triangle.
21. sin0
23. sec 0
=
25. tan 0
=
27. sec 0
29.
30.
31.
32.
104
=
12
13
—
5
7
=
—
Ifcos75°
Ifcos75° =
If tan(ir/8) =
If tan(ir/8) =
—
—
—
—
22. cos0
=
24. csc 0
=
26. cot 0
=
28. tan 0
=
ViO
—
7
3
v’), find the exact value of sinl5°.
V’),findtheexactvalueofsec75°.
1, find the exactvalue of cot(3ir/8).
1,findtheexactvalueoftan(3’ff/8).
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
I.
In Problems 33-46, use the identities of this section to find the exact value of the given
trigonometric expression. Do not use a calculator.
33.
3sin-
+ 3cos
—
2
34.
+
sin35°
sin
5
2
5°
35. 1 + 2
sin 18°
cos
l
8° + 2
33°
2
36. 1 + tan
37.
38. 2
—4csc
1
13°
2
3° + 4cot
tan
—
sec2
—
sec
3
2
3°
sin 100
sin 100
sin 80°
cos 100
41. 5cot4l°cot49°
42. cosl1°sec11°
43. sin 28° cot 28° csc 62°
44. lOsin—cot—sec—
45. sin 10°cos 80° + cos 10°sin80°
46. tan 30°cot60°
40. sec20°
—
csc70°
—
.T
IT
3
3
IT
—
3
sec30°csc60°
[n Problems 47—54, given that cos 300 = V/2. Use the identities of this section to
lind the exact value of the given trigonometric function. Do not use a calculator.
47. sin 30°
48. cos 60°
50. cot 30°
52. csc 30°
54. tan 30° + cot 60°
49. tan 60°
51. sec 30°
53. cos 30°tan 30°
Trigonometric Functions of
SpeciaL AngLes
E Introduction The angles 30°(r/6 radian),45 °(/4 radian), and 60°(/3 radians)
ire considered to be special angles because they occur so often in the study of trigonom
try and its use in calculus. Thus, it is to your advantage to know the exact values of the
;ine and cosine of each of these angles. In the discussion that follows, we derive these
ialues using some results from plane geometry.
J VaLues of sin 45° and cos 45° To find the values of the sine and cosine functions
a 45° angle, we consider the isosceles right triangle with two equal sides of length 1
;hown in FIGURE 2.3.1. From plane geometry we know that the acute angles in this triangle
ire equal; therefore, each acute angle measures 45°. We can find the length of the
lypotenuse by using the Pythagorean theorem:
2
(hyp)
=
(1)2 + (1)2
=
2
gives
hyp=
hyp
Ehus from(1) of Section 2.2 we obtain
Isosceles
right triangle
FIGURE 2.3.1
sin45
opp
hyp
1
\/
V
2
cos45
adj
hyp
1
\/
2
(1)
(2)
2.3 Trigonometric Functions of Special Angles
105
D Values of sin 300 and cos 30° To find the values of the trigonometric functions
of 300 and 60° angles, we consider the equilateral triangle AOB with sides of length 2
shown in FIGURE 2.3.2(a). From plane geometry we know that the three angles of an
equilateral triangle each measure 60°. As shown in Figure 2.3.2(b), if we bisect the
angle at 0, then CO is the perpendicular bisector of AB. It follows that
AC
LAOB
LAOC
=
AB
(2)
=
=
1
i(60°)
=
and
=
30°,
LACO
=
90c.
A
A
2
2
2
2
B
B
(b)
(a)
FIGURE 2.3.2 Equiaterat triange in (a); two congruent right trianges in (b)
If we apply the Pythagorean theorem to the red right triangle ACO in Figure 2.3.2(b),
we get ()2 + 12 = 22. Solving for CO we get CO = \/. Therefore, from the right
triangle ACO and(l) of Section 2.2, we obtain the following values:
1
PP
2
hyp
adj
=
cos30° =
hyp
2
sin30 0
D VaLues of sin 60° and cos 60° Now using the 60° angle in the red right triangle
2.
ACO in Figure 2.3.2(b) we identify opp = V’, adj = 1, and hyp
Therefore
sin 60° =
cos60 0
hyp
adj
hyp
=
-
2
1
2
=—=—.
U Cofunctions We did not have to use a right triangle to derive the values in (5) and
(6). Recall, from Section 2.2 we showed that cofunctions of complementary angles are
equal. Thus, (5) and (6) follow immediately from the results in (3) and (4):
sin 60° = cos 30°
cos60°
106
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
sin 30°
——
—.
EXAMPLE 1
Values of the Other Trigonometric Functions
‘ind the values of tan(IT/6), cot(ir/6), sec(ir/6), and csc(ir/6).
;olution The angle 30° is equivalent to r/6 radian. Using the quotient and reciprocal
lentities in Section 2.2 along with the results in (3) and (4) we get
=
an
6
iT
cot—=
6
iT
sec-=
sin(r/6)
cos(/6)
1
tan(ir/6)
1
cos(ir/6)
1/2
—
—
=
1
—
/2
—
3
—
—
=-‘;
1/v’
1
2
2V”i
3
V/2
iT
csc—=
6
sin(ir/6)
1/2
We leave finding the values tan 0, cot 0, sec 0, and csc 0 for 0 = 7T/4 and 0 = ir/3
s exercises. See Problems 1 and 2 in Exercises 2.3.
Table 2.3.1 summarizes the values of the sine, cosine, and tangent functions that we
avejust determined for the special angles 30°, 45°, and 60°. As mentioned in the intro
uction to this section, these function values are used so frequently that we feel that
ey should be committed to memory. Knowing these values and the fundamental
Ientities we discussed earlier will enable you to determine any of the trigonometric
inctions for these special angles.
TABLE2.3.1
0
0
(degrees)
(radians)
sin 0
cosU
1
2
\/
2
iT
30°
—
6
—
\/
—
3
v/
iT
45°
tan0
1
—
4
2
2
1
IT
60°
3
EXAMPLE 2
Find the Exact Values
md the exact value of the given trigonometric expression.
i)
sin2
—
cos
(b) cos30°tan60°
iT
(c) 2 + 4sin—
3
iT
—
6cos—
6
lution In each case we will use the information in Table 2.3.1.
sin—
—
-r
cos—
) cos30° tan60°
=
—
—
±_LJ_ I
2422
=
2
=
)
2.3 Trigonometric Functions of Specia Anges
107
0 Use of a CalcuLator Approximations to the values of the trigonometric functions
can be obtained using a scientific calculator. But before using a calculator to find trigono
metric function values of angles measured in radians, you must set the calculator in
the radian mode. If the angles are measured in degrees, then you must select the degree
mode before making your calculations. Also, if the angles are given in degrees, minutes,
and seconds, they must be converted to decimal form first. Scientific calculators have
and 5j for computing the values of these functions. To obtain
keys labeled
keys with the reciprocal
or
the values of csc, sec, or cot, we can use the
The following example illustrates the process.
key
,
.
EXAMPLE 3
Using a CaLcuLator
Use a calculator to approximate each of the following.
(a) sin45°
(c) sec 0.23
(b) cos8°l5’
(d) cot
Solution (a) First we make sure that the calculator is set in degree mode. Then we
key to obtain
enter 45 and use the
sin 45°
0.707 1068,
which is a seven-decimal-place approximation to the exact value \//2 given in (1).
(b) Since the angle is given in degrees and minutes, we must first convert it to decimal
= 8.25°. Now with the calculator set in degree mode, we
form: 8°15’= 80 +
enter 8.25 and use the j key to obtain
()
cos 8°15’= cos 8.25°
0.98965 14.
(c) Since degrees are not indicated, we recognize that this angle is measured in radians.
To evaluate sec 0.23, we will use the fundamental identity sec 0 = 1/cos 0. With the
key, and then take the reciprocal
calculator set in radian mode, we enter 0.23, use the
key. Thus we have
of the result by pressing the
1
cos 0.23
sec0.23
1.0270458.
(d) We observe that this angle is measured in radians and set the calculator accordingly.
key to obtain
key, and then the
We first enter r, divide by 7, use the
cot—
7
I
Exercises
=
—
tan
2.0765214.
—
7
Answers to seected odd-numbered probLems
begin on page ANS-7.
In Problems I and 2, use the results of this section to find the values of tan 0, cot0,
sec 0, and csc 0 for the given angle.
1. 45°
108
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
2. ir/3
In Problems 3—22, find the exact value of the given trigonometric expression. Do not
use a calculator.
IT
cos
—
3. 2
3
5. sec45°csc45°
•IT
7. sin
—
cot
4. 2
tan
—
6
6. sin60°cos30°
IT
8. 6 sec
IT
—
csc
IT
—
10. tan 600 cot 30°
9. 9 sec 450 csc 450
11. sincos + cosjsin
12. coscos
13. 6 tan 30° + 7 tan 60°
14. 3 sin
15. tan45°
16. sec2 + 4csc2
4
3
2
\/sin(ir/4)
18
17
—
cot45°
8sin(Ir/4)
tan(ir/4)
sinsin
5 cos
—
sec(ir/3)
cos(r/4)
19. sin
3O° + cos
2
45°
2
21
—
—
—
tan(Ir/6)
1 + tan(w/4)tan(n-/6)
20. 2 + cot
3O°
2
22
—
3O°
2
lOcsc
tan(7T/3) + tan(ir/4)
1
—
tan(ir/3)tan(w/4)
In Problems 23—32, use a calculator to find the approximate values of the six trigonometric
functions of the given angle. Round your answer to four decimal places.
23 17°
25 143°
27. 71°30’15”
24 82°
26 3475°
28. 46°15’8”
29.
30.
31. 0.6725
32. 1.24
-
For Discussion
33. Without a calculator, find the exact value of the product
2w
3ir
tan— tan— tan—
180
180
180
89w
tan—.
180
Trigonometric Functions of
GeneraL Angles
Introduction Until now we have defined the trigonometric functions only for
acute angles. However, many applications of trigonometry involve angles that are not
acute. Consequently, it is necessary to extend the definition of the six trigonometric
functions in (1) of Section 2.2 to general angles. Naturally, we want the extended def
inition to agree with the earlier definition whenever the angle is acute. To accomplish
this we proceed in the following manner.
2.4 Trigonometric Functions of Generat Anges
109
y
P(x, Y)
r
Let 0 be an acute angle in standard position_and choose a point P(x, y) on the ter
Vx2 + y
,weseeinFIGIJPE2.4.1 thatx,y,and
2
minalsideofO.Ifweletr = d(O,P)
hyp,
r are the lengths of the sides of a right triangle. With y = opp, x = adj, and r
2.2.1,
andwehavefromDefinition
sinO
FIGURE 2.4.1
An acute ange
and
cosO
=
tan0
(I)
.
The expressions in (1) provide us with a model on which to base our extended defini
tion for any angle 0 in standard position, such as those illustrated in FIGURE 2.4.2.
y
y
P(x, y)
x
x
r
x
y
P(x, y)
P(x, y)
(b)
(a)
‘*.
(c)
FIGURE 2.4.2 Ang’es that are not acute
We now have the following definition of the trigonometric functions of a gen
eral angle.
DEFINITION 2.4.1
Trigonometric Functions
Let 0 be any angle in standard position,_and let P(x, y) be any point other than (0, 0)
2 is the distance between (0, 0) and P(x, y),
on the terminal side of 0. If r = V’x2 + y
then the trigonometric functions are defined to be
y
sinO=—
x
cosO=—
tan0=
cot0=
sec0=—
x
csc0=—
y
(2)
provided no denominator is 0.
It can be shown by using similar triangles that the values of the six trigonometric
functions depend only on the angle 0 and not on which point P(x, y) is chosen on the
terminal side of 0. The justification of this statement is like the one made for acute
angles on page 99.
D Domains A trigonometric function defined in (2) will be undefined if its denom
The angles are odd multiples of ir/2.
110
2 is never zero. Thus the
(0, 0), then r = Vx2 + y
inator is zero. Since P(x, y)
domains of the sine and the cosine functions consist of all angles 0. However, the tan
gent and the secant functions are undefined if the terminal side of 0 lies on the y-axis
because thenx = 0. Therefore, the domains of tan 0 and sec 0 consist of all angles 0 except
those having radian measure ±r/2, ±3ir/2, ±5’rr/2, and soon. Using set notation,
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
nd the fact that an odd integer can be written as 2n + 1, n an integer, the domains of
ie tangent and the secant functions are
(2n +
{o(o
(2n + l)r/2,n
{ojo
=
1)900,11
0, ±1, ±2,...
}
0, ±1, ±2,..
‘he cotangent and the cosecant functions are not defined for angles with their terminal
ides on the x-axis because then y = 0. Thus the domains of cot 0 and csc 0 consist of
11 angles 0 except those having radian measure 0, ± r, ± 2, ± 3, and so on; that is,
n7T,n = 0, ±1, ±2,.. .}or{00
00
180°n,n = 0, ±1, ±2,...}.
Since r
x/rI
1 and
i/ri
it follows that
Ix
r and
1. Therefore, as before
lsinOI
imilarly, because I r/x I
1
IcosOl
and
1 and r/y
cscoi
1
II
The angles are integer multiples of if.
r, or equivalently,
1.
(3)
1 we have
and
IsecOl
1.
(4)
‘he inequalities in (3) and (4) hold for every 0 in the domain of each of these functions.
EXAMPLE 1
Values of the Trigonometric Functions
md the exact values of the six trigonometric functions of the angle 0 if 0 is in standard
osition and the terminal side of 0 contains the point P( —3, 1).
olution The terminal side of the obtuse angle 0 is sketched in
lentificationsx
—3,y = 1, and
FIGURE 2.4.3.
With the
yi
=
P(-3, I)
‘e have from (2),
x
Vi
=
sec0
=
tanO
=
sin0
cos0
-3
=
-
=
—3
3
—3
=
—j-—
=
csc0
—,
3
3Vi
10
FIGURE 2.4.3
Example 1
—_____
=
=
cot0
—-—
v
i
1
d
=
—3
EXAMPLE 2
Angle 0 in
—3,
=
Values of the Trigonometric Functions
md the values of the six trigonometric functions of 0 if 0
—7r/2.
=
lution First we place 0 in standard position as shown in FIGURE 2.4.4. According to
efinition 2.4.1, we can choose any point P(x,
the terminal
on
of 0. For con
y)
side
v2 + y
4
mience,weselectP(0, —1) so thatx = O.y = —1,andr = \
2 = l.Thus,
.Ir
=
coti
=
sini
—1
1
—
—l
—1
cos
0
=
—
owever, the expressions tan0
ncex = 0.
csc
v/x and sec0
(
(
x
0
0
2)
1T\
=
P(O, —1)
rtv are undefined forD
—/2
FIGURE 2.4.4 Angle
0
in
Example 2
Algebraic Signs Depending on the quadrant in which the
terminal_side
of 0 lies,
e or both coordinates of P(x, v) may be negative. Since r = Vx2 + y
2 is always
2.4 Trigonometric Functions of GeneraL Angles
111
positive, each of the six trigonometric functions of 0 has negative as well as positive val
y/r is positive if the terminal side of 0 lies in quadrants I or
ues. For example, sinO
II (where y is positive), and sin 0 = v/r is negative if the terminal side of 0 lies in quad
—
rants III or IV (where)’ is negative). FIGURE 2.4.5 summarizes the algebraic signs of the
six trigonometric functions defined in (2). For convenience, if the terminal side of 0
lies in quadrant II, we will refer to 0 as a quadrant II angle or say that 0 is in quadrant II.
We will use similar terminology when we refer to angles with terminal sides in quad
rants I, Ill, or IV.
I
II
cos8<O sin9>O cos6>O sinO>O
tanO<O cot8<O tan6>O cot8>O
secO<O csc8>O secO>O csc>O
cos6<O sin9<O cosO>O sin8<O
tanO>O cot8>O tanO<O cotO<O
secB<O cscO<O sec8>O cscO<O
IV
III
FIGURE 2.4.5 Agebraic signs of the six
trigonometric functions
EXAMPLE 3
Using Figure 2.4.5
In which quadrant does the terminal side of 0 lie if sin 0 > 0 and tan 0 < 0?
Solution From Figure 2.4.5 we see that the sine function is positive for angles in quad
rants I and II and the tangent function is negative in quadrants II and IV the terminal
side of 0 must lie in quadrant II.
S
P(, )
r
FIGURE 2.4.6
An arbitrary
The reciprocal, quotient, and Pythagorean
2.2
also hold for general angles. For exam
Section
given
in
acute
angles
identities for
be
ple, to derive the Pythagorean identities let 0 any angle in standard position. As showr
in FIGURE 2.4.6, we let P(x, y) be any point other than the origin on the terminal side ol
.Dividingbotl
2
2 = r
0.Ifweagainletr = d(O, P) = Vy2,thenwehavex2 + y
gives
2
r
by
sides of the last equation
Pythagorean Identities—Revisited
U
—+‘--=
r
Recognizing that x/r
=
r
1
cos 0 and y/r
\
fy\
/X\
I—I +1—I =1.
\rJ
J
or
=
/
\
/
ange 0
sin 0, we obtain the basic Pythagorean identit3
2
0 + cosO
2
sin
=
1.
(5
2 0 be written first. If both sides o
In (5) we have followed the convention that sin
obtain
0
we
2
sin
and
0
2
cos
turn,
by
in
divided,
(5) are
1 + tan
0
2
and
0 + 1
2
cot
=
O
2
sec
(6
=
.
csc
6
2
(7
Formulas (5), (6), and (7) are identical to (10), (11), and (12) in Section 2.2. But unlik
the latter formulas, the trigonometric functions in (5), (6), and (7) are
• valid for all angles for which the functions are defined, and
• the values of the functions can have negative values.
We will encounter the Pythagorean identities one more time (in Chapter 3) whei
we show that each of the trigonometric functions can be defined for real numbers instea
of angles.
112
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
EXAMPLE 4
Using (5)
liven that cos 0
and that 0 is a quadrant IV angle, find the exact values of the
maining five trigonometric functions of 0.
olution Substituting cos 0
into (5) gives
=
(1)2
+
sinO
=
1 — — =
99
=
—.
ince the terminal side of 0 is in quadrant IV sin 0 is negative. Therefore, we must
]ect the negative square root of:
sinO
=
[
—
(See Figure 2.4.5.
2\/
—1•
=
low using
sin 0
tan0
secO
=
cotO
cos 0
1
cos 0
1
=
csc0
tanO
=
sin0
e find that the values of the remaining four functions are
tanO
=
=
secO
=
—2,
cot0
3,
1
=
csc0
EXAMPLE 5
=
-2V
1
—,
4
3V
—2v’/3
4
Using (6)
iven tanG = —2 and sinO
inctions of 0.
lution Letting tanG
0, find the exact values of the remaining five trigonometric
>
—2 in the identity I + tan
20
O
2
sec
=
1 +
(_2)2
=
2 0, we find
sec
=
rice tanG is negative in quadrants II and IV and sinG is positive in quadrants I and II,
e terminal side of 0 must lie in quadrant II. Thus we must take
sec0
:om sec 0
=
=
—\/..
1/cos 0, it follows that
1
1
cos0=—=
—\/
secO
sing tanG
=
sin 0/cos 0, we obtain
sinG
ien
5
=
cosG tanG
1
csc0
=
cotO
=
2\/
=
1
=
I
—
tanG
—-----)(_2)
=
2V3)5
1
—2
=
=
1
—
2.4 Trigonometric Functions of GeneraL AngLes
113
In Section 2.3 we found exact values for the six trigonometric functions of the spe
cial angles of 30°, 45°, and 60° (or 71/6, 71/4, and 71/3, respectively, in radian measure).
These values can be used to determine the exact trigonometric function values of cer
tain nonacute angles by means of a reference angle.
DEFINITION 2.4.2
Reference Angle
Let 0 be an angle in standard position such that its terminal side does not lie on a
coordinate axis. The reference angle 0’ for 0 is defined to be the acute angle formed
by the terminal side of 0 and the x-axis.
FIGURE 2.4.7
illustrates this definition for angles with the terminal sides in each of the
four quadrants.
-V
V
8
8=0’
>Jç
e5
(a)
(d)
(c)
(b)
FIGURE 2.4.7 An angle B (red) and its reference angle 0’ (bLue)
EXAMPLE 6
Reference AngLes
Find the reference angle for each angle 0.
(a) 0
=
(b) 0
40°
271
(c) 0
=
210°
971
(d) 0
=
Solution (a) From FIGURE 2.4.8(a) we see that 0’ 40°.
271/3 = 71/3.
0 =
(b) From Figure 2.4.8(b), 0’
180°
30°.
180° = 210°
(c) FromFigure2.4.8(c),0’= 0
—971/4 is coterminal with
(d) Since 0
—
—
—
—
971
—-a-
+ 2ir
71
=
we find that 0’ = 71/4. See Figure 2.4.8(d).
y
V
9,
(a)
FIGURE 2.4.8
114
1’
(b)
Reference angLes in ExampLe 6
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
(c)
(d)
x
I
1 Property of Reference AngLes The usefulness of reference angles in evaluat
ing trigonometric functions is a result of the following property:
4
The absolute value of any trigonometric function of an angle 0 equals the value
ofthatfimnction for the reference angle 0’.
Forinstance, sin0 = sinO’, cosol = cosO’, and so on.
We will verify the foregoing property for the sine function. If the terminal side of
9 lies in quadrant I, then 0
0’ and sinO is positive, so
sin0’= sin0
=
sin0.
rom FIGURE 2.4.9, we see that if 0 is a quadrant II, III, or IV angle, then we have
sin0’= --
=
=
sin0,
Nhere P(x, v) is any point on the terminal side of 0 and r
y
y
y
P(x, v)
r
y=IyII
x
..
(a)
FIGURE 2.4.9
(b)
(c)
Reference angles
We can now describe a step-by-step procedure for determining a trigonometric
unction value of any angle 0.
FINDING THE VALUE OF A TRIGONOMETRIC FUNCTION
Suppose 0 represents any angle.
(i) Find the reference angle 0
(ii) Determine the value of the trigonometric function for 0
(iii) Select the correct algebraic sign of the value in (ii) by considering
in which quadrant the terminal side of the angle 0 lies.
EXAMPLE 7
Finding VaLues Using Reference AngLes
md the exact values of sin 0, cos 0. and tan 0 for each of the following angles.
27r
97T
(b) 0 = 210°
(c) 0 =
) 0 =
4
olution We use the procedure just discussed along with Table 2.3.1 of Section 2.3.
i) In part (b) of Example 6 we found the reference angle for0 = 2r/3 to be 0’ = ir/3.
low we know that sin(7r/3) = v/2, cos(/3) = 1/2, and tan(r/3) =
ecause 0 = 2r/3 is a quadrant II angle, where the sine is positive but the cosine and
ie tangent are negative, we conclude
\/
2r
s1n——
=
2
cos——
=
—i-,1
and
2ir
tan—
3
=
—\/.
2.4 Trigonometric Functions of General Angles
115
(b) Referring to part (c) of Example 6, we see that the reference angle is 0’ = 30°.
Using the property of reference angles and the fact that the terminal side of 0’ 210°
lies in quadrant III, we get
sin2lO°
=
—sin30°
cos 210°
—cos 30°
tan 210°
tan 30°
=
——,
see Figure 2.4.5 for the
correct algebraic signs
—
2
=
(c) From part (d) of Example 6 we know that the reference angle 0’ =
0 = —9ir/4 is a quadrant IV angle, it follows that
9irN
1
.f
sini
I
\ 4/
—-———
cos_T)
—sin—
4
=
—
2
‘r
=
cos
=
—tan-a-
=
—1.
Finding AngLes
Find all angles 0 satisfying 0°
y
Since
1T
=
iT
EXAMPLE 8
iT/4.
0 < 3600 such that sin 0
=
Solution From what we know about the special angles 30°, 60°, and 90°, we know that
= 30° is one solution. Using 30° as a reference angle in the second quadrant, as shown
in FIGURE 2.4.10, we find 0 = 150° as a second solution. Since the sine function is neg
ative for angles in quadrants III and IV, there are no additional solutions satisfying
0°0<360°.
o
FIGURE 2.4.10
Solutions in
Example 8
.iEXAMPLE 9
Finding AngLes
Find all angles 0 satisfying 0
0 < 2ir such that cos 0
=
—
Solution Since the given value of the cosine function is negative, we first determine the
reference angle 0’ such that cos 0’ = \//2. From Section 2.3 we know that 0’ = ir/4.
Since the cosine function is negative for angles in quadrants II and III, we position the
reference angle 0= ir/4 as shown in FIGURE 2.4.11. We then obtain 0 = 3iT/4 and
0 = 5ir/4 as solutions.
3,r
4
(b)
(a)
FiGURE 2.4.11
116
Solutions in Example 9
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
NOTES FROM THE CLASSROOM
In this section we have purposely avoided using calculators. For a
complete understanding of trigonometry, it is essential that you
master the concepts and be able to perform without the aid of a
calculator the types of calculations and simplifications we have dis
cussed. The exercise set that follows should be worked without the
use of a calculator.
Exercises
P1
Answers to se’ected odd-numbered probems
begin on page ANS-7.
e recommend that you do not use a calculator in solving any of the problems thatfohlow.
7
4
n Problems 1—10, evaluate the six trigonometric functions of the angle 0 if 0 is in stanlard position and the terminal side of 0 contains the given point.
1.
3.
5.
7.
9.
(6, 8)
(5, —12)
(0,2)
(—2, 3)
(—v’,—l)
2.
4.
6.
8.
10.
(—1,2)
(—8, —15)
(—3,0)
(5,—I)
(V/)
n Problems 11—18, find the quadrant in which the terminal side of an angle 0 lies if 0
atisfies the given conditions.
Li.
.3.
.5.
.7.
sin0 <Oandtan0 >0
tanO <0 and sec0 <0
cotO > 0 and sinO > 0
sinO > Oandcos0 <0
12.
14.
16.
18.
I
cos9 > Oandsin0 <0
secO <Oandcsc0 <0
csc0 > Oandcot0 <0
tan0 <Oandcsc0 >0
n Problems 19—28, the value of one of the trigonometric functions of an angle 0 is
iven. From the given value and the additional information, determine the values of the
ive remaining trigonometric functions of 0.
.9.
1.
3.
5.
7.
sinO
tan 0
csc 0
sinO
tan0
9.
0.
1.
2.
3.
4.
If cos 0 =
find all possible values of sin 9.
If sin 0 =
find all possible values of cos 0.
If 2 sinO
cos 0 = 0, find all possible values of sin 0 and cos 0.
Ifcot0 = find all possible values ofcsc0.
If sec 0 = —5, find all possible values of sinO and cos 0.
If 3 cos 0 = sin 0, find all possible values of tan 0, cot 0, sec 0, and csc 0.
=
=
=
=
=
0 is in quadrant II
3, 0 is in quadrant III
—10, 0 is in quadrant IV
cosO > 0
8, sec0 > 0
,
—,
20.
22.
24.
26.
28.
cos 0
cot 0
sec 0
cos0
sec0
=
=
=
0 is in quadrant II
2, 0 is in quadrant III
3, 0 is in quadrant IV
sin0 <0
—4, csc0 <0
—,
—,
=
,
—,
—
,
2.4 Trigonometric Functions of Genera Anges
117
35. Complete the following table.
0 (degrees)
0 (radians)
sin 0
cos 6
0°
0
0
1
300
IT/6
1/2
450
ir/4
\//2
\//2
60°
/3
‘.//2
1/2
90°
ir/2
1
0
120°
2/3
\//2
—1/2
135°
3ir/4
150°
5ir/6
180°
7T
210°
7i-/6
—1/2
—V’/2
225°
5/4
240°
4r/3
270°
37T/2
300°
5!3
315°
7ir/4
330°
11IT/6
360°
2
csc 6
sec 0
tan 0
—
—v’s
36. Complete the following table.
0 (degrees)
0 (radians)
0°
0
30°
ir/6
45°
T/4
60°
Tr/3
2V’/3
90°
/2
I
120°
2ir/3
135°
3r/4
150°
5ir/6
180°
118
210°
7/6
225°
Sir/4
240°
4ir/3
270°
3rr/2
300°
5ir/3
315°
7-/4
330°
1l7r/6
360°
27r
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
—
2
1
cot 0
—
2\//3
2
—
v/3
0
In Problems 37—52, find the exact value of the given expression.
37. cos5r
38.
13r
39. cot—
40. tan—
41. sin(_)
23ir
42. cos—-—
43.
45.
47.
49.
51.
9r
csc (—i)
&
t
23ir
44. tan—T
sec(—120°)
sin 1500
tan 405°
cot (—720°)
46.
48.
50.
52.
In Problems 53—58, find all angles 0, where 0
condition.
0
53. tanO
=
54. sin0
=
55. coso
=
56. sec6
=
57. csc0
=
58. cotO
=
2
—1
In Problems 59—64, find all angles 0, where 0
59. sinOO
61. sec0 = —\/
63. cot0 = —V
h
csc495°
cos(—45°)
sin3l5°
sec(—300°)
0
<
60.cos0
62. csc0
64. tan0
<
360°, satisfying the given
——
2\/
3
1
2r, satisfying the given condition.
=
=
=
—1
2
1
MisceLLaneous AppLications
65. Free Throw Under certain conditions the maximum height y attained by a
basketball released from a height h at an angle a measured from the horizontal
with an initial velocity v
0 is given by
y
=
h + (v sin
a)/2g,
2
where g is the acceleration due to gravity. Compute the maximum height reached
0 = 8 m/s, a = 64.47°, and g = 9.81 rn/s
by a free throw if h = 2.15 m, v
.
2
66. Putting the Shot The range of a shot put released from a height h above
the ground with an initial velocity v
0 at an angle 0 to the horizontal can be
approximated by
cosS
0
v
R
=
g
0 sin +
(v
Free throw
2 + 2gh),
\/vö sin
where g is the acceleration due to gravity.
, compare the ranges
2
(a) If v
0 = 13.7 m/s, 4 = 40°, and g = 9.81 m/s
achieved for the release heights h = 2.0 m and h = 2.4 m.
(b) Explain why an increase in h yields an increase in R if the other parameters
are held fixed.
(c) What does this imply about the advantage that height gives a shot-putter?
2.4 Trigonometric Functions of General Angles
119
67. Acceleration Due to Gravity Because of its rotation, the Earth bulges at the
equator and is flattened at the poles. As a result, the acceleration due to gravity
actually varies with latitude 0. Satellite studies have shown that the acceleration
due to gravity sat is approximated by the function
1
g
978.0309
+
(a) Find g
1 at the equator (0
latitude.
y
Line through
origin in Probem 71
FIGURE 2.4.12
CONCEPTS REVIEW
0
2
5.18552 sin
=
0.
sin
2
0.00570 2
0°), (b) at the north pole, and (c) at 45° north
For Discussion
68. Is there an angle 0 such that cos 0 = ? Explain.
1? Explain.
69. Is there an angle 0 such that 2 csc 0
without the aid of a calculator that both
to
determine
possible
it
how
is
70. Discuss
sin 4 and cos 4 are negative.
71. Let L be a nonvertical line that passes through the origin and makes an angle 0
measured counterclockwise from the positive x-axis. Prove that the slope in of
the line L is tan 0. See FIGURE 2.4.12.
You should be able to give the meaning of each ofthefollowing concepts.
Straight angle
Quadrantal angle
Supplementary angles
Right angle
Arc length
Conversion:
degrees to radians
radians to degrees
Reference angle
Right triangles:
side adjacent
side opposite
Initial side of an angle
Tenninal side of an angle
Standard position of an angle
Coterminal angles
Minutes
Seconds
Degree measure of an angle
Central angle
Radian measure of an angle
Acute angle
Complementary angles
Obtuse angle
CHAPTER 2
hypotenuse
Trigonometric functions:
of acute angles
of general angles
Quotient identities
Reciprocal identities
Pythagorean identities
Cofunctions
Cofunction identities
Review Exercises
A. True/False
In Problems 1—10, answer true or false.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
120
—
sin(-rr/6)
cos(/3)
sin(r/2) = sin(5ir/2)
sin = 30°__
tan7r = 0
1
cscOl
(90° —0) = 1
2
O + sin
2
sin
The angles 120° and —240° are coterminal.
2 and cos 0 = 5.
If tan 0 = then sin 0
If sec (3 = \/7, then cos 0 =
30’ is equivalent to 0.5°.
=
,
CHAPTER 2 RIGHT TRIANGLE TRIGONOMETRY
Answers to selected odd-numbered
problems begin on page ANS-8.
15.
27. sinO
29. 3(\/
rathans
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
55.
57.
59.
61.
63.
65.
67.
69.
71.
75.
77.
79.
81.
19. 5.17°
10.6547°
23. 30048360
210°46’48”
27. r/4
ir/18
31. —23/18
3ir/2
400
35. 1200
39. 177.62°
225°
43. 110°
155°
47. 7ir/4
—205°
4
2.28
51. 2
1.3r
—-/4
(b) 131.75°
(a) 41.75°
(a) The given angle is greater than 900.
(b) 81.60
(b) 3r/4
(a) /4
than
greater
is
Tr/2.
given
angle
(a) The
(b) ir/3
(b) —1845°; —10.25r
(a) 216°; 1.2ir
because the hour hand moves counterclockwise: —60°, —7r/3
(b) 2h
(a) 16h
(b) 15
(a) 9
(b) 85.94°
(a) 1.5
(b) 3.074641 km/s
(a) 0.000072921 rad/s
miles
1.15 statute
(a) 3irrad/s
IT cm/s
0
(b) 30
(a) 711.1 rev/mm
(b) 4468 rad/min
—
103
*696
1. sinO
=3,coso
3. sinO
=
3,tanO ‘3,cot9 = 3,seco =4,csco
= ,
tanO = 3, cotO
cosO
secO =
—
33.3
37. —1
41. 5
45.1
49. \/
53.
=3
3.
°,
3,
Page 108
Exercises 2.3
V’, csc45° =
1, sec45°
7.
5.2
3.3
ii. 3(’v’ +
9. 18
17. 2\/
15. 0
13. 9V’
21. 2—V’
19.3
23. sinl7° = 0.2924,cosl7° = 0.9563,tanl7° = 0.3057,
3.4203
cotl7° = 3.2709,secl7° = l.0457,cscl7°
25. sin 14.3° = 0.2470, cos 14.3° = 0.9690, tan 14.30 = 0.2549,
4.0486
3.9232. sec 14.3° = 1.0320, csc 14.3°
cot 14.30
27. sin7l.504l7° = 0.9483,cos7l.50417° = 0.3172,
tan7l.50417° = 2.9894,cot7l.50417° = 0.3345,
sec7l.504l7° = 3.1522,csc7l.50417° = 1.0545
29. sin() = 0.5878, cos3) = 0.8090, tan() = 0.7265,
cot() = 1.3764, sec() = 1.2361, csc3) = 1.7013
0.6229, cos0.6725 = 0.7823,
31. sinO,6725
1.2558,
0.7963.cot0.6725
tan0.6725
1.6053
secO.6725 = l.2783,csc0.6725
1. tan45°
1, cot45°
=
11. tan6
=
4, cotO 4, secO
=
cot&
=
,
13. tanO 5
15. cosO
17. sinO
19. sinO
21. cos6
=
=
=3
,secB =-,csc
=
cotO
,cscO =
=
8, secO
=
,
csc8
=
Exercises 2.4 rPage 117
= ,
=
tanO
tanO
2V, cotO
= ,
cosO
=
=
—-f, cotG
9. sinG
=
—*,cosG
=
secG
=
_E6,cotO
=
cotO
=
=
fs,
-.
cotO
secO
=
=
4. secO
—vT
21. sinG
=
—°,cosG
cotO
=
3
23. sinG
=
=
cotG
=
3
27. sinG
=
=
34
=
cscb
, cscO
tanG
=
=
\/,cscO
secO = ,
=
19. cosG
cotG
cotG
=
—4
=
tanG
=
csc6
=
13. II
17. II
25. cosO
=
=
3
3,
3,
4,
4,coto =
1. sinG = cosO = tanG = cscO = secO
7
3
7
5
3. sinG = —5,cosO = y,tanG = —°,cscO = —js,
secO = ,cotO =
5. sinG = 1, cosG = 0, tanG is undefined, cscG = 1, secO is
undefined, cotG = 0
cosO = —‘,tanG = —4, cscO =
7. sinG =
V’
3, cosO 3, tanO 3, cotO 3
=
23. sinO =4,coso =o2,tanO =,cotO
25. sinO
oo,
,cscO
secO=—-
=
11. III
15. I
,tanO =,cotO
_,cosQ
=
cscO
1
—
3
secO
5. sinO =4,coso =‘,tanO =cot0
secO = 7] cscO = 4
= ,
tanO = , cotO = 2V,
7. sinO = cosO
3
cscO
secO =
9. sinO
=
coto
—
=
V’Tö, cscO
=
31. v’
35.2
39. 0
43.
47.
3,
Exercises 2.2
4, tano
cos6 =
—,
=
—,cscG
—-f3,cscG
- tanG
1
j
4
34
,
cosG =
=
=
4, secO
=
,secG
—,
secG
=
—‘s/iô
=
—3vTi
tanG
—-j,
=
cscO
—5, secO
cscO =,secG
=
,
cos6 5
=
=
V,cotG
=
4
29. ±
2
=
31. sinG
33. cosO
=
=
±, cosG
—3,
sinG
ANSWERS TO SELECTED ODD-NUMBERED PROBLEMS
=
±
±6
ANS-7