Download Chapters 4-5 - Mobile Christian School

C In Chapter 3, you
studied exponential
and logarithmic
functions, which are
two types of
transcendental
functions.
C In Chapter 4, you will:
■ Use trigonometric functions to
solve right triangles.
■ Find values of trigonometric
functions for any angle.
C SATELLITE NAVIGATION Satellite navigation systems operate by
receiving signals from satellites in orbit, determining the distance to
each of the satellites, and then using trigonometry to establish the
location on Earth's surface. These techniques are also used when
navigating cars, planes, ships, and spacecraft.
■ Graph trigonometric and
inverse trigonometric functions.
PREREAD Use the prereading strategy of previewing to make two
or three predictions of what Chapter 4 is about.
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eGlossary
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NewVocabulary
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Textbook Option
Take the Quick Check below.
QuickCheck
Find the missing value in each figure.
(Prerequisite Skill)
3.
4.
Determine whether each of the following could represent the measures
of three sides of a triangle. Write yes or no. (Prerequisite Skill)
4 ,8 ,1 2
5.
7.
6. 1 2,15 ,18
ALGEBRA The sides of a triangle have lengths x, x + 17, and 25. If
the length of the longest side is 25, what value of x makes the
triangle a right triangle? (Prerequisite Skill)
Find the equations of any vertical or horizontal asymptotes.
8. f(x) =
10. f(x) =
12. h(x) =
2
x '-4
x2
+ 8
x (x
- 1 )2
(x — 2)(x + 4)
x 2 + x - 20
x+ 5
9. h(x) =
11. g(x) =
13. f(x) =
(Lesson 2-5)
x 3 - 27
x+ 5
x+ 5
( x - 3 ) ( x - 5)
Espanol
trignom etric functions
p. 220
funciones trigonom etricas
sine
p. 220
seno
cosine
p. 220
coseno
tangent
p. 220
tangente
cosecant
p. 220
cosecant
secant
p. 220
secant
cotangent
p. 220
funcion reciproca
reciprocal function
p. 220
cotangente
inverse sine
p. 223
seno inverso
inverse cosine
p. 223
coseno inverso
inverse tangent
p. 223
tangente inversa
radian
p. 232
radian
coterminal angles
p. 234
angulos coterminales
reference angle
p. 244
angulo de referencia
unit circle
p. 247
circuio de unidad
circular function
p. 248
funcion circular
period
p. 250
periodo
sinusoid
p. 256
sinusoid
am plitude
p. 257
am plitud
frequency
p. 260
frecuencia
phase shift
p. 261
cambio de fase
Law of Sines
p. 291
ley de senos
Law of Cosines
p. 295
ley de cosenos
ReviewVocabulary
reflection p. 48 reflexion the mirror image of the graph of a function
with respect to a specific line
dilation p. 49 homotecia a nonrigid transformation that has the
effect of compressing (shrinking) or expanding (enlarging) the graph of a
function vertically or horizontally
2 x 2 + 5x — 12
2x — 3
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219
• You evaluated
functions.
(Lesson 1-1)
Large helium-filled balloons are a tradition of many holiday parades.
Long cables attached to the balloon are used by volunteers to lead
the balloon along the parade route.
Find values of
trigonometric
functions for acute
angles of right
triangles.
Suppose two of these cables are attached to a balloon at the same
point, and the volunteers holding these cables stand so that the ends
of the cables lie in the same vertical plane. If you know the measure
of the angle that each cable makes with the ground and the distance
between the volunteers, you can use right triangle trigonometry to
find the height of the balloon above the ground.
: Solve right triangles.
NewVocabulary
trigonometric ratios
trigonometric functions
sine
cosine
tangent
cosecant
secant
cotangent
reciprocal function
inverse trigonometric
function
inverse sine
inverse cosine
inverse tangent
angle of elevation
angle of depression
solve a right triangle
Values of Trigonometric Ratios
The word trigonometry m eans triangle measure. In this
chapter, you will study trigonometry as the relationships among the sides and angles of
triangles and as a set of functions defined on the real num ber system. In this lesson, you will study
right triangle trigonometry.
1
Using the side m easures of a right triangle and a reference angle labeled 9 (the Greek letter theta),
we can form the six trigonometric ratios that define six trigonom etric functions.
KeyConcept Trigonometric Functions
9 be an acute angle in a right triangle and the abbreviations
9, the
length of the side adjacent to 9, and the length of the hypotenuse,
Let
opp, adj, and hyp refer to the length of the side opposite
\h y p
opp
respectively.
Then the six trigonometric functions of
e\
1
9 are defined as follows.
adj
sine (9) = Sin 9 = ^
cosecant (9) = CSC 9 =
hyp
opp
cosine (9) = COS 9 = 7—!-
secant (9) = sec 9 =
tangent (9) = tan 9 =
cotangent (9) = COt 9 = —
hyp
adj
opp
V
J
The cosecant, secant, and cotangent functions are called reciprocal functions because their ratios
are reciprocals of the sine, cosine, and tangent ratios, respectively. Therefore, the following
statements are true.
csc 9 =
„
sin 9
sec 9 = — cos 9
cot 9 = • 1
tan 9
From the definitions of the sine, cosine, tangent, and cotangent functions, you can also derive the
following relationships. You will prove these relationships in Exercise 83.
,
n
tan 9 =
220
Lesson 4-1
SUl U
j
and
. n
cot 9 =
COS
s in 9
Memorizing Trigonometric Ratios
The mnemonic device
SOH-CAH-TOA is most commonly
used to remember the ratios for
Find the exact values of the six trigonom etric functions of 9.
The length of the side opposite 9 is 8, the length of the side
adjacent to 9 is 15, and the length of the hypotenuse is 17.
sine, cosine, and tangent.
—
opp =
sine = ^ P
8 and hyp = 17
hyp
csc i
opp
hyp
COS (
_ adj
’ hyp
opp
tan 6
adj
1
Find Values of Trigonometric Ratios
StudyTip
COS I
adj
15
°P P
8
tan 9 = —— or —
ad
15
opp =
8
a
hyp
17
sec 9 = —— or —
adj
15
adj = 15 and hyp = 17
hyp 01 17
17
or —
8 and adj = 15
adj
cot i
opp
15
or —
8
p GuidedPractice
1A.
1B.
13
12
Consider sin 0 in the figure.
Using A ABC: sin i
S'
BC
AB
Using A AB'C': sin 9 = -^p-
Notice that the triangles are sim ilar because they are two right triangles
that share a com m on angle, 9. Because the triangles are similar, the ratios
BC
B'C
of the corresponding sides are equal. So, — =
Therefore, sin 9 has the same value regardless of the triangle used. The values of the functions are
constant for a given angle measure. They do not depend on the size of the right triangle.
Use One Trigonometric Value to Find Others
If cos 9 — —, find the exact values of the five rem aining trigonom etric functions for the
acute angle 9.
Begin by drawing a right triangle and labeling one acute angle 9.
WatchOut!
Common Misconception
Because cos f =
hypotenuse 5.
= j , label the adjacent side 2 and the
In Example 2, the adjacent side of
the triangle could also have been
labeled 4 and the hypotenuse 10.
This is because cos 8 = | gives
By the Pythagorean Theorem, the length of the leg opposite
9 is V 5 2 - 2 2 or V z f.
the ratio of the adjacent side and
hypotenuse, not their specific
°P P
measures.
hyp
hyp
opp
V21
5
tan 9 =
5
5V21
-= o r-
cot 9 =
\Jl1
21
°P P
V21
adj
2
adj
°P P
sec
2
2 V2T
-p r r O r -------V5T
21
p GuidedPractice
2. If tan 9 = 1 , find the exact values of the five rem aining trigonom etric functions for the
acute angle 9.
You will often be asked to find the trigonom etric functions of specific acute angle measures.
The table below gives the values of the six trigonom etric functions for three com m on angle
measures: 30°, 45°, and 60°. To remember these values, you can use the properties of 30°-60°-90°
and 45°-45°-90° triangles.
KeyConcept Trigonometric Values of Special Angles
30°-60°-90° Triangle
o
CO
B
V3
I
I
2
45°
60°
V2
2
V |
2
V2
2
1
2
1
tan e
y r
csc o
w rw i
■
N
2 \3
M
M
3
45°-45°-90° Triangle
4$°
V3
Vs
V2
2V 3
3
V2
2
1
V3
3
You w ill verify some of these values in Exercises 57-62.
Solving Right Triangles Trigonometric functions can be used to find missing side lengths
and angle measures of right triangles.
H Q 2 5 J J 3 3 ^ f in d a Missing Side Length
TechnologyTip
Find the value o f x. Round to the nearest tenth, if necessary.
Degree Mode To evaluate a
trigonometric function of an angle
measured in degrees, first set the
calculator to
degree mode by
selecting DEGREE on the MODE
Because you are given an acute angle measure and the length of the
hypotenuse of the triangle, use the cosine function to find the length
of the side adjacent to the given angle.
feature of the graphing calculator.
sci
Eno
cos 6 ■
(' i : J >t E b ? B 5
FiftD iflri m
PftR
B
j l
PPL
HDRIZ
SET CLOCK
cos 42° = ^
°SIH U L
•3+bl-
adj
hyp
l-
18 cos 42° = x
Cosine function
0 = 42°, adj = x, and hyp = 18
Multiply each side by 18.
|]-T
13.4 ~ x
Use a calculator.
Therefore, x is about 13.4.
CHECK You can check your answer by substituting x = 13.4 into cos 42° = — .
cos 42° =
18
13.4
cos 42° =
18
0.74 = 0.74 ✓
► GuidedPractice
222
Lesson 4-1
R ight Triangle T rig o n o m e try
x = 13.4
Simplify.
^ ;^ B V m p liy '!iT 'l'll'll.-i Finding a Missing Side Length
TRIATHLONS A com petitor in a triathlon is running along the
course shown. Determine the length in feet that the runner
must cover to reach the finish line.
A n acute angle m easure and the opposite side length are given,
so the sine function can be used to find the hypotenuse.
sin 9 =
sin 63° =
opp
Sine function
hyp
200
0 = 63°, opp = 200, and hyp = x
x
x sin 63° = 200
The Ironman Triathlon held in
X:
Kailua-Kona Bay, Hawaii, consists
of three endurance events,
including a 2.4-mile swim, a
200
s in 6 3 c
Multiply each side by
or about 224.47
x.
Divide each side by sin 63°.
So, the com petitor m ust run about 224.5 feet to finish the triathlon.
112-mile bike ride, and a
26.2-mile marathon.
Source: World Triathlon Corporation
t GuidedPractice
4. TRIATHLONS Suppose a com petitor in the
swim m ing portion of the race is swimming
along the course shown. Find the distance
the com petitor must swim to reach the shore.
W hen a trigonometric value of an acute angle is known, the corresponding inverse trigonometric
function can be used to find the measure of the angle.
ReadingMath
Inverse Trigonometric Ratios
KeyConcept Inverse Trigonometric Functions
In v e r s e S in e
is the measure of angle 9. That is, if sin 9 = x, then sin-1 x = 9.
The expression sin-1 x is read the
inverse sine ofx. Be careful not to
If 9 is an acute angle and the sine of 9 is x then the inverse sine of x
In v e r s e C o s in e
confuse this notation with the
If 9 is an acute angle and the cosine of 9 is x then the inverse cosine of x is the measure of
angle 9. That is, if cos 9 = x, then cos-1 x = 9.
notation for negative exponents:
sin-1 x
- J —. Instead, this
sin x
notation is similar to the notation
In v e r s e T a n g e n t
If 9 is an acute angle and the tangent of 9 is x, then the inverse tangent of x is the measure
of angle 9. That is, if tan 9 =
for an inverse function, f~ 1(x).
.............
x then tan-1
........................................................
x = 9.
................................................
...
J
( 2 £ E E S 3 R nc* a Missing Angle Measure
Use a trigonometric function to find the measure of 6. Round to
the nearest degree, if necessary.
Because the measures of the sides opposite and adjacent to 0 are
given, use the tangent function.
tan 9 =
tan 9 =
°PP
Tangent function
adj
26
opp = 26 and adj = 11
11
1= tan 1 ^
or about 67°
Definition of inverse tangent
5B.
E
connectED.mcgraw-hill.com |
223
Some applications of trigonometry use an angle of elevation or depression. An angle of elevation
is the angle formed by a horizontal line and an observer's line of sight to an object above. An
angle of depression is the angle formed by a horizontal line and an observer's line of sight to
an object below.
__________________
J
angle o f —
depression /
/
/
line of sight
angle of elevation
In the figure, the angles of elevation and depression are congruent because they are alternate
interior angles of parallel lines.
Real-World Example 6 Use an Angle of Elevation
AIRPLANES A ground crew w orker who is 6 feet tall is directing a plane on a runway. If the
w orker sights the plane at an angle of elevation of 32°, w hat is the horizontal distance from
the worker to the plane?
Because the worker is 6 feet tall, the vertical distance from the worker to the plane is 150 — 6, or
144 feet. Because the measures of an angle and opposite side are given in the problem , you can
use the tangent function to find x.
Real-W orldCareer
opp
tan 9 = — —
adj
Airport Ground Crew
Tangent function
Ground crewpersons operate
tan 32° = - ^ -
ramp-servicing vehicles, handle
0 = 32°, opp = 144, and adj = x
cargo/baggage, and marshal or
x tan 32° = 144
tow aircraft. Crewpersons must
have a high school diploma,
x =
a valid driver’s license, and
a good driving record.
x
/
144
Multiply each side by x.
—™
Divide each side by tan 32°.
230.4
Use a calculator.
tan 32°
So, the horizontal distance from the w orker to the plane is approxim ately 230.4 feet.
GuidedPractice
6. CAMPING A group of hikers on a cam ping trip clim b to the top of a 1500-foot mountain.
W hen the hikers look dow n at an angle of depression of 36°, they can see the campsite in
the distance. W hat is the horizontal distance betw een the cam psite and the group to the
nearest foot?
V
224
| Lesson 4-1 | R ight T riangle T rig o n o m e try
Angles of elevation or depression can be used to estim ate the distance betw een two objects, as well
as the height of an object w hen given two angles from two different positions of observation.
Real-World Example 7 Use Two Angles of Elevation or Depression
StudyTip
BALLOONING A hot air balloon that is m oving above a neighborhood has an angle of
depression of 28° to one house and 52° to a house down the street. If the height of the
balloon is 650 feet, estimate the distance between the two houses.
Indirect Measurement When
calculating the distance between
two objects using angles of
depression, it is important to
Draw a diagram to model this situation. Because
the angle of elevation from a house to the balloon
is congruent to the angle of depression from the
balloon to that house, you can label the angles
of elevation as shown. Label the horizontal distance
from the balloon to the first house x and the distance
betw een the two houses y.
remember that the objects must
lie in the same horizontal plane.
From the sm aller right triangle, you can use the
tangent function to find x.
opp
tan 9 = — —
adj
Tangent function
tan 52° = ^
e = 52°- °PP = 650’ and adi = x
x tan 52° = 650
x =
Multiply each side by x.
650
Divide each side by tan 52°.
tan 52
From the larger triangle, you can use the tangent function to find y .
°P P
tan 6 = — —
adj
tan 28° =
Tangent function
0 = 28°, opp = 650, and adj = x + y
x -+- y
(x + y) tan 28° = 650
x + y =
TechnologyTip
J
Using Parentheses When
650
expression using a graphing
calculator, be careful to close
parentheses. While a calculator
+ v =
tan 52° T y
evaluating a trigonometric
>
u =
y
Divide each side by tan 28°.
tan 28
tan 28°
65P
■
tan 28°
y ~ 7 1 4 .6
returns the same value for the
Multiply each side by x + y.
Substitute
------- 65Q—
tan 52°
Subtract
x=
650
850 .
tan 52°
tan 52°
from each side.
Use a calculator.
expressions tan(26 and tan(26),
it does not for expressions
V
tan(26 + 50 and tan(26) + 50.
Therefore, the houses are about 714.6 feet apart.
______________
p- GuidedPractice
7.
BUILDINGS The angle of elevation from a car to the top of an apartm ent building is 48°. If the
angle of elevation from another car that is 22 feet directly in front of the first car is 64°, how
tall is the building?
Trigonometric functions and inverse relations can be used to solve a right triangle, w hich m eans to
find the measures of all of the sides and angles of the triangle.
ReadingMath
Labeling Triangles
Throughout this chapter,
a capital letter will be used to
1 9 1
Solve a Right Triangle
Solve each triangle. Round side lengths to the nearest tenth and angle m easures to the nearest
degree.
a.
represent both a vertex of a
triangle and the measure of the
angle at that vertex. The same
letter in lowercase will be used
to represent both the side
opposite that angle and
Y
the length of that side.
Find x and z using trigonom etric functions.
tan 35° = ^
10 tan 35° = x
7.0 ^ x
cos 35° = —
Substitute.
Substitute.
z
z cos 35° = 10
Multiply.
Multiply.
10
Use a calculator.
2= •
cos 35°
Divide.
2 ~ 12.2
Use a calculator.
Because the measures of two angles are given, Y can be found by subtracting X from 90°.
35° + Y = 90°
Y = 55°
Angles Xand Kare complementary.
Subtract.
Therefore, Y = 55°, x = 7.0, and z
12.2.
Because two side lengths are given, you can use the Pythagorean Theorem to find that
k = V l8 5 or about 13.6. You can find J by using any of the trigonom etric functions.
tan/ = f
Substitute.
/ = ta n -1 ^ -
Definition of inverse tangent
/ ~ 53.97°
Use a calculator.
Because / is now known, you can find L by subtracting / from 90°.
53.97° + L « 90°
L « 36.03°
Angles Jand L are complementary.
Subtract.
Therefore, J ~ 54°, L ~ 36°, and k ~ 13.6.
^ GuidedPractice
8B.
226
| Lesson 4-1 i R ight T riangle T rig o n o m e try
Exercises
= Step-by-Step Solutions begin on page R29.
Find the exact values of the six trigonom etric fu nctions o f 0.
(2 7 ) MOUNTAIN CLIMBING A team of climbers m ust determine
the w idth o f a ravine in order to set up equipm ent to
cross it. If the clim bers w alk 25 feet along the ravine from
their crossing point, and sight the crossing point on the
far side of the ravine to be at a 35° angle, how wide is the
ravine? (Example 4)
(Example 1)
28. SNOWBOARDING Brad built a snow boarding ramp w ith a
height of 3.5 feet and an 18° incline. (Example 4)
a. Draw a diagram to represent the situation.
VT65
b. Determ ine the length of the ramp.
7.
29. DETOUR Traffic is detoured from Elwood Ave., left
0.8 mile on M aple St., and then right on Oak St., which
intersects Elw ood Ave. at a 32° angle. (Example 4)
10
a. Draw a diagram to represent the situation.
b. Determ ine the length of Elw ood Ave. that is detoured.
Use the given trigonom etric function value of the acute
angle 9 to find the exact values of the five rem aining
trigonom etric fu nction values of 6. (Example 2)
9. s in 0 =
f
10,)
30. PARACHUTING A paratrooper
encounters stronger winds than
anticipated while parachuting
from 1350 feet, causing him to
drift at an 8° angle. H ow far
from the drop zone will the
paratrooper land? (Example 4)
COS 0 = y
11 . tan 9 = 3
13.
COS 0 = jr7
15. cot 0 = 5
17. sec 0 = |
%
Find the m easure o f angle 9. Round to the nearest degree, if
necessary. (Example 5)
Find the value o f x. Round to the n earest tenth, if necessary.
(Example 3)
31.
19.
33.
35.
^connectED^Ticgrav\MTilLcorTj|
227
39. PARASAILING Kayla decided to try parasailing. She was
strapped into a parachute towed by a boat. An 800-foot
line connected her parachute to the boat, w hich w as at a
32° angle of depression below her. H ow high above the
water was Kayla? (Example 6)
3 3 ) LIGHTHOUSE Two ships are spotted from the top of a
156-foot lighthouse. The first ship is at a 27° angle of
depression, and the second ship is directly behind the
first at a 7° angle of depression. (Example 7)
a. Draw a diagram to represent the situation.
b.
Determ ine the distance betw een the two ships.
MOUNT RUSHMORE The faces of the presidents at Mount
Rushm ore are 60 feet tall. A visitor sees the top of George
W ashington's head at a 48° angle of elevation and his chin
at a 44.76° angle of elevation. Find the height of M ount
Rushm ore. (Example 7)
OBSERVATION WHEEL The London Eye is a 135-meter-tall
observation wheel. If a passenger at the top of the wheel
sights the London Aquarium at a 58° angle of depression,
what is the distance betw een the aquarium and the
London Eye? (Example 6)
135 m
( '41, ROLLER COASTER On a roller coaster, 375 feet of track
ascend at a 55° angle of elevation to the top before the
first and highest drop. (Example 6)
Solve each triangle. Round side lengths to the nearest tenth
and angle measures to the nearest degree. (Example 8)
a. Draw a diagram to represent the situation.
b.
Determine the height of the roller coaster.
42. SKI LIFT A com pany is installing a new ski lift on a
225-meter-high mountain that will ascend at a
48° angle of elevation. (Example 6)
a. Draw a diagram to represent the situation.
b.
Determine the length of cable the lift requires to
extend from the base to the peak of the mountain.
43. BASKETBALL Both Derek and Sam are 5 feet 10 inches tall.
Derek looks at a 10-foot basketball goal w ith an angle of
elevation of 29°, and Sam looks at the goal w ith an angle
of elevation of 43°. If Sam is directly in front of Derek,
how far apart are the boys standing? (Example 7)
55. BASEBALL M ichael's seat at a game is 65 feet behind home
plate. His line of vision is 10 feet above the field.
a. Draw a diagram to represent the situation.
44. PARIS A tourist on the first observation level of the
Eiffel Tower sights the M usee D 'O rsay at a 1.4° angle of
depression. A tourist on the third observation level
located 219 meters directly above the first, sights the
Musee D 'O rsay at a 6.8° angle of depression. (Example 7)
b.
W hat is the angle of depression to home plate?
56. HIKING Jessica is standing 2 miles from the center of the
base of Pikes Peak and looking at the sum m it of the
m ountain, which is 1.4 m iles from the base.
a. Draw a diagram to represent the situation.
a. Draw a diagram to represent the situation.
b.
b.
228
Determine the distance betw een the Eiffel Tower and
the Musee D'Orsay.
Lesson 4-1
R ight Triangle T rig o n o m e try
With w hat angle of elevation is Jessica looking at
the sum m it of the mountain?
Find the exact value of each expression w ithout using
a calculator.
57. sin 60°
cot 30°
59. sec 30°
60. cos 45°
61. tan 60°
62) csc 45°
82.
a. GRAPHICAL Let P(x, y) be a point in the first quadrant.
Graph the line through point P and the origin. Form a
right triangle by connecting the points P, (x, 0), and the
origin. Label the lengths of the legs of the triangle in
terms of x or y. Label the length of the hypotenuse as r
and the angle the line m akes with the x-axis 9.
Without using a calculator, find the measure of the acute
angle 9 in a right triangle that satisfies each equation.
63. tan 9 = 1
65. cot
( ^ .
cos 9 = ^y
V3
3
67. csc 9 = 2
MULTIPLE REPRESENTATIONS In this problem , you will
investigate trigonom etric functions of acute angles and
their relationship to points on the coordinate plane.
b. ANALYTICAL Express the value of r in terms of x and y.
V2
C. ANALYTICAL Express sin 9, cos 9, and tan 9 in terms of
x, y, and /or r.
sec 9 = 2
d. VERBAL U nder what condition can the coordinates of
point P be expressed as (cos 9, sin 9)1
Without using a calculator, determine the value of x.
e. ANALYTICAL W hich trigonom etric ratio involving 9
corresponds to the slope of the line?
f. ANALYTICAL Find an expression for the slope of the line
perpendicular to the line in part a in terms of 9.
71. SCUBA DIVING A scuba diver located 20 feet below the
surface of the water spots a shipwreck at a 70° angle of
depression. After descending to a point 45 feet above the
ocean floor, the diver sees the shipwreck at a 57° angle of
depression. Draw a diagram to represent the situation,
and determine the depth of the shipwreck.
Find the value of cos 6 if 9 is the measure of the smallest
angle in each type of right triangle.
72. 3-4-5
73. 5-12-13
74. SOLAR POWER Find the total area of the solar panel
shown below.
H.O.T. Problem s
83.
Use Higher-Order Thinking Skills
PROOF Prove that if 9 is an acute angle of a right triangle,
then tan 9 =
cos 8
and cot 9 =
84. ERROR ANALYSIS Jason and N adina know the value of
sin 9 = a and are asked to find csc 9. Jason says that this
is not possible, bu t N adina disagrees. Is either of them
correct? Explain your reasoning.
85.
WRITING IN MATH Explain why the six trigonometric
functions are transcendental functions.
86.
CHALLENGE W rite an expression in terms of 9 for the area
of the scalene triangle shown. Explain.
3.5 ft
W ithout using a calculator, insert the appropriate symbol > ,
< , or = to complete each equation.
75. sin 45°
cot 60°
76. tan 60°
cot 30‘
77. cos 30°
csc 45°
78. cos 30°
sin 60‘
79. sec 45°
csc 60°
80. tan 45°
sec 30'
81. ENGINEERING Determine the depth of the shaft at the large
end d of the air duct shown below if the taper of the duct
is 3.5°.
(87) PROOF Prove that if 9 is an acute angle of a right triangle,
then (sin 9)2 + (cos 9)2 = 1.
REASONING If A and B are the acute angles of a right triangle
and m Z A < m Z B , determ ine w hether each statement is true
or false. If false, give a counterexam ple.
88. sin A < sin B
89. cos A < cos B
90. tan A < tan B
91. WRITING IN MATH N otice on a graphing calculator that
there is no key for finding the secant, cosecant, or
cotangent of an angle measure. Explain why you think
this m ight be so.
&
connectEDTrncgraw-hiH^
229
Spiral Review
92. ECONOMICS The Consumer Price Index (CPI) measures inflation. It is based on the average
Year
prices of goods and services in the United States, w ith the annual average for the years
1982-1984 set at an index of 100. The table show n gives some annual average CPI values
from 1955 to 2005. Find an exponential model relating this data (year, CPI) by linearizing the
data. Let x = 0 represent 1955. Then use your model to predict the CPI for 2025. (Lesson 3-5)
Solve each equation. Round to the nearest hundredth. (Lesson 3-4)
93. e5x = 24
94.2 ex“ 7 — 6 = 0
CPI
1955
26.8
1965
31.5
1975
53.8
1985
107.6
1995
152.4
2005
195.3
Source: Bureau of Labor
Statistics
Sketch and analyze the graph of each function. Describe its domain, range, intercepts,
asymptotes, end behavior, and where the function is increasing or decreasing. (Lesson 3-1)
95.
96.f{x ) = 2 3j - 4 + 1
f( x ) = - 3 X~ 2
97. fix ) = -4 ~
Solve each equation. (Lesson 2-5)
98 . -----* 2 — 16------ = - A ------------ 1—
(x + 4)(2x —1)
101.
x+4
99 . -------------
2x - 1
(x + l)(x —5)
= _ J l _ + _ JL _
x+ 1
x —5
100.
2x2 + —
3x2 + 5x + 2
3x + 2
x+ 1
NEWSPAPERS The circulation in thousands of papers of a national newspaper is
shown. (Lesson 2-1)
Year
2002
2003
2004
2005
2006
2007
2008
Circulation
(in thousands)
904.3
814.7
773.9
725.5
716.2
699.1
673.0
a. Let x equal the number of years after 2001. Create a scatter plot of the data.
b. Determine a pow er function to model the data.
c. Use the function to predict the circulation of the newspaper in 2015.
Skills Review for Standardized Tests
104. A person holds one end of a rope
102. SAT/ACT In the figure,
what is the value of z?
2y+ 5
Mote: Figure not drawn to scale.
A 15
D 30V2
B 15V 2
E 30V3
C 15V3
103. REVIEW Joseph uses a ladder to reach a window
10 feet above the ground. If the ladder is 3 feet away
from the wall, how long should the ladder be?
F
9.39 ft
that runs through a pulley and has
a w eight attached to the other end.
Assum e that the weight is at the
same height as the person's hand.
W hat is the distance from the
person's hand to the weight?
A 7.8 ft
B 10.5 ft
C 12.9 ft
D 14.3 ft
105. REVIEW A kite is being flow n at a 45° angle. The string
of the kite is 120 feet long. How high is the kite above
the point at w hich the string is held?
G 10.44 ft
F 60 ft
H 11.23 ft
G 60V 2 ft
J 12.05 ft
H 60V 3 ft
J 120 ft
230
| Lesson 4-1 I R ight Triangle T rig o n o m e try
12 f t
I
** 1 ?
Degrees and Radians
4
•Then
• You used the
measures of acute
angles in triangles
given in degrees.
(Lesson 4-1)
j / y NewVocabulary
vertex
initial side
terminal side
standard position
radian
coterminal angles
linear speed
angular speed
sector
Now
1
Why?
Convert degree
measures of angles
to radian measures,
and vice versa.
) Use angle measures
i to solve real-world
problems.
In Lesson 4-1, you worked only with acute
angles, but angles can have any real number
measurement. For example, in skateboarding,
a 540 is an aerial trick in which a skateboarder
and the board rotate through an angle of 540°,
or one and a half complete turns, in midair.
A ngles and Their M easu res
From geometry, you m ay recall an angle being defined as
two noncollinear rays that share a com m on endpoint know n as a vertex. An angle can also be
thought of as being formed by the action of rotating a ray about its endpoint. From this dynamic
perspective, the starting position of the ray forms the initial side of the angle, while the ray's
position after rotation forms the angle's term inal side. In the coordinate plane, an angle with its
vertex at the origin and its initial side along the positive x-axis is said to be in standard position.
1
A ngle
The m easure of an angle describes the am ount and direction of rotation necessary to move from
the initial side to the term inal side of the angle. A positive angle is generated by a counterclockwise
rotation and a negative angle by a clockwise rotation.
P ositive A ngle
The most com m on angular unit of measure is
the degree (°), which is equivalent to
of a full
rotation (counterclockwise) about the vertex.
From the diagram shown, you can see that 360°
corresponds to 1 com plete rotation, 180° to a
1 rotation, 90° to a v rotation, and so on, as
2
4
marked along the circum ference of the circle.
StudyTip
Base 60 The concept of degree
Degree measures can also be expressed using a decimal degree form or a degree-minute-second
^(DMS) form where each degree is subdivided into 60 minutes (') and each m inute is subdivided
into 60 seconds (").
measurement dates back to the
ancient Babylonians, who made
early astronomical calculations
J O S H E S ] Convert Between DMS and Decimal Degree Form
using their number system, which
Write each decimal degree measure in DMS form and each DMS measure in decimal degree
form to the nearest thousandth.
was based on 60 (sexagesimal)
rather than on 10 (decimal) as we
do today.
a.
56.735°
First, convert 0.735° into minutes and seconds.
56.735° = 56° + 0.735°
= 5 6 ° + 44.1'
1° = 60'
Simplify.
Next, convert 0 .1 'in to seconds.
56.735° = 56° + 4 4 '+ 0.1'
= 5 6 ° + 4 4 '+ 6 "
1'=60"
Simplify.
Therefore, 56.735° can be written as 56° 44' 6".
TechnologyTip
b.
32°
5 '2 8 "
1
1
1
Each m inute is — of a degree and each second is — of a minute, so each second is ——
r
DMS Form You can use some
calculators to convert decimal
60
of a degree.
degree values to degrees, minutes,
and seconds using the D M S
32- 5' 28" = 32” + 5' ( J j ) + 2 8 " (J £ . )
function under the A n g le menu.
3600
^
6u
V . ± < n ana I ' . ^
= 32° + 0.083 + 0.008
Simplify.
= 32.091°
Add.
Therefore, 32° 5 ' 28" can be written as about 32.091°.
^ GuidedPractice
1B.
1A. 213.875°
30 56' 7"
V___________________
Measuring angles in degrees is appropriate w hen applying trigonometry to solve many real-world
problems, such as those in surveying and navigation. For other applications w ith trigonometric
functions, using an angle measured in degrees poses a significant problem. A degree has no
relationship to any linear measure; inch-degrees or - :*nc^t has no m eaning. M easuring angles
in radians provides a solution to this problem.
KeyConcept Radian Measure
Words
The measure 8 in radians of a central angle of a circle is
equal to the ratio of the length of the intercepted arc s to the
radius r of the circle.
Symbols
8 = j , where 8 is measured in radians (rad)
Example
A central angle has a measure of 1 radian if it intercepts an
arc with the same length as the radius of the circle.
v.
V 7
8 = 1 radian when s = r.
_____
, , , ...............
Notice that as long as the arc length s and radius r are measured using the same linear units, the
ratio j is unitless. For this reason, the word radian or its abbreviation rad is usually om itted when
writing the radian measure of an angle.
232
| Lesson 4-2
Degrees and Radians
y
StudyTip
Degree-Radian Equivalences
From the equivalence statement
shown, you can determine
The central angle representing one full rotation counterclockwise about a vertex corresponds to an
>arc length equivalent to the circum ference of the circle, 2ixr. From this, you can obtain the following
radian measures.
-1
1
1
1
1
1
2ttt
—rotation = •2 tt
—rotation = — •2tt
1 rotation
—rotation = — •2 tc
r
6
6
4
4
that 1° = 0.017 rad and
= 77 rad
= 2 tt rad
1 rad k 57.296°.
■rad
= f rad
Because 2tt radians and 360° both correspond to one com plete revolution, you can write 360° = 2tt
radians or 180° = it radians. This last equation leads to the follow ing equivalence statements.
ReadingMath
i° =
180
Angle Measure If no units of
angle measure are specified,
radians
1 radian : m \ °
\ it
and
Using these statements, we obtain the follow ing conversion rules.
radian measure is implied. If
degrees are intended, the degree
symbol (°) must be used.
KeyConcept Degree/Radian Conversion Rules
1. To convert a degree measure to radians, multiply by
2. To convert a radian measure to degrees, multiply by
•k radians
180° '
180°
irradians'
Convert Between Degree and Radian Measure
Write each degree measure in radians as a m ultiple of tv and each radian measure in degrees,
a. 120°
1 ?n° = 1 ?n° 171radians'\
180°
{
radians or
=
b.
180° ’
Simplify.
I tv r a d i a n s \
180°
Multiply by
-nr radians
I
= —^ ra d ia n s or —^
Simplify.
^ radians
6
Multiply by
5 tt
6
=
d.
i r radians
-4 5 °
-45° = -45°
c.
Multiply by
I
6
180°
■k
radians (— ^
) or 150°
\7T radians /
radians
Simplify.
3 ir
radians
=
2
Multiply by
radians (
18,°°
) or -2 7 0 °
\7T radians/
180°
i t ra dians'
Simplify.
p GuidedPractice
2A. 210°
2B. -60°
2C.
f
2D.
conncctED.m cgraw-hill.co m j |
233
ReadingMath
Naming Angles In trigonometry,
By defining angles in terms of their rotation about a vertex, two angles can have the same initial
, and terminal sides but different measures. Such angles are called coterminal angles. In the figures
>below, angles a and (3 are coterminal.
angles are often labeled using
P ositive and N egative
C oterm inal A ngles
Greek letters, such as a (alpha),
/3 (beta), and 6 (theta).
P ositive
C oterm inal A ngles
The two positive coterm inal angles shown differ by one full rotation. A given angle has infinitely
many coterm inal angles found by adding or subtracting integer m ultiples of 360° or 27r radians.
KeyConcept Coterminal Angles
Degrees
If a is the degree measure of an angle, then all angles
measuring a + 360n°, where n is an integer, are coterminal
Radians
If a is the radian measure of an angle, then all angles measuring
a + 2/77v, where n is an integer, are coterminal with a .
with a .
and Draw Coterminal Angles
Identify all angles that are coterm inal with the given angle. Then find and draw one positive
and one negative angle coterm inal with the given angle.
a. 45°
" I
All angles measuring 45° + 360n° are
coterm inal w ith a 45° angle. Let n = 1
and —1.
All angles measuring —— + 2mr are
coterm inal w ith a — ^ angle. Let n = 1
and —1.
45° + 360(1)° = 45° + 360° or 405°
►GuidedPractice
3A. - 3 0 °
234
Lesson 4 -2
Degrees and Radians
3B.
3t t
4
2
Applications w ith Angle Measure
Solving 9 = j for the arc length s yields a
convenient formula for finding the length of an arc of a circle.
KeyConcept Arc Length
If 9 is a central angle in a circle of radius r, then the length of the intercepted
arc s is given by
1
s = r9,
lv »
1
'
where 9 is measured in radians.
^
N --------/
J
J
1
W hen 9 is m easured in degrees, you could also use the equation s =
incorporates the degree-radian conversion.
Find Arc Length
StudyTip
Find the length of the intercepted arc in each circle with the given central angle measure and
radius. Round to the nearest tenth.
Notice in Example 4a that when
r = 5 centimeters and 9 = ^
Sir centimeters,4
Sir
4
not ^ centimeter-radians. This
>
a. -tv r = 5 cm
4
r e
Arc length
/ 7T\
4
is because a radian is a unitless
ratio.
I
r = 5 and 9 = ^
4
5( f )
5ir
4
l
The length of the intercepted arc is ^
b.
/
h
5 cm
J
Simplify.
or about 3.9 centimeters,
60°, r = 2 in.
Method 1
Convert 60° to radian measure, and then use s = r9
to find the arc length.
60° = 6 0 ° (™
d0ioa n s )
Multiply by
7T
Simplify.
3
II
Substitute r = 2 and 0
Arc length
s = rO
Method 2
■
k radians
180° '
- 2( i )
r=2and 0 =
H
Operating with Radians
radians, s =
w hich already
Simplify.
Use s =
loU
7u 9
180°
tt (2)(60°)
180°
2tt
3
to find the arc length.
Arc length
r = 2 and 0 = 60°
Simplify.
2-it
The length of the intercepted arc is — or about 2.1 inches.
►GuidedPractice
4A. | , r = 2 m
4B. 135°, r = 0.5 ft
235
&
The formula for arc length can be used to analyze circular motion. The rate at which an object
moves along a circular path is called its linear speed. The rate at w hich the object rotates about a
fixed point is called its angular speed. Linear speed is measured in units like m iles per hour, while
angular speed is measured in units like revolutions per minute.
KeyConcept Linear and Angular Speed
Suppose an object moves at a constant speed along a circular path of radius
r.
s ' • ' ' ’" ' - x
If s is the arc length traveled by the object during time
ReadingM ath
speed v is given by v=
Omega The lowercase Greek
If
t, then the object’s linear
then the
/
|
N'
'
J------ -
\ >r
6 is the angle of rotation (in radians) through which the object moves during time t,
letter omega u> is usually used to
denote angular speed.
/
1
|
1
angular speed u j of the object is given by
j
V
Find Angular and Linear Speeds
BICYCLING A bicycle messenger rides the bicycle shown.
a. During one delivery, the tires rotate at a rate of
140 revolutions per m inute. Find the angular speed
of the tire in radians per minute.
Because each rotation m easures 2tt radians,
140 revolutions correspond to an angle of rotation
9 of 140 x 27V or 2807V radians.
lo =
=
In some U.S. cities, it is
—
Angular speed
2807V ra d ia n s
— ^— :— -—
1 m in u te
„
0
„„„
,.
.
= 280 5r radians and
t —
. . . .
1minute
possible for bicycle
Therefore, the angular speed of the tire is 280ir or about 879.6 radians per minute.
messengers to ride an average of
30 to 35 miles a day while making
30 to 45 deliveries.
Source: New York Bicycle Messenger
Association
b.
On part of the trip to the next delivery, the tire turns at a constant rate of 2.5 revolutions
per second. Find the linear speed of the tire in miles per hour.
A rotation of 2.5 revolutions corresponds to an angle of rotation 9 of 2.5 x 27V or 57V.
v = j-
Linear speed
15(5tv ) in c h e s
= —
3
1 secon d
7517 in c h e s
— or —
r — 15 inches, 0 = 5 it radians, and
—
1 secon d
t —
1 second
Use dim ensional analysis to convert this speed from inches per second to miles per hour.
7 5 tv i o A e r ~ x
1 seeerrrd"
60 se c e fn fe - x
1 m » w tr
6 0 m itttr f e s - x
1 hour
L ie e t12 in e h e s"
1 m i l e _____ 1 3 .4 m ile s
5 2 8 0 fe e t-
hour
Therefore, the linear speed of the tire is about 13.4 miles per hour.
p GuidedPractice
MEDIA Consider the DVD shown.
5A. Find the angular speed of the DVD in radians per second if the
disc rotates at a rate of 3.5 revolutions per second.
5B. If the DVD player overheats and the disc begins to rotate at a slower
rate of 3 revolutions per second, find the disc's linear speed in
meters per minute.
236
| Lesson 4 -2
Degrees and Radians
Recall from geom etry that a sector of a circle is a region bounded by a central angle and its
intercepted arc. For exam ple, the shaded portion in the figure is a sector of circle P . The ratio of the
area of a sector to the area of a whole circle is equal to the ratio of the corresponding arc length to
the circum ference of the circle. Let A represent the area of the sector.
A
_ le n g th o f QRS
irr2
A
area of sector _
arc length
area of circle — circum ference o f circle
2m'
_
■nr2
rO
The length of QRS is rO.
2~nr
Solve for A
A = j r 29
KeyConcept Area of a Sector
The area A of a sector of a circle with radius rand central angle 9 is
0
/
A = \ r2e,
where 9 is measured in radians.
w
^ ---------^
p r T fT n fffT h Find Areas of Sectors
a. Find the area of the sector of the circle.
77T
The measure of the sector's central angle 9 is — , and the radius
is 3 centimeters.
Area of a sector
A = - | r 20
M
, 63 tt
f ) '
16
r = 3 and 0 ■-
7-k
Therefore, the area of the sector is -y -- or about 12.4 square centimeters.
b.
WIPERS Find the approximate area swept by
the w iper blade shown, if the total length of
the windshield w iper mechanism is 26 inches.
The area swept by the wiper blade is the
difference betw een the areas of the sectors with
radii 26 inches and 26 — 16 or 10 inches.
Convert the central angle measure to radians.
130° = 130°
Real-W orldLink
( 7r radians j _ 13tt
18
180°
Then use the radius of each sector to find the area swept. Let A 1 = the area of the sector with
a 26-inch radius, and let A 2 = the area of the sector w ith a 10-inch radius.
A typical wipe angle for a
front windshield wiper of a
Swept area
■A,
A = A,
passenger car is about 67°.
Windshield wiper blades are
= }(2 6 )2
- h
Area of a sector
generally 12-30 inches long.
Source: Car and Driver
_ 2197-rr
325 tt
9
9
= 2087V or about 653.5
Simplify.
Simplify.
Therefore, the swept area is about 653.5 square inches.
GuidedPractice
Find the area of the sector of a circle with the given central angle 0 and radius r.
6A. 9 = ^ - , r = 1.5 ft
4
6B. 9 = 50°, r = 6 m
connectED.mcgraw-hill.com~H
237
* 5 5 /5 (0
Exercises
= Step-by-Step Solutions begin on page R29.
Write each decimal degree measure in DMS form and each
DMS measure in decimal degree form to the nearest
thousandth. (Example 1)
1. 11.773°
2. 58.244°
3. 141.549°
4. 273.396°
5. 87° 53' 10"
7. 45° 2 1 '2 5 "
Find the length of the intercepted arc with the given central
angle measure in a circle with the given radius. Round to the
nearest tenth. (Example 4)
28. ■yv r = 3 in.
6. 126° 6 '3 4 "
29. % r = 4 yd
30. 105°, r = 18.2 cm
8. 301°
31. 45°, r = 5 mi
'8 "
9. NAVIGATION A sailing enthusiast uses a sextant, an
instrument that can measure the angle betw een two
objects with a precision to the nearest 10 seconds, to
measure the angle betw een his sailboat and a lighthouse.
If his reading is 17° 37' 50", what is the measure in
decimal degree form to the nearest hundredth? (Example 1)
CO
N3
27. f , r = 2.5 m
4 2
*£ x
150°, r = 79 mm
33) AMUSEMENT PARK A carousel at an am usem ent park
rotates 3024° per ride. (Example 4)
a. H ow far would a rider seated 13 feet from the center of
the carousel travel during the ride?
b. H ow m uch farther would a second rider seated 18 feet
from the center of the carousel travel during the ride
than the rider in part a?
Find the rotation in revolutions per minute given the
angular speed and the radius given the linear speed and the
rate of rotation. (Example 5)
35. w = 135tt
Write each degree measure in radians as a multiple of tv and
each radian measure in degrees. (Example 2)
12. -1 6 5 °
13.
14.
16.
2 tv
15.
3
_7T
I
11. 225°
Ul
o
10. 30°
5-jt
2
17. _7 rr
4
6
Identify all angles that are coterminal with the given angle.
Then find and draw one positive and one negative angle
coterminal with the given angle. (Example 3)
18. 120°
19.
20. 225°
21.
22.
24.
TV
23.
3
7T
25.
12
3tt
’ 4
377
2
26. GAME SHOW Sofia is spinning
There are 20 values in equ al-sized spaces around the
circum ference of the w heel. The value that Sofia needs
to w in is two spaces above the space w here she starts
her spin, and the w heel m ust m ake at least one full
rotation for the spin to count. D escribe a spin rotation
in degrees that w ill give Sofia a w inning result.
(Example 3)
Space Sofia
needs to land on
Start of
Sofia’s spin
238
Lesson 4-2
Degrees and Radians
36.
lo
= 104tt
rad
ra d
37. v = 82.3 — , 131 -=^rs
38. v = 144.2 - 4 - , 10.9 —
mm
mm
mm
39. w = 5 5 3 ^ , 0 . 0 9 ^
h
mm
40. MANUFACTURING A com pany m anufactures several
circular saws w ith the blade diam eters and motor
speeds show n below. (Example 5)
Blade
Motor
Diameter (in.)
Speed (rps)
3
2800
5
5500
4
4500
5500
5000
a. Determ ine the angular and linear speeds of the blades
in each saw. Round to the nearest tenth.
b. How m uch faster is the linear speed of the 6-i-inch
saw com pared to the 3-inch saw?
41. CARS On a stretch of interstate, a vehicle's tires range
betw een 646 and 840 revolutions per minute. The
diam eter of each tire is 26 inches. (Example 5)
a. Find the range of values for the angular speeds of the
tires in radians per minute.
b. Find the range of values for the linear speeds of the
tires in miles per hour.
^ g n °
1. v i
42. TIME A wall clock has a face diam eter of 8 - j inches, th i
length of the hour hand is 2.4 inches, the length of the
minute hand is 3.2 inches, and the length of the second
hand is 3.4 inches. (Example 5)
V r o
55. Describe the radian m easure betw een 0 and 27r of an
angle 9 that is in standard position w ith a terminal side
that lies in:
c. Q uadrant I I I
d. Quadrant IV
a. Q uadrant I
b. Q uadrant II
56. If the term inal side of an angle that is in standard position
lies on one of the axes, it is called a quadrantal angle. Give
the radian m easures of four quadrantal angles.
a. Determine the angular speed in radians per hour and
the linear speed in inches per hour for each hand.
b. If the linear speed of the second hand is 20 inches per
m inute, is the clock running fast or slow? How much
time would it gain or lose per day?
57. GEOGRAPHY Phoenix, Arizona, and Ogden, Utah, are
located on the same line of longitude, w hich means that
Ogden is directly north of Phoenix. The latitude of
Phoenix is 33° 26' N, and the latitude of Ogden is
41° 12' N. If Earth's radius is approxim ately 3963 miles,
about how far apart are the two cities?
110 W
120 W
GEOMETRY Find the area of each sector. (Example 6)
Find the measure of angle 9 in radians and degrees.
48.
61.
49. GAMES The dart board shown is divided into twenty
equal sectors. If the diameter of the board is 18 inches,
what area of the board does each sector cover? (Example 6)
62. TRACK The curve of a standard 8-lane track is semicircular
as shown.
50. LAWN CARE A sprinkler waters an area that forms
one third of a circle. If the stream from the sprinkler
extends 6 feet, what area of the grass does the sprinkler
water? (Example 6)
The area of a sector of a circle and the measure of its central
angle are given. Find the radius of the circle.
511 A = 29 ft2, 9 = 68°
53. A = 377 in2, 9 = ^
52. A = 808 crti2, 9 = 210°
54. A = 75 m 2, 0 =
37V
4
1.22 m
a. W hat is the length of the outside edge of Lane 4 in
the curve?
b. How m uch longer is the inside edge of Lane 7 than
the inside edge of Lane 3 in the curve?
r-p
&
c o n n e c tE D .m c g ra w -h ill.c o m
| 239
63. DRAMA A pulley with radius r is being used to remove
part of the set of a play during intermission. The height of
the pulley is 12 feet.
a. If the radius of the pulley is 6 inches and it rotates
180°, how high will the object be lifted?
b. If the radius of the pulley is 4 inches and it rotates
900°, how high will the object be lifted?
72. SKATEBOARDING A physics class conducted an experiment
to test three different wheel sizes on a skateboard with
constant angular speed.
a. W rite an equation for the linear speed of the
skateboard in terms of the radius and angular speed.
Explain your reasoning.
b. Using the equation you wrote in part a, predict the
linear speed in meters per second of a skateboard
w ith an angular speed of 3 revolutions per second
for w heel diam eters of 5 2 ,5 6 , and 60 millimeters.
C. Based on your results in part b, how do you think
wheel size affects linear speed?
H.O.T. Problem s
64. ENGINEERING A pulley like the one in Exercise 63 is being
used to lift a crate in a warehouse. Determ ine w hich of
the following scenarios could be used to lift the crate a
distance of 15 feet the fastest. Explain how you reached
your conclusion.
I. The radius of the pulley is 5 inches rotating at
65 revolutions per minute.
Use Higher-Order Thinking Skills
73. ERROR ANALYSIS Sarah and M ateo are told that the
perim eter of a sector of a circle is 10 times the length of
the circle's radius. Sarah thinks that the radian measure
of the sector's central angle is 8 radians. M ateo thinks
that there is not enough inform ation given to solve the
problem. Is either of them correct? Explain your
reasoning.
74. CHALLENGE The two circles show n are concentric. If the
length of the arc from A to B m easures 87T inches and
DB = 2 inches, find the arc length from C to D in terms
of 7T.
II. The radius of the pulley is 4.5 inches rotating at
70 revolutions per minute.
III. The radius of the pulley is 6 inches rotating at
60 revolutions per minute.
GEOMETRY Find the area of each shaded region.
REASONING D escribe how the linear speed would change for
each param eter below. Explain.
75. a decrease in the radius
76. a decrease in the unit of time
67. CARS The speedom eter shown measures the speed of a
car in miles per hour.
77. an increase in the angular speed
78. PROOF If y - =
prove that 9^ = 62-
79. REASONING W hat effect does doubling the radius of a
circle have on each of the follow ing measures? Explain
your reasoning.
a. If the angle betw een 25 m i/h and 60 mi/h is 81.1°,
about how many miles per hour are represented by
each degree?
b. If the angle of the speedom eter changes by 95°, how
much did the speed of the car increase?
Find the complement and supplement of each angle,
if possible. If not possible, explain your reasoning.
68. 2 *5
240
69. —
6
L esson 4 -2
70. —
8
D e g re e s a n d R a d ia n s
71.
3
a. the perim eter of the sector of the circle w ith a central
angle that m easures 9 radians
b. the area of a sector of the circle w ith a central angle
that measures 9 radians
80. WRITING IN MATH Compare and contrast degree and radian
m easures. Then create a diagram sim ilar to the one on
page 231. Label the diagram using degree measures on
the inside and radian m easures on the outside of the
circle.
Spiral Review
Use the given trigonometric function value of the acute angle 6 to find the exact values of the
five remaining trigonometric function values of 0. (Lesson 4-1)
81.
4 V7
82.sec6> =
sin 0 = ^
^
83.cot 0 =
84. BANKING A n account that H ally's grandm other opened in 1955
earned continuously com pounded interest. The table shows the
balances of the account from 1955 to 1959. (Lesson 3-5)
a. Use regression to find a function that m odels the am ount in
the account. Use the num ber of years after Jan. 1 ,1 9 5 5 , as the
independent variable.
b. Write the equation from part a in terms of base e.
C.
W hat was the interest rate on the account if no deposits or
withdrawals were m ade during the period in question?
Date
Balance
1
Jan. 1,1955
$2137.52
2
Jan. 1,1956
$2251.61
3
Jan. 1,1957
$2371.79
4
Jan. 1,1958
$2498.39
5
Jan. 1,1959
$2631.74
Express each logarithm in terms of In 2 and In 5. (Lesson 3-2)
85. I n #
86.
16
87.
I n 250
10
In
25
List all possible rational zeros of each function. Then determ ine which, if any, are
zeros. (Lesson 2-4)
88.
89.g (x ) = x 3 — 5x2 - 4x + 20
f(x) = x 4 - x 3 - 12x — 144
90.g (x) = 6x4 + 35x3 — x 2 — 7x — 1
Write each set of num bers in set-builder and interval notation if possible. Lesson 1-3)
91 . f( x ) = Ax5 + 2x4 - 3x - 1
92.g(x)= - x 6 + x4 - 5x2 + 4
93.h(x) = —
+ 2
Describe each set using interval notation. (Lesson 0-1)
94. n > - 7
96. y < 1 or y > 11
95.- 4 < x < 10
Skills Review fo r Standardized Tests
97. SAT/ACT In the figure, C and D are the centers of the
two circles w ith radii of 3 and 2, respectively. If the
larger shaded region has an area of 9, what is the area
of the smaller shaded region?
99. REVIEW If sec
25
, then sin 0 =
7_
A
25
B
24
25
C — or - 2 i
25
25
D
25
7
100. W hich of the follow ing radian m easures is equal
Note: Figure not drawn to scale.
A 3
C 5
B 4
D 7
E 8
98. REVIEW If cot 0 = 1 , then tan 0 =
F -1
GO
to 56°?
F -2L
15
G —
45
H ^
45
I -3
J
H I
J 3
241
Trigonometric Functions
on the Unit Circle
Then
•
You found values
of trigonometric
functions for acute
angles using ratios
in right triangles.
Now
* 1
(Lesson 4-1)
NewVocabulary
quadrantal angle
reference angle
unit circle
circular function
periodic function
period
Why?
Find values of
trigonometric functions
for any angle.
2
A blood pressure of 120 over 80, measured in millimeters of mercury,
means that a person’s blood pressure oscillates or cycles between
20 millimeters above and below a pressure of 100 millimeters of
mercury for a given time fin seconds. A complete cycle of this
oscillation takes about 1 second.
Find values of
trigonometric
functions using the
unit circle.
If the pressure exerted by the blood at time t = 0.25 second
is 120 millimeters of mercury, then at time t — 1.25 seconds
the pressure is also 120 millimeters of mercury.
Trigonometric Functions of Any Angle
In Lesson 4-1, the definitions of the six
trigonometric functions were restricted to positive acute angles. In this lesson, these
definitions are extended to include any angle.
1
KeyConcept Trigonometric Functions of Any Angle
Let 9 be any angle in standard position and point P{x, y) be a point on the terminal side
of 9. Let /represent the nonzero distance from Pto the origin.
That is, let r= \/x2 + y2 ± 0. Then the trigonometric functions of 9 are as follows.
sin 6 = j
csc0 = y,y # O
cos 9 = j
sec 9 =
ta n 0 = i
xj= 0
cot 9 = ^,yj= 0
xj=0
Evaluate Trigonometric Functions Given a Point
Let (8, —6) be a point on the terminal side of an angle 9 in standard position. Find the exact
values of the six trigonometric functions of 9.
Use the values of x and y to find r.
y
Pythagorean Theorem
r = \]x2 + y 2
.— - x 61
= V s 2 + (—6 )2
x = 8 a n d y = —6
= V lO O or 1 0
Take the positive square root.
d
*
(8, —6 ) V
Use x
=
8,
y =
—6, and r
sin 0
= — = — ppor
csc 9
=
10
r
—
y
=
-6
or
—
—
5
3
=
1 0 to write the six trigonom etric ratios.
cos 0
= — =
sec 9
= — = ip
r
x
10
8
or
or
5
j
4
tan 9
= — =
x
8
or —
3
4
4
cot 9 = — =
or —
3
y
-6
►GuidedPractice
The given point lies on the terminal side of an angle 0 in standard position. Find the values
of the six trigonometric functions of 9.
1A. (4 ,3 )
a
242
L esson 4-3
1B. ( - 2 , - 1 )
In Exam ple 1, you found the trigonom etric values of 9 w ithout know ing the measure of 9. Now
we will discuss methods for finding these function values w hen only 9 is known. Consider
trigonom etric functions of quadrantal angles. W hen the term inal side of an angle 9 that is in
standard position lies on one of the coordinate axes, the angle is called a quadrantal angle.
KeyConcept Common Quadrantal Angles
StudyTip
y
Quadrantal Angles There are
y
y
y
(0,
infinitely many quadrantal angles
r)‘
that are coterminal with the
quadrantal angles listed at the
0
0
right. The measure of a quadrantal
X
0
(r, 0) x
angle is a multiple of 90° or y .
*(—r, 0)
X
0
\
X
0
(0 ,-r )
9 = 0° or 0 radians
6=
90° or y radians
e = 180°
or t radians
0 = 270° or -y - radians
I
You can find the values of the trigonom etric functions of quadrantal angles by choosing a point on
the term inal side of the angle and evaluating the function at that point. Any point can be chosen.
However, to sim plify calculations, pick a point for w hich r equals 1.
H ^ J J S S E iE v a lu a t e Trigonometric Functions of Quadrantal Angles
Find the exact value of each trigonom etric function, if defined. If not defined, write
u n defin ed.
a.
sin (-1 8 0 ° )
The terminal side of —180° in standard position lies on
the negative x-axis. Choose a point P on the term inal side
of the angle. A convenient point is (—1, 0) because r = 1.
Sine function
sin (-1 8 0 ° ) = f
y = 0 a n d r= 1
= y or 0
b.
tan
The terminal side of
in standard position lies on the
negative y-axis. Choose a point P(0, —1) on the term inal
side of the angle because r = 1.
t
3tx V
ta n - = I
= —jy or undefined
Tangent function
y = —1 and x —0
C. s e c 4iv
The term inal side of 4tt in standard position lies on the
positive x-axis. The point (1, 0) is convenient because r = 1.
sec 4 tt = ■
= y or 1
Secant function
r= 1 a n d x = l
p GuidedPractice
2A. cos 270°
2B. c s c f
2C. cot ( - 9 0 °
connectED.mcgraw-hill.com
I
243
To find the values of the trigonometric functions of angles that are neither acute nor quadrantal,
consider the three cases shown below in w hich a and b are positive real numbers. Compare the
values of sine, cosine, and tangent of 9 and 9 '.
StudyTip
Quadrant II
Quadrant III
y
y
i \
Reference Angles Notice that in
-.0
<
theta
a
prime) are the same, in other
y
r
some cases, the three trigonometric
values of 0 and 0 ' (read
Quadrant IV
O
'
*
a
i,—^ /jA
0 ^ 0
X
X
-
C V 0
V
0i / r
cases, they differ only in sign.
- b)
(a, - 6 ) ^
sin 9 = j
sin 9' = j
sin 0 = - j
sin 9' = j
sin 0 = —j
sin 9' = j
cos 9 = - j
cos 0 ' = |
cos 0 = —j
cos 9' = j
cos 0 = 7
cos 0 ' = j
tan 9 = - |a
tan 0 ' = |a
tan 0 = 4a
tan 0 ' = -|a
tan 0 = — 4a
tan 0 ' = a
This angle 9 ', called a reference angle, can be used to find the trigonometric values of any angle I
KeyConcept Reference Angle Rules
if
9 is an angle in standard position, its reference angle 0 ' is the acute angle formed by the terminal side of 0 and the x-axis. The
O'for any angle 0, 0° < 6 < 360° or 0 < 9 < 2ir, is defined as follows.
reference angle
Quadrant II
\y
o0
9’
9'=9
9' =
180° —
9
0 '= i r - 0
9'= 0 - 1 8 0 °
9' = 9 - tv
0' =
9
360° -
0'= 2tt - 9
To find a reference angle for angles outside the interval 0° < 9 < 360° or 0 < 9 < 2 n , first find a
corresponding coterm inal angle in this interval.
H 2 E H 3 3 D Find Reference Angles
Sketch each angle. Then find its reference angle.
a.
b.
300°
The terminal side of 300° lies in
Q uadrant IV. Therefore, its reference angle
is 9' = 360° - 300° or 60°.
27T
A coterminal angle is 2 tt —
terminal side of — lies in Q uadrant III, so
J
.
its reference angle is
— tt or .
p GuidedPractice
3A. ^
244
3B. - 2 4 0 °
| Lesson 4 -3 I T r ig o n o m e tric F u n c tio n s o n th e U n it C irc le
or 4 ^ . The
3C. 390°
B ecause the trig on o m etric valu es of an an gle and its referen ce an gle are equ al or differ only
in sign, you can use the fo llow in g steps to fin d the valu e o f a trig on o m etric fu n ction of any
angle 9.
KeyConcept Evaluating Trigonometric Functions of Any Angle
ETHTn
Find the reference angle 9'.
y
PT7!TIW Find the value of the trigonometric function for 9'.
ETTBtil Using the quadrant in which the terminal side of 0 lies, determine the sign
"" i
The signs of the trigonom etric functions in each quadrant can
be determ ined using the function definitions given on page 242.
it follows that sin 9 is negative
when y < 0, w hich occurs in Quadrants III and IV. Using this
same logic, you can verify each of the signs for sin 9, cos 9, and
tan 9 shown in the diagram. N otice that these values depend
only on x and y because r is always positive.
StudyTip
Memorizing Trigonometric Values
To memorize the exact values of
sine for 0°, 30°, 45°, 60°, and 90°,
consider the following pattern.
sin 0° =
Quadrant II
Quadrant 1
sin 9: +
cos 9: tan 9: -
sin 9: +
cos 9: +
tan 9: +
Quadrant III
Quadrant IV
sin ft —
cos ft +
tan ft -
sin 9: cos 9: tan 9: +
Because you know the exact trigonom etric values of 30°, 45°, and 60° angles, you can find the exact
trigonometric values of all angles for w hich these angles are reference angles. The table lists these
values for 9 in both degrees and radians.
9
or 0
30° o r £
6
45° o r £
4
1
V2
2
V3
2
V2
2
1
2
1
V3
sin 9
sin 30° = ^ , or±
cos 9
Vz
sin 45° = y f
tan 9
sin 60° = ^
sin 90° =
\ ~ 'l
y
y
For exam ple, because sin 9 =
9
^
2
V3
2
V3
3
60° or y
or 1
A similar pattern exists for the
cosine function, except the values
are given in reverse order.
H 2 J y U J 3 3 Use Reference Angles to Find Trigonometric Values
Find the exact value of each expression,
a.
(
cos 120°
l
,
°
)
Because the terminal side of 9 lies in Q uadrant II,
the reference angle 9' is 180° — 120° or 60°.
cos 120° = —cos 60°
In Quadrant II, cos 0 is negative.
( 0
cos 60° = -1
b.
tan
0-
6
Because the terminal side of 9 lies in Q uadrant III,
th e r e fe r e n c e a n g le
tan
6
= tan ^
6
y/3
3
'TT
9' i s -
— iv
i t or
or
TV
—.
In Quadrant III, tan 0 is positive.
V3
tanf = -3
C=
&
connectED.mcgraw-hill.com |
245
C. CSC
15tt
A coterm inal angle of 9 is — ------ 27T or — , w hich lies in
77T
tv
Q uadrant IV. So, the reference angle 9' is 27T — — ° r —.
Because sine and cosecant are reciprocal functions and
sin 9 is negative in Q uadrant IV, it follows that csc 9 is
also negative in Q uadrant IV.
15 tv
csc -
In Quadrant IV, csc 0 is negative.
-csc ~
4
csc 0 = - 1
sin 0
sin —
4
=
7 =- or
V2
— \/2
sin i
4
= — ■
2
2
CHECK You can check your answer by using a graphing calculator.
157T
csc —— ' -1 .4 1 4 ✓
4
-y /2 :
-1 .4 1 4 ✓
p GuidedPractice
Find the exact value of each expression.
4A. tan
4B. sin
4C. sec(—135°)
6
If the value of one or more of the trigonom etric functions and the quadrant in which the terminal
side of 9 lies is known, the remaining function values can be found.
E S ! I 3 3 Use One Trigonometric Value to Find Others
_5
12
Let tan 9 = rrr, where sin 6 < 0. Find the exact values of the five rem aining trigonometric
functions of 9.
To find the other function values, you m ust find the coordinates of a point on the terminal side
of 9. You know that tan 9 is positive and sin 9 is negative, so 9 must lie in Q uadrant III. This
m eans that both x and y are negative.
y
5
Because tan 9 = —or — , use the point (—12, —5) to find r.
' = \fx 2 ^ y 2
Pythagorean Theorem
= V ( - 1 2 ) 2 + (—5)2
x = — 12 and y = —5
= V l6 9 or 13
Take the positive square root.
Use x = —12, y = —5, and r = 13 to write the five rem aining trigonom etric ratios.
sin 9 = —or —-^r
r
13
csc 9 = — or ■
Rationalizing the Denominator Be
sure to rationalize the denominator,
if necessary.
L esson 4-3
sec i
12
13
cot 9 = ^ or y
-X° r ~ r12!
p GuidedPractice
WatchOut!
246
13
' 5
cos V = —or
r
Find the exact values of the five rem aining trigonom etric functions of 6.
5A. sec 9 = V 3 , tan 9 < 0
T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle
5B. sin 9 =
cot 9 > 0
Real-World Example 6 Find Coordinates Given a Radius and an Angle
ROBOTICS As part of the range of motion category in a high
school robotics com petition, a student program m ed a
20-centim eter long robotic arm to pick up an object at point C
and rotate through an angle of exactly 225° in order to release it
into a container at point D. Find the position of the object at
point D, relative to the pivot point O.
225“
W ith the pivot point at the origin and the angle through which
the arm rotates in standard position, point C has coordinates
(20, 0). The reference angle 9' for 225° is 225° — 180° or 45°.
Let the position of point D have coordinates (x, y). The definitions of sine and cosine can then be
used to find the values of x and y. The value of r, 20 centim eters, is the length of the robotic arm.
Since D is in Q uadrant III, the sine and cosine of 225° are negative.
cos 9 = —
cos 225° =
RoboCup is an international
competition in which teams
compete in a series of soccer
matches, depending on the size
and intelligence of their robots. The
aim of the project is to advance
artificial intelligence
and robotics research.
Source: RoboCup
—cos 45° =
_V 2
2
x
20
x
20
x
■
20
-1 0 V 2 = :
Cosine ratio
Sine ratio
0 = 225° and r = 20
cos 225° = —cos 45°
cos 45° =
2
Solve for x.
y_
sin 225°
20
sin 225° = - s i n 45°
- s i n 45° =
y_
_V 2.
2
20
-10V 2 =
9 = 225° and r = 20
y
sin 45'
2
Solve for y.
The exact coordinates of D are ( —I 0 V 2 , —I 0 V 2 ). Since 10\/2 is about 14.14, the object is about
14.14 centim eters to the left of the pivot point and about 14.14 centim eters below the pivot point.
p GuidedPractice
6. CLOCKWORK A 3-inch-long m inute hand on a clock
shows a time of 45 m inutes past the hour. W hat is the
new position of the end of the m inute hand relative
to the pivot point at 10 m inutes past the next hour?
2
Trigonometric Functions on the Unit Circle
a
unit circle is a circle of radius 1 centered at the origin.
Notice thaton a unit circle, the radian m easure of a central
angle 9 = y or s, so the arc length intercepted by 9 corresponds
exactly to the angle's radian measure. This provides a way
of mapping a real number input value for a trigonometric
function to a real num ber output value.
StudyTip
Wrapping Function The
association of a point on the
number line with a point on a circle
is called the wrapping function,
w(t). For example, if w(t)
associates a point f on the number
line with a point P(x, y) on the unit
circle, then w(ir) = ( - 1 , 0 ) and
w(2ir) = (1,0).
Consider the real num ber line placed vertically tangent to the unit circle at (1, 0) as shown below.
If this line were wrapped about the circle in both the positive (counterclockwise) and negative
(clockwise) direction, each point t on the line would map to a unique point P(x, y) on the circle.
Because r = 1, we can define the trigonom etric ratios of angle t in terms of just x and y.
P o s itiv e V a lu e s o f t
N e g a tiv e V a lu e s o f t
KeyConcept Trigonometric Functions on the Unit Circle
Let f be any real number on a number line and let P(x, y) be the point on t
when the number line is wrapped onto the unit circle. Then the trigonometric
functions of fare as follows.
P (x,y) or
P (c o s t, sin t)
cos t = x
ta n f= |,x ^ 0
c s c f= ly = jtO
s e c t = j,x j= 0
c o tf= |,y ^ 0
Therefore, the coordinates of P corresponding to the angle f can be written as
P(cos t, sin t).
^ - <
Ac
V
x
sin f = y
y
)
X
L
J
Notice that the input value in each of the definitions above can be thought of as an angle measure
or as a real number t. W hen defined as functions of the real num ber system using the unit circle,
the trigonometric functions are often called circular functions.
Using reference angles or quadrantal angles, you should now be able to find the trigonometric
function values for all integer m ultiples of 30°, or ^ radians, and 45°, or ^ radians. These special
values wrap to 16 special points on the unit circle, as show n below.
16-Point Unit Circle
StudyTip
16-Point Unit Circle You have
already memorized these values in
the first quadrant. The remaining
values can be determined using the
x-axis, y-axis, and origin symmetry
of the unit circle along with the
signs of xand yin each quadrant.
Using the (x, y ) coordinates in the 16-point unit circle and the definitions in the Key Concept Box
at the top of the page, you can find the values of the trigonom etric functions for com m on angle
measures. It is helpful to m em orize these exact function values so you can quickly perform
calculations involving them.
P E E E 0 3 2 Find Trignometric Values Using the Unit Circle
Find the exact value of each expression. If undefined, w rite u n d efin ed.
a.
s in f
•j corresponds to the point (x, y) = |y, ^ - J on the unit circle.
sin t = y
7T V3
sin y = “ 2 “
248
Definition of sin t
V3 .
. Tt
y=-y-when f = y -
| Lesson 4 -3 j T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle
b.
co s 135°
135° corresponds to the point (x, y) =
Definition ofcost
cos t = x
cos 135°=
C.
° n the unit circle.
~ ~ ^ r when *= 135°'
ta n 2 7 0 °
270° corresponds to the point (x, y) = (0, —1) on the unit circle.
y
Definition of tan t
tan t = j
x = 0 and y = -1 , when t= 270°.
tan 270° =
Therefore, tan 270° is undefined.
d.
c s c ^
l l i v __________ .
\ (V 3
1
corresponds to the point (x, y) = {——, —i j on the unit circle.
Definition of csc t
csc t = —
' y
c s c -^ t— = ~^r
6
i
y = —-J-whenf
2
2
= —2
Simplify,
p GuidedPractice
7A. cos ~
4
StudyTip
Radians vs. Degrees While we
could also discuss one wrapping
as corresponding to an angle
measure of 360°, this measure is
not related to a distance. On the
unit circle, one wrapping
corresponds to both the angle
measuring 2 ir and the distance
2 tt around the circle.
7B.sin 120°
7C.cot 210°
7D.sec ~
4
As defined by wrapping the num ber line around the unit circle, the dom ain of both the sine and
cosine functions is the set of all real numbers (—o o , o o ). Extending infinitely in either direction, the
number line can be wrapped m ultiple times around the unit circle, m apping more than one f-value
to the same point P(x, y) with each wrapping, positive or negative.
Because cos t = x, sin t = y, and one w rapping corresponds to a distance of 2tt,
cos (t + 2«7r) = cos t
and
sin (t + 2 mr) = sin t,
for any integer n and real num ber t.
71
connectED.mcgraw-hill.com
§
249
StudyTip
Periodic Functions The other
three circular functions are also
periodic. The periods of these
functions will be discussed in
Lesson 4-5.
The values for the sine and cosine function therefore lie in the interval [—1 ,1 ] and repeat for every
integer m ultiple of 2-tt on the number line. Functions w ith values that repeat at regular intervals are
called periodic functions.
V
K eyConcept Periodic Functions
A function y = f(t) is periodic if there exists a positive real number c such that f(t + c) = f[t) for all values of t in the domain of f.
The smallest number c for which f is periodic is called the period of f.
The sine and cosine functions are periodic, repeating values after 27T, so these functions have a
period of 2 t t . It can be show n that the values of the tangent function repeat after a distance of t t on
the number line, so the tangent function has a period of tt and
ta n t = t a n ( t 4- H7r),
for any integer n and real number f, unless both tan t and tan (t + ntt) are undefined. You can use
the periodic nature of the sine, cosine, and tangent functions to evaluate these functions.
w
m
Use the Periodic Nature of Circular Functions
m
Find the exact value of each expression.
_
H it
a. cos ——4
cos —r — = cos ( + 2 tt I
4
V 4
= cos ~
I
Rewrite -US- as the sum of a number and 2-re.
4
SE and ^
4
4
4
+ 2it map to the same point (x, y) = [ —— ,
\
1on
2
2 /
the unit circle.
= —^ r 2
b. sin
cos f = xand x = w h e t ? f = - ^ ,
2
4
(-f)
sin f —
= sin
+ 2 (—l)T rj
= s in 4 ?
Rewrite
as the sum of a number and an integer multiple of 2x.
4 ? and ™ - 2(—1)ir map to the same point (x, y) = ( - ~
J
J
\
C.
C. /
on
the unit circle.
= —
„
,
si nf = y a n d y = —^ p w h e n ? = 4 p
19-jt
6
C. tan ——
ta n
o
= ta n (^- + 37t)
\o
/
= tan —
6
Rewrite
6
and
6
6
as the sum of a number and an integer multiple of it.
-f 3 it map to points on the unit circle with the same
tangent values.
= - 7= o r - ^ V3
3
tan t = ~ ; x = ~ and y = ~r when t —^ .
x’
2
’
2
6
~T
^ GuidedPractice
8A. sin
8B.cos
8C.t a n ~ -
Recall from Lesson 1-2 that a function/is even if for every x in the dom ain o t f f ( —x) = f( x ) and odd
if for every x in the domain o f f , f ( —x) = —f(x ). You can use the unit circle to verify that the cosine
function is even and that the sine and tangent functions are odd. That is,
cos (—f) = cos t
250
L esson 4 -3
T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle
sin (—t) = —sin f
tan (—f) = —tan t.
Exercises
m
The given point lies on the term inal side of an angle 9 in
standard position. Find the values o f the six trigonom etric
functions of 0. (Example 1)
1. (3,4)
21 ( - 6 , 6 )
3. ( - 4 , - 3 )
4. (2 ,0 )
5. (1, - 8 )
6. (5, - 3 )
7. ( - 8 ,1 5 )
8. ( - 1 , - 2 )
4 1 ) CAROUSEL Zoe is on a carousel at the carnival. The
diam eter of the carousel is 8 0 feet. Find the position of
her seat from the center of the carousel after a rotation
of 2 1 0 ° . (Example 6)
J
Find the exact value o f each trigonom etric fu nction, if
defined. If not d efined, w rite u n d efin ed. (Example 2)
9. s i n y
11. cot (-1 8 0 °)
13. cos (-2 7 0 °)
©
tan 27T
csc 270°
sec 180°
15. tan tt
SCC( - f
Sketch each angle. T h en find its reference ;
17. 135°
19.
18. 210°
©
7tr
12
I I tv
3
22. - 7 5 °
21. -4 0 5 °
s>
13tt
6
5ir
23.
6
Find the exact value o f each expression. (Example 4)
4tv
25. cos —
,
7ir
26. tan —
6
27. s i n ^ f
4
28. cot (—45°)
29. csc 390°
30. sec (-1 5 0 ° )
, IItt
31. tan ——
6
32. sin 300°
Find the exact values o f the five rem aining trigonom etric
functions of 0. (Example 5)
33. tan 9 = 2, where sin 9 > 0 and cos 9 > 0
34. csc 9 = 2, where sin 9 > 0 and cos 9 < 0
35. sin 9 = ——, where cos 9 > 0
5
36. cos 9 = —j j , where sin 9 < 0
37. sec 9 = \[?>, where sin 9 < 0 and cos 9 > 0
38. cot 9 = 1 , where sin 9 < 0 and cos 9 < 0
39. tan 0 = —1, where sin 0 < 0
40. cos 9 -
42. COIN FUNNEL A coin is dropped into a funnel where it
spins in sm aller circles until it drops into the bottom of
the bank. The diam eter of the first circle the coin makes is
24 centimeters. Before com pleting one full circle, the coin
travels 150° and falls over. W hat is the new position of the
coin relative to the center of the funnel? (Example 6)
1
, where sin 9 > 0
Find the exact value o f each expression. If un defined, write
u n defin ed. (Examples 7 and 8)
43. sec 120°
44. sin 315°
45. c o s ^
46. tan (-5 2 1 )
47. csc 390°
48. cot 510°
49. csc 5400°
50. sec
5tt\
17tv
51. cot
(-? )
52. csc
53. tan
5tt
54. s e c -^p
6
55. sin —
56. cos
57. tan 14tt
58. cos
(-¥ )
7tt
(-t
)
59. RIDES Jae and Anya are on a ride at an amusement park.
After the first several sw ings, the angle the ride makes
with the vertical is m odeled by 9 = 22 cos 7rf, w ith 9
m easured in radians and t measured in seconds.
Determ ine the m easure of the angle in radians for t = 0,
0 .5 ,1 ,1 .5 ,2 , and 2.5. (Example 8)
77.
Complete each trigonometric expression.
cos 60° = s in
60.
61. tan
= s in ___
TIDES The depth y in m eters of the tide on a beach varies
as a sine function of x, the hour of the day. On a certain
day, that function was y = 3 sin
62. sin ~
3
= cos
63. cos ^
sin (-4 5 ° ) = c o s
64.
65. cos
6
= s in ___
-(x ~ 4) 4- 8, where
x = 0 ,1 , 2, ..., 24 corresponds to 12:00 midnight, 1:00
2:00 a . m ., ..., 12:00 m idnight the next night.
= s in _
a
. m .,
a. W hat is the m axim um depth, or high tide, that day?
66. ICECREAM The m onthly sales in thousands of dollars
for Fiona's Fine Ice Cream shop can be modeled by
TT(t - 4)
y = 71.3 + 59.6 s in
>where t = 1 represents
b. At w hat time(s) does the high tide occur?
78.
January, t = 2 represents February, and so on.
MULTIPLE REPRESENTATIONS In this problem , you will
investigate the period of the sine function.
a.
a. Estimate the sales for January, March, July, and
October.
b.
Describe why the ice cream shop's sales can be
represented by a trigonometric function.
o
67. cos(—9) = y -; cos 9 = 1 ; sec 9 = 1
69. sec 9 =
7
iiiii
•77
'K
6
4
7T
3
27T
sin0 jjj|||j||
sin20 lllll
Use the given values to evaluate the trigonometric functions.
68. sin(—9) =
TABULAR Copy and com plete a table similar to the one
below that includes all 16 angle measures from the
unit circle.
sin40 jjlllll
sin 9 = 1; csc 9 = ?
cos 9 = 1 ; cos(—9) = 1
b.
VERBAL After w hat values of 9 do sin 9, sin 29, and
sin 49, repeat their range values? In other words, what
are the periods of these functions?
C.
VERBAL M ake a conjecture as to how the period of
y = sin n9 is affected for different values of n.
70. csc 9 = -jy; sin 9 = 1 ; sin(—9) = 1
71. GRAPHS Suppose the terminal side of an angle 9 in
standard position coincides with the graph of y = 2x
in Quadrant III. Find the six trigonometric functions of 9.
H.O.T. Problem s
Use Higher-Order Thinking Skills
(79) CHALLENGE For each statem ent, describe n.
Find the coordinates of P for each circle with the given
radius and angle measure.
a. cos (n •y j = 0
b.
csc In • y j is undefined.
REASONING Determ ine w hether each statement is true or
fa l s e . Explain your reasoning.
80.
81.
If cos 9 = 0.8, sec 9 — cos (—0)= 0.45.
Since tan (—f) = —tan t, the tangent of a negative angle is
a negative number.
82. WRITING IN MATH Explain why the attendance at a
year-round them e park could be m odeled by a periodic
function. W hat issues or events could occur over time to
alter this periodic depiction?
REASONING Use the unit circle to verify each relationship.
83. sin (—f) = —sin t
84. cos (—f) = cos t
85. tan (—t) = —tan t
76. COMPARISON Suppose the terminal side of an angle 9 } in
standard position contains the point (7, —8), and the
terminal side of a second angle 92 in standard position
contains the point (—7, 8). Compare the sines of 9 ^ and 92.
252
Lesson 4 -3 | T r ig o n o m e tric F u n c tio n s o n th e U n it C ircle
86. WRITING IN MATH Make a conjecture as to the periods of
the secant, cosecant, and cotangent functions. Explain
your reasoning.
Spiral Review
Write each decimal degree measure in DM S form and each DMS measure in decimal degree
form to the nearest thousandth. (Lesson 4-2)
87.168.35°
88.27.465°
89. 1 4 ° 5 '2 0 "
90. 173° 2 4 '3 5 "
91. EXERCISE A preprogrammed w orkout on a treadmill consists of intervals walking
at various rates and angles of incline. A 1% incline m eans 1 unit of vertical rise
for every 100 units of horizontal run. (Lesson 4-1)
a. At what angle, w ith respect to the horizontal, is the treadm ill bed
when set at a 10% incline? Round to the nearest degree.
b. If the treadmill bed is 40 inches long, what is the vertical rise w hen set
at an 8% incline?
Evaluate each logarithm. Lesson 3-3)
92. log8 64
93. log 1 2 5
'
94. log2 32
95. log4 128
List all possible rational zeros of each function. Then determine which, if any, are zeros. (Lesson 2-4)
96. f( x ) = x3 - Ax2 + x + 2
97. g(x) = x 3 + 6x2 + lOx + 3
98. h(x) = x4 — x 2 + x — 1
99. h(x) = 2 x 3 + 3 x 2 — 8x + 3
100./(x) = 2 x 4 + 3x3 - 6 x 2 - l l x -
3
101. g(x) = 4 x 3 + x 2 + 8x + 2
i
I
102.
NAVIGATION A global positioning system (GPS) uses satellites to allow a user to determ ine
his or her position on Earth. The system depends on satellite signals that are reflected to
and from a hand-held transmitter. The tim e that the signal takes to reflect is used to
determine the transm itter's position. Radio waves travel through air at a speed of
299,792,458 meters per second. Thus, d(t) = 299,792,458/: relates the time t in seconds to
the distance traveled d(t) in meters. (Lesson 1-1)
J
a. Find the distance a radio wave will travel in 0.05, 0 .2 ,1 .4 , and 5.9 seconds.
^
b.
i
If a signal from a GPS satellite is received at a transm itter in 0.08 second, how far from
the transmitter is the satellite?
Skills Review for Standardized Tests
103. SAT/ACT In the figure, AB and AD are tangents to
circle C. W hat is the value of m l
105.
REVIEW Find the angular speed in radians per second
of a point on a bicycle tire if it com pletes 2 revolutions
in 3 seconds.
F
H
104. Suppose 6 is an angle in standard position w ith
sin 9 > 0. In which quadrant(s) could the terminal
side of 9 lie?
A I only
C I and III
B I and II
D I and IV
J
106.
2 tt
3
4tt
3
REVIEW W hich angle has a tangent and cosine that are
both negative?
A 110°
B 180°
C 210°
D 340°
253
Graphing Technology Lab
oooo
oooo
oooo
Graphing the Sine Function
Parametrically
o o o
H H H BI
Use a graphing calculator
and parametric equations
to graph the sine function
and its inverse.
As functions of the real number system, you can graph trigonometric functions on the coordinate plane and apply the
same graphical analysis that you did to functions in Chapter 1. As was done in Extend 1-7, parametric equations will
be used to graph the sine function.
Activity 1
Parametric Graph of y = sin
x
G raph x = t , y = sin t.
CT7?fn Set the mode. In the I MODE j menu, select RADIAN, PAR, and SIMUL. This allows the
equations to be graphed simultaneously. N ext, enter the param etric equations. In
param etric form, X ,T ,9 ,n | will use f instead of x.
SCI
P loti Plots PlotJ
Eri'3
0i23HS67B9
DEGREE
F'DL SEU
DDT__
1~EQLJE TlTIM L S O T
\ X itB T
V it B s i n ( T )
\ X :t =
V ;t =
n X jt =
VJt=
\X ht =
af;ii a+bi,
H H
HDRIZ
G-T
SET CLOCK
W INDOW
ET7SBW Set the x- and f-values to range from 0 to 2tt. Set
Tm i n = 0
Tstep and x-scale to y^-. Set y to [—1 ,1 ] scl: 0.1. The
T n a x = 6 . 2 8 3 1 8 5 3 ..
T s t e p = . 2 6 1 7 9 9 3 ..
calculator automatically converts to decimal form.
X n in = 0
X n a x = 6 . 2 8 3 1 8 5 3 ..
X s c l = . 2 6 1 7 9 9 3 8 ..
■ W m n= ‘ 1
Graph the equations. Trace the function to
identify points along the graph. Select Trace
and use the right arrow to m ove along the curve.
Record the corresponding x- and y-values.
T:.E2SE?B70
N=.£23E9B7B V=.S ■
[0, 2ir] scl: ^ by [ - 1 ,1 ] scl: 0.1
t: [0, 2it]; fstep
The table shows angle measures from 0° to 180°, or 0 to tv , and the corresponding values
for sin t on the unit circle. The figures below illustrate the relationship betw een the graph
and the unit circle.
Degrees
StudyTip
Decimal Equivalents Below are
the decimal equivalents of
common trigonometric values.
~
= 0.866
- j - * 0.707
0
30
Radians
0
0.52
y = sin t
0
0.5
( - 0 . 5 , 0.9)
( - 0 . 7 , 0.7)
-0 .9 , 0.5)
( - 1 , 0)
Lesson 4 -4
135
150
180
1.571
2.094
2.356
2.618
3.14
1
0.866
0.707
0.5
0
90
0.79
1.05
0.707
0.866
(1, 0)
(0.9, -0 .5 )
(0.7, -0 .7 )
(0.5, -0 .9 )
(0, - 1)
120
60
(0.5, 0.9)
(0.7, 0.7)
(0.9, 0.5)
-0 .9 , -0 .5 )
(-0 .7 , -0 .
( - 0 . 5 , -0 .9 )
254
45
Exercises
Graph each function on [0,2 tt].
1.
x
= t ,y = c o s t
2. x = f,y = s in 2 t
3.
x
= f, y = 3 cos t
4. x = t, y = 4 sin t
5.
x
= t ,y = cos (f + it)
6. x = t, y = 2 sin
—- j j
By definition, sin f is the y-coordinate of the point P(x, y) on the unit circle to which the real number f on the number
line gets wrapped. As shown in the diagram on the previous page, the graph of y = sin t follows the y-coordinate of
the point determined by t as it moves counterclockwise around the unit circle.
The graph of the sine function is called a sine curve. From Lesson 4-3, you know that the sine function is periodic with
a period of 2tt. That is, the sine curve graphed from 0 to 2tt would repeat every distance of 2tt in either direction,
positive or negative. Parametric equations can be used to graph the inverse of the sine function.
Activity 2
Graph an Inverse
Graph x + t, y + sin t and its inverse. Then determ ine a domain for which y — sin t is
one-to-one.
P T T m Inverses are found by switching x and y. Enter
the given equations as X l T and Y lT . To graph the
inverse, set X 2T = Y i t and Y 2T = X lT . These are
found in the VARS 1menu. Select Y-VARS,
parametric, Y u . Repeat for X u .
StudyTip
Tstep If your graph appears to be
pointed, you can change the tstep
to a smaller value in order to get a
smoother curve.
Graph the equations. Adjust the window so that
both of the graphs can be seen, as shown. You may
need to set the tstep to a smaller value in order to
get a sm ooth curve.
x = t, y = sin f
[ —3 ir, 3 it] s c l : b y [ - 1 0 , 1 0 ] scl: 2
t: [—3tt, 3it]; fstep
Because the sine curve is periodic, there are an infinite num ber of domains for w hich the
curve will pass the horizontal line test and be one-to-one. One such domain is
TT 3TT
2
'
2
Exercises
Graph each function and its inverse. Then determ ine a domain for which each function is
one-to-one.
7. x = t ,y = cos 21
9. x = t, y = 2 cos t
11. x = t, y = 2 cos (f — t t )
8. x =
10.
X =
12.
X =
connectED.mcgraw-hill.com |
255
•
(Lesson 1-5)
2
NewVocabulary
sinusoid
amplitude
frequency
phase shift
vertical shift
midline
As you ride a Ferris wheel, the height that you are above
the ground varies periodically as a function of time. You
can model this behavior using a sinusoidal function.
Graph transformations
of the sine and cosine
functions.
You analyzed
graphs of functions.
Use sinusoidal
functions to solve
problems.
Transformations of Sine and Cosine Functions As show n in Explore 4-4, the graph
y = sin t follows the y-coordinate of the point determ ined by t as it m oves around the unit
circle. Similarly, the graph of y = cos t follow s the x-coordinate of this point. The graphs of these
functions are periodic, repeating after a period of 2tt. The properties of the sine and cosine
functions are sum marized below.
1
KeyConcept Properties of the Sine and Cosine Functions
Cosine Function
Sine Function
D o m a in :
R ange:
( - 00, 00)
y -in te r c e p t:
[—1 ,1 ]
x -in te r c e p ts :
rnr, n e Z
R a n g e : [—1,1]
(— 00, 00)
y -in te r c e p t:
0
x -in te rc e p ts :
D o m a in :
1
y n, n e Z
C o n tin u ity :
continuous on ( - 00, 00)
C o n tin u ity :
continuous on (— 00, 00)
S y m m e try :
origin (odd function)
S y m m e try :
y-axis (even function)
E x tre m a :
E x tre m a :
maximum of 1 at
X = y + 2 n it, 0 6 Z
minimum of - 1 at x = i t + 2 n-rc,
minimum of - 1 at
neZ
3ir
E nd B e h a v io r:
O s c illa tio n :
lim
X — * — OC
sin
maximum of 1 at x = 2m:,
os Z
X -+00
between - 1 and 1
E nds B
x and lim
inehavior:
/d o not exist. lim
X— * — 0 0
O s c illa tio n :
cos xand lim cos xdo not exist.
X— »oo
between - 1 and 1
The p o rtio n o f each graph on [0, 2tt] rep resents one p eriod or cy cle o f the fu nction . N otice
th at the cosine graph is a h o rizo n tal tra n sla tio n o f the sine graph. A n y tran sform atio n o f a
sin e fu nction is called a sin u so id . T he g eneral form o f the sin u soid al fu n ction s sine and
cosine are
y = a sin (bx + c) + d
and
where a, b, c, and d are constants and neither a nor b is 0.
256
| L esson 4 -4
y = a cos (bx + c) + d
Notice that the constant factor a in y = a sin x and y = a cos x expands the graphs of y = sin x and
y = cos x vertically if \a\ > 1 and com presses them vertically if \a\ < 1.
StudyTip
Dilations and x-intercepts
Notice that a dilation of a
sinusoidal function does not affect
where the curve crosses the
x-axis, at its x-intercepts.
Vertical dilations affect the amplitude of sinusoidal functions.
KeyConcept Amplitudes of Sine and Cosine Functions
Words
M odel
The am plitude of a sinusoidal function is
half the distance between the maximum and
minimum values of the function or half the
height of the wave.
Symbols
V
; ..........
r"
amplitude
\
/
Fo ry = asin (bx + c ) + (/and
y = a cos (fix + c) + d, amplitude = lal.
/
jrp p h tu d e
... J
To graph a sinusoidal function of the form y = a sin x or y = a cos x, plot the x-intercepts of the
parent sine or cosine function and use the am plitude \a\ to plot the new m axim um and minimum
points. Then sketch the sine wave through these points.
P E 3 J 0 3 J j ^ raPtl Vertical Dilations of Sinusoidal Functions
D escribe how the graphs of f i x ) — sin x and gix ) = j sin x are related. T h en find the
am plitude of g ix ), and sketch two periods o f both fu nctions on the sam e coordinate axes.
The graph of g ix ) is the graph of fi x ) com pressed vertically. The am plitude of g(x ) is |-j |or i .
Create a table listing the coordinates of the x-intercepts and extrem a for/(x) = sin x for one
period on [0, 2tt]. Then use the amplitude of g ix ) to find corresponding points on its graph.
Function
x-intercept
f(x) = sin x
g(x) =
(0 ,0 )
^
sin x
(0 ,0 )
M axim um
(f’1)
I2 ' 4>1
fir
11
x-intercept
M inim um
( iv , 0 )
(¥■-’)
(iv, 0)
I'31 2ir ’
1 '
4,
x-intercept
(2 tv, 0)
(2 -tv , 0 )
Sketch the curve through the indicated points for each function. Then repeat the pattern
suggested by one period of each graph to com plete a second period on [2 tt, 4 tt]. Extend each
curve to the left and right to indicate that the curve continues in both directions.
StudyTip
Radians Versus Degrees
You could rescale the x-axis in
terms of degrees and produce
sinusoidal graphs that look similar
to those produced using radian
measure. In calculus, however,
you will encounter rules that
depend on radian measure.
So, in this book, we will graph all
trigonometric functions in terms
of radians.
p GuidedPractice
D escribe how the graphs o i f i x ) and gix ) are related. T h en find the am plitude of gix), and
sketch two periods o f b oth fu nctions on the same coordinate axes.
1A. fi x )
COS X
1
g ix ) = - c o s x
1B. f i x ) = sin x
gix ) = 5 sin x
1C. f i x ) = cos
X
gix ) = 2 cos x
257
If a < 0, the graph of the sinusoidal function
is reflected in the x-axis.
Graph Reflections of Sinusoidal Functions
Describe how the graphs of f i x ) — cos x and g ix ) = —3 cos x are related. Then find the
amplitude of g ix ), and sketch two periods of both functions on the same coordinate axes.
WatchOut!
The graph of g(x) is the graph of /(x) expanded vertically and then reflected in the x-axis.
The am plitude of g(x) is |—3| or 3.
Amplitude Notice that Example 2
does not state that the amplitude
of <7(x) = - 3 cos x is -3 .
Amplitude is a height and is not
directional.
Create a table listing the coordinates of key points of /(x) = cos x for one period on [0, 2tv].
Use the am plitude of g(x) to find corresponding points on the graph of y = 3 cos x.
Then reflect these points in the x-axis to find corresponding points on the graph of gix).
f(x )
y
=
=
Extremum
/-in te rc e p t
Extremum
/-in te rc e p t
Extremum
(0.1)
i[?•»)
(TV, - 1 )
( f- o )
(2 tv , 1)
(0,3)
iI f )
(tv, - 3 )
1(f . o )
(2tv, 3)
(0, - 3 )
i( f ° )
(TV, 3)
1I f ' 0)
CO
I
£
CNJ,
Function
cos X
3 cos x
g (x ) =
- 3 cos x
Sketch the curve through the indicated points for each function. Then repeat the pattern
suggested by one period of each graph to com plete a second period on [27V, 4 tv], Extend each
curve to the left and right to indicate that the curve continues in both directions.
w
GuidedPractice
Describe how the graphs o f/(.r) a n d g (x ) are related. Then find the amplitude of g i x
sketch two periods of both functions on the same coordinate axes.
2A. f{ x ) = cos x
g ( x ) = —| -co sx
),
and
2B. f( x ) = sin x
g (x) = —4 s in x
In Lesson 1-5, you learned that if g (x ) = f( b x ) , th eng(x) is the graph of/(x) com pressed horizontally
if \b\ > 1 and expanded horizontally if \b\ < 1. H orizontal dilations affect the period of a sinusoidal
function— the length of one full cycle.
258
| L esson 4 -4 | G ra p h in g S ine a n d C o s in e F u n c tio n s
Determining Period When
determining the period of a
periodic function from its graph,
remember that the period is the
smallest distance that contains all
values of the function.
KeyConcept Periods of Sine and Cosine Functions
Words
Symbols
The period of a sinusoidal function is the
distance between any two sets of repeating
points on the graph of the function.
Model
V
F o r y = a s i n ( / w + c ) + (fand
y = a cos (bx + c) + d, where b=f= 0,
period = f y
period
To graph a sinusoidal function of the form y = sin bx or y = cos bx, find the period of the function
period
and successively add — - — to the left endpoint of an interval w ith that length. Then use these
values as the x-values for the key points on the graph.
J E S 3 H 1 3 & GraP*1 Horizontal Dilations of Sinusoidal Functions
Describe how the graphs of f i x ) = cos x and g (x ) = cos j are related. Then find the period
of g ix ), and sketch at least one period of both functions on the same coordinate axes.
y
1
Because cos — = cos —x, the graph of g (x) is the graph of f( x ) expanded horizontally.
The period of g(x) is t— or 6tt.
I 3|
Because the period of g(x) is 6tv, to find corresponding points on the graph of g(x), change the
x-coordinates of those key points on/(x) so that they range from 0 to 67V, increasing by
increments of
or 4 p
Function
M axim um
II
O
O
cn
WatchOut!
g(x) = cos |
(0 ,1 )
(0 ,1 )
x-intercept
(w,-1)
( f ’ °)
(t
’O
M inim um
)
(3ir, —1)
x-intercept
(?■ “ ]
(¥■ “ )
M axim um
(2ir, 1)
(6 tv,1 )
Sketch the curve through the indicated points for each function, continuing the patterns to
com plete one full cycle of each.
§►GuidedPractice
Describe how the graphs of fi x ) an d g (x ) are related. Then find the period o i g ix ), and sketch
at least one period of each function on the same coordinate axes.
3A. f( x ) = cos x
g (x ) = cos -y
3B. f( x ) = sin x
g (x ) = sin 3x
3C. f( x ) = cos x
g (x ) = COS-jX
Horizontal dilations also affect the frequency of sinusoidal functions.
KeyConcept Frequency of Sine and Cosine Functions
Words
The frequency of a sinusoidal function
is the number of cycles the function completes
in a one unit interval. The frequency is the
reciprocal of the period.
Symbols
F o r y = a s i n ( / w + c ) + dand
y = a cos (bx + c) + oI,
Model
frequency = — V -r or — .
period
m
mIttiSE
- ... '
Because the frequency of a sinusoidal function is the reciprocal of the period, it follows that the
period of the function is the reciprocal of its frequency.
j
Real-World Example 4 Use Frequency to Write a Sinusoidal Function
J
>
Real-WorldLink
In physics, frequency is measured
in hertz or oscillations per second.
For example, the number of sound
waves passing a point A in one
second would be the wave’s
frequency.
Source:
Science World
2tt
MUSIC M usical notes are classified by frequency. In the equal tem pered scale, middle C
has a frequency of 262 hertz. Use this inform ation and the inform ation at the left to w rite an
equation for a sine function that can be used to model the initial behavior of the sound wave
associated with middle C having an amplitude of 0.2.
The general form of the equation will be y = a sin bt, where t is the time in seconds.
Because the am plitude is 0.2, \a\ = 0.2. This means that a = ± 0 .2 .
The period is the reciprocal of the frequency or y y . Use this value to find
period = ~
1
262
2 tv
\b\
\b\ = 2tt(262) or 524tt
b = ±5247f
b.
Period formula
period =
1
262
Solve for |6j.
Solve for b.
By arbitrarily choosing the positive values of a and b, one sine function that models the initial
behavior is y = 0.2 sin 5247rt.
►GuidedPractice
4.
MUSIC In the same scale, the C above middle C has a frequency of 524 hertz. W rite an
equation for a sine function that can be used to model the initial behavior of the sound wave
associated with this C having an am plitude of 0.1.
A phase of a sinusoid is the position of a wave relative to som e reference point. A horizontal
translation of a sinusoidal function results in a phase shift. Recall from Lesson 1-5 that the
graph of y = f ( x + c) is the graph of y = f( x ) translated or shifted |c| units left if c > 0 and
|c| units right if c < 0.
260
Lesson 4 -4
G ra p h in g S ine a n d C o sin e F u n c tio n s
KeyConcept Phase Shift of Sine and Cosine Functions
Words
The phase shift of a sinusoidal function
Model
is the difference between the horizontal
position of the function and that of an
otherwise similar sinusoidal function.
Symbols
For y = a sin (bx + c) + d and
y = a cos (bx+ c) + a1, where b =/= 0,
phase shift =
phase shift
1*1
You will verify the formula for phase shift in Exercise 44.
StudyTip
Alternative Form The general
forms of the sinusoidal functions
can also be expressed as
y = a sin b(x- h) + k
and y = a cos b(x—h) + k.
In these forms, each sinusoid has
a phase shift of h and a vertical
translation of /(in comparison to
the graphs of y = a sin bx and
y = a cos bx.
To graph the phase shift of a sinusoidal function of the form y = a sin (bx + c) 4- d or
y = a cos (bx + c) + d, first determ ine the endpoints of an interval that corresponds to one cycle
of the graph by adding —j to each endpoint on the interval [0, 27t] of the parent function.
B J 2 H uIEI33E! GraPh Horizontal Translations of Sinusoidal Functions
State the amplitude, period, frequency, and phase shift of y — sin
two periods of the function.
— y j . Then graph
In this function, a = 1, b = 3, and c = —
Period: # f = ^—o or
r —
^
|3|
3
Am plitude: \a\ = |1| or 1
Frequency:
|3|
2-k ° r 2-k
2tt
Phase shift: —
\b\
= — rr-o r^
|3|
To graph y = sin ^3x— y j , consider the graph of y = sin 3x. The period of this function is -y-.
2tc
Create a table listing the coordinates of key points of y = sin 3x on the interval 0, — .
To account for a phase shift of ~ , add y to the x-values of each of the key points for the
graph of y = sin 3x.
x-intercept
Function
y = sin 3 /
y = sin ( 3 * - y )
(0 ,0 )
m
Sketch the graph of y = sin
com plete two cycles.
M axim um
/-in te rc e p t
Minimum
i
I( ] M i
( f - 1)
It ')
I( f ° ) i
(T - ~ ' . I
/-in te rc e p t
1
(t
-° )
— y j through these points, continuing the pattern to
►GuidedPractice
State the amplitude, period, frequency, and phase shift of each function. Then graph two
periods of the function.
5A. y = c o s (| + ^ )
5B. y = 3 sin ^2x — y j
—-.
fHconnectED.mcgraw-hill.com |
261
StudyTip
Notation
sin ( / + d) j= sin x -t- d
The first expression indicates
a phase shift, while the second
expression indicates a
vertical shift.
The final w ay to transform the graph of a sinusoidal function is through a vertical translation or
Nvertical shift. Recall from Lesson 1-5 that the graph of y = f ( x ) + d is the graph of y = f( x )
translated or shifted \d\ units up if d > 0 and \d\ units dow n if d < 0. The vertical shift is the average
of the m axim um and m inimum values of the function.
The parent functions y = sin x and y = cos x oscillate about
the x-axis. After a vertical shift, a new horizontal axis known
as the midline becom es the reference line or equilibrium
point about w hich the graph oscillates. For exam ple, the
midline of y = sin x + 1 is y = 1, as shown.
In general, the midline for the graphs of y = a sin (bx + c) + d and y = a cos (bx + c) + d is y = d.
Graph Vertical Translations of Sinusoidal Functions
State the amplitude, period, frequency, phase shift, and vertical shift of y = sin (x + 2tt) — 1.
Then graph two periods of the function.
In this function, a = 1, b = 1, c = 2ir, and d = —1.
Am plitude: |<i| = |1| or 1
Period: ^
i>
Phase shift:
Vertical shift: d or —1
lfel
—r r = —2tt
1
or 27r
Frequency: Jp- = Jp- or -y1
J 2tt
2tt
2tt
Midline: y = d o r y ■
First, graph the midline y = —1. Then graph y = sin x shifted 27T units to the left and
1 unit down.
Notice that this transformation is equivalent to a translation 1 unit dow n because the phase shift
was one period to the left.
f
GuidedPractice
State the amplitude, period, frequency, phase shift, and vertical shift of each function.
Then graph two periods of the function.
6A. y = 2 cos x + l
The characteristics of transform ations of the parent functions y : : sin x and y = cos x are
sum marized below.
TechnologyTip
Zoom Trig When graphing a
trigonometric function using your
graphing calculator, be sure you are
in radian mode and use the ZTrig
selection under the zoom feature to
change your viewing window from
the standard window to a more
appropriate window of [ - 2 i r , 2ir]
scl: -ir/2 by [ - 4 , 4 ] scl: 1.
>
ConceptSummary Graphs of Sinusoidal Functions
The graphs of y = a sin (bx + c) + c/and y = a cos (bx + c) + d, where a # 0 and b
have the following characteristics.
0,
Amplitude: lal
Period: | j L
Frequency: — or
Phase shift: -
Vertical shift: d
Midline: y = d
2 6 2 | L esson 4 -4 i G ra p h in g S ine a n d C o sin e F u n c tio n s
lol
2i t
Period
Applications of Sinusoidal Functions M any real-world situations that exhibit periodic
mm behavior over time can be m odeled by transform ations of y = sin x or y = cos x.
Real-World Example 7 Modeling Data Using a Sinusoidal Function
METEOROLOGY Use the inform ation at the left to w rite a sinusoidal function that models
the num ber of hours of daylight for New York City as a function of time x, where x = l
represents January 15, x = 2 represents February 15, and so on. Then use your model to
estimate the num ber of hours of daylight on Septem ber 30 in New York City.
R TW n M ake a scatter plot of the data and choose a model.
The graph appears wave-like, so you can use a
sinusoidal function of the form y = a sin (bx + c) + d
or y = a cos {bx + c) + d to model the data. We will
choose to use y = a cos (bx + c) + d to model the data.
0 ,1 2 ] scl: 1 by [0, 20] scl: 2
Find the m axim um M and m inim um m values of the
data, and use these values to find a, b, c, and d.
The m axim um and m inim um hours of daylight are 15.07 and 9.27, respectively.
The am plitude a is half of the distance betw een the extrema.
Real-WorldLink
a = | (M - m) = ± (15.07 - 9.27) or 2.9
The table shows the number of
daylight hours on the 15th of each
month in New York City.
Month
The vertical shift d is the average of the m axim um and m inim um data values.
Hours of
Daylight
January
9.58
February
10.67
March
11.9
April
13.3
May
14.43
June
15.07
July
14.8
August
13.8
September
12.48
October
11.15
November
9.9
December
9.27
Source: U.S. Naval Observatory
d = | (M + m) = | (15.07 + 9.27) or 12.17
A sinusoid com pletes half of a period in the tim e it takes to go from its maxim um to its
minim um value. One period is twice this time.
: December 15 or month 12 and
Period = 2(xmax - x mw) = 2(12 - 6) or 12
i June 15 or month 6
Because the period equals -^f, you can write \b\ =
Therefore, \b\ = —J , or
|b| J
Period
11
12
6
The m axim um data value occurs when x = 6. Since y = cos x attains its first maximum
when x = 0, we m ust apply a phase shift of 6 — 0 or 6 units. Use this value to find c.
Phase shift = ——
|b|
6
=
Phase shift formula
Phase shift = 6 and |i>[ = -
-~ w
6
C =
Solve for c.
— 7T
ETfnFl Write the function using the values for a, b, c, and d. Use b = ^~.
6
y = 2.9 cos ^ x — 7t| + 12.17 is one model for the hours of daylight
Graph the function and scatter plot in the same view ing window, as in Figure 4.4.1.
To find the num ber of hours of daylight on Septem ber 30, evaluate the model for x = 9.5.
y = 2.9 cos (^ (9 .5 ) — 7rj + 12.17 or about 11.42 hours of daylight
p GuidedPractice
METEOROLOGY The average m onthly tem peratures for Seattle, W ashington, are shown.
Figure 4.4.1
Month
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sept
Oct
Nov
Dec
Temp. (°)
41
44
47
50
56
61
65
66
61
54
46
42
7A. W rite a function that models the m onthly tem peratures, using x = 1 to represent January.
7B. According to your m odel, what is Seattle's average monthly temperature in February?
connectED.mcgraw-hill.com ]
263
Exercises
= Step-by-Step Solutions begin on page R29.
Describe how the graphs of fix ) and g(x) are related. Then
find the amplitude of gix), and sketch two periods of both
functions on the same coordinate axes. (Examples 1 and 2)
21. TIDES The table shown below provides data for the first
high and low tides of the day for a certain bay during one
day in June. (Example 7)
Tide
2. f(x ) = cos x
1. f(x ) = sin x
g(x) = ~
gix) = -i sin x
first high tide
cos x
first low tide
4. fix ) = sin x
3. f[x ) = cos x
gix) = —8 sin x
gix) = 6 cos x
Describe how the graphs of fix ) and gix) are related. Then
gix) = sin 0.25x
22. METEOROLOGY The average monthly temperatures for
Boston, M assachusetts are shown. (Example 7)
9. VOICES The contralto vocal type includes the deepest
female singing voice. Some contraltos can sing as low as
the E below middle C (E3), w hich has a frequency of
165 hertz. Write an equation for a sine function that
models the initial behavior of the sound wave associated
with E3 having an amplitude of 0.15. (Example 4)
Month
Write a sine function that can be used to model the initial
behavior of a sound wave with the frequency and amplitude
given. (Example 4}
10.
/ = 440, a = 0.3
11. / = 9 3 2 ,0 = 0.25
12.
/ = 1245,0 = 0.12
13. / = 623, a = 0.2
(v ^ y = 3 s m ( x - f j
17. y = sin 3 x — 2
T8> y = cos ^x —
19.,, y = sin (x +
—1
+ 4
VACATIONS The average number of reservations R
that a vacation resort has at the beginning of each
month is shown. (Example 7)
Month
R
Month
R
Jan
200
May
121
Feb
173
Jun
175
Mar
113
Jul
198
Apr
87
Aug
168
a. Write an equation of a sinusoidal function that models
the average num ber of reservations using x = 1 to
represent January.
b. According to your model, approximately how many
reservations can the resort anticipate in November?
L esson 4 -4
Month
Temp. (°F)
Jan
29
Jul
74
Feb
30
Aug
72
Mar
39
Sept
65
Apr
48
Oct
55
May
58
Nov
45
Jun
68
Dec
34
b. Write an equation of a sinusoidal function that
models the m onthly temperatures.
C. According to your model, w hat is Boston's average
tem perature in August?
( j ^ . y = cos (| +
16. y = 0.25 cos x + 3
Temp. (°F)
a. Determ ine the am plitude, period, phase shift, and
vertical shift of a sinusoidal function that models
the m onthly tem peratures using x = 1 to represent
January.
State the amplitude, period, frequency, phase shift, and
vertical shift of each function. Then graph two periods of
the function. (Examples 5 and 6)
264
10:55 a.m.
8. fix ) = sin x
gix) = cos 0.2x
20.
2.02
C. According to your m odel, what was the height of the
tide at 8:45 p . m . that night?
gix) = cos 2x
7. fix ) = cos x
4:25 a.m.
b. Write a sinusoidal function that models the data.
6. fix ) = cos x
gix) = sin 4x
Time
12.95
a. Determ ine the amplitude, period, phase shift, and
vertical shift of a sinusoidal function that models the
height of the tide. Let x represent the number of hours
that the high or low tide occurred after midnight.
find the period of g(x), and sketch at least one period of both
functions on the same coordinate axes. (Example 3)
5. fix ) = sin x
Height (ft)
G ra p h in g S ine a n d C o sin e F u n c tio n s
GRAPHING CALCULATOR Find the values of x in the interval
x < 77that m ake each equation or inequality true.
(Hint: Use the intersection function.)
-tv <
23.
—sin x = cos x
24. sin x — cos x = 1
25.
sin x + cos x = 0
26. cos x < sin x
27.
sin x cos x > 1
28. sin x cos x < 0
ffl) CAROUSELS A wooden horse on a carousel moves up and
dow n as the carousel spins. W hen the ride ends, the horse
usually stops in a vertical position different from where it
started. The position y of the horse after t seconds can be
modeled by y = 1.5 sin (2 1 + c), where the phase shift
m ust be continuously adjusted to com pensate for the
different starting positions. If during one ride the horse
77T
reached a m axim um height after — seconds, find the
equation that models the horse's position.
30. AMUSEMENT PARKS The position y in feet of a passenger
cart relative to the center of a Ferris wheel over f seconds
is shown below.
39. $ MULTIPLE REPRESENTATIONS In this problem , you will
investigate the change in the graph of a sinusoidal
function of the form y = sin x or y = cos x when
m ultiplied by a polynom ial function.
a. GRAPHICAL Use a graphing calculator to sketch the
graphs of y = 2x, y = —2x, and y = 2 x cos x on the
same coordinate plane, on the interval [—20, 20].
b. VERBAL Describe the behavior of the graph of
y = 2 x cos x in relation to the graphs of y = 2x and
y = - 2x.
c. GRAPHICAL Use a graphing calculator to sketch the
graphs of y = x 2, y = —x 2, and y = x2 sin x on the same
coordinate plane, on the interval [—20, 20].
Side view of Ferris wheel overtime interval [0, 5.5]
d. VERBAL Describe the behavior of the graph of
y = x 2 sin x in relation to the graphs of y = x 2 and
2
y = - x z.
t= 0
f= 2
f = 3.75
t= 5.5
a. Find the time t that it takes for the cart to return to
y = 0 during its initial spin.
e. ANALYTICAL M ake a conjecture as to the behavior of the
graph of a sinusoidal function of the form
y = sin x or y = cos x w hen multiplied by polynomial
function of the form y = f( x ) .
b. Find the period of the Ferris wheel.
C. Sketch the graph representing the position of the
passenger cart over one period.
d. Write a sinusoidal function that models the position of
the passenger cart as a function of time t.
H.O.T. Problem s
Use Higher-Order Thinking Skills
40. CHALLENGE W ithout graphing, find the exact coordinates
of the first m axim um point to the right of the y-axis for
y = 4 sin ( f * - f ) .
Write an equation that corresponds to each graph.
REASONING D eterm ine w hether each statement is true or
fa ls e . Explain your reasoning.
41. Every sine function of the form y = a sin (bx + c) + d can
also be w ritten as a cosine function of the form
y = a cos (bx + c) + d.
42. The period of f i x ) = cos 8x is equal to four times the
period of g(x) = cos 2x.
(43) CHALLENGE How m any zeros does y = cos 1500x have on
the interval 0 < x < 2ir?
44. PROOF Prove the phase shift formula.
45. WRITING IN MATH The Pow er Tower ride in Sandusky,
Ohio, is show n below. Along the side of each tower is a
string of lights that send a continuous pulse of light up
and down each tower at a constant rate. Explain why the
distance d of this light from the ground over time t cannot
be represented by a sinusoidal function.
Write a sinusoidal function with the given period and
amplitude that passes through the given point.
35. period: 7r; amplitude: 5; point:
|-j
36. period: 4tt; amplitude: 2; point: (tt, 2)
37. period:
amplitude: 1.5; point:
38. period: 37T; amplitude: 0.5; point: |tt,
c o n n e c tiB 'm c g r a w ^ L r o m l
265
Spiral Review
The given point lies on the terminal side of an angle 9 in standard position. Find the values
of the six trigonometric functions of 9. (Lesson 4-3)
(- 4 ,4 )
46.
47. ( 8 , - 2 )
48. ( - 5 , - 9 )
49. (4 ,5 )
Write each degree measure in radians as a multiple of tt and each radian measure
in degrees. (Lesson 4-2)
50. 25°
-4 2 0 °
51.
53. ^
52.
54. SCIENCE Radiocarbon dating is a m ethod of estimating the age of an organic material by
calculating the amount of carbon-14 present in the material. The age of a m aterial can be
calculated using A = t •
where A is the age of the object in years, t is the half-life of
carbon-14 or 5700 years, and R is the ratio of the am ount of carbon-14 in the sam ple to the
am ount of carbon-14 in living tissue. (Lesson 3-4)
a. A sample of organic material contains 0.000076 gram of carbon-14. A living sam ple of the
same material contains 0.00038 gram. About how old is the sample?
b.
A specific sample is at least 20,000 years old. W hat is the m axim um percent of carbon-14
remaining in the sample?
State the num ber of possible real zeros and turning points of each function. Then determine
all of the real zeros by factoring. (Lesson 2-2)
55. f i x ) = x 3 + 2 x 2 — 8x
56. f( x ) = x 4 - W x2 + 9
f( x ) = x5 + 2 x 4 - 4 x 3 - 8x2
57.
58. f( x ) = x 4 - 1
Determine w h e th e r/h a s an inverse function. If it does, find the inverse function and state
any restrictions on its domain. (Lesson 1-7)
59.
60. f( x )
f( x ) = - x - 2
61. f i x ) = (x - 3)2 - 7
62. fi x ) =
Skills Review fo r Standardized Tests
63.
SAT/ACT If x + y = 90° and x and y are both
Identify the equation represented by the graph.
65.
nonnegative angles, what is equal to ~^ ~ ?
A 0
D 1.5
C 1
E Cannot be
determ ined from
the inform ation
given.
BI
64.
10
REVIEW If tan x = — in the figure below , what are sin
24
x and cos x?
A
B
C
D
y = 21 sin 4x
y = 4i sin 2x
y = 2 sin 2x
y = 4 sin h
66. REVIEW If cos 9 = — and the terminal side of the
G sin x = i p and cos x =
26
26
angle is in Q uadrant IV, what is the exact value of
sin 6?
15
TT
26
J
26
H sin x = — and cos x = —
] sin x = i^r and cos x = —r
J
26
24
266
Lesson 4 -4 | G ra p h in g S ine a n d C osine F u n c tio n s
C
17
G “ 15
H
JJ
17
15
—
Graphing Technology Lab
oooo
oooo
oooo
Sums and Differences
of Sinusoids
CDOO
The graphs of the sums and differences of two sinusoids will often have different periods than the
•
Graph and examine the
periods of sums and
differences of sinusoids.
graphs of the original functions.
Activity 1
mm
Sum of Sinusoids
Determine a com m on interval on which b o th /(x ) = 2 sin 3x and g ix ) = 4 cos j complete a
whole num ber of cycles. Then graph h(x) = f i x ) + g ix ), and identify the period of the function.
ETHTn Enter/(x) for Y i and g(x) for Y 2 . Then adjust the window
until each graph com pletes one or m ore w hole cycles on
the same interval. One interval on w hich this occurs is
[0, 4tt]. On this interval, g (x) com pletes one w hole cycle
and/(x) com pletes six whole cycles.
R7!TTO To graph 7z(x) as Y3, under the I VARS I menu, select Y-VARS,
function, and Y i to enter Y i. Then press I + I and select
Y-VARS, function, and Y 2 to enter Y 2 .
W
l Graph/(x), g(x) and h(x) on the same screen. To m ake the graph of hix) stand out, scroll
to the left of the equals sign next to Y3, and press Ie n t e r |. Then graph the functions using
the same w indow as above.
pioti
Pi*t3
^ iB 2 s in < 3 X )
\ V 2B 4 c o s ( X / 2 )
V ^ B Y i +Vz
\ Y h=
\Y s =
\Y f i=
\Y?=
TechnologyTip
Hiding Graphs Scroll to the
equals sign and select enter to
make a graph disappear.
CTTSflW By adjusting the x-axis from [0, 4-tt] to [0, 8tt]
to observe the full pattern of hix), we can see
that the period of the sum of the two sinusoids
is 4 - tv .
Period - [0, 8 ir] scl: 2 tt by [ - 6 , 6] scl: 1
Exercises
Determine a com m on interval on which b o th /(x ) and gix) complete a whole num ber of cycles.
Then graph aix) = fix ) + gix) and b(x) = f(x ) — gix), and identify the period of the function.
1. f i x ) = 4 sin 2x
gix ) = —2 cos 3x
2.
4. f i x ) = j sin 4x
5- / M =
g(x) = 2 sin [x - y )
f i x ) = sin 8x
gix ) = cos 6x
\
cos f
g ix ) = - 2 cos (x - y j
3. f i x ) = 3 sin (x — 7v)
gix ) = —2 cos 2x
6. fi x ) = —j sin 2x
g(x) = 3 cos 2x
7. MAKE A CONJECTURE Explain how you can use the periods of two sinusoids to find the period
of the sum or difference of the two sinusoids.
^j^^^^^E^ricgra^iinxoiJ 267
Mid-Chapter Quiz
Lessons 4-1 through 4-4
■ ■ ■ ■ ■ ■ ■ ■ ■ ■
Find th e e x a c t v a lu e s of th e six trig o n o m e tric
12. TRAVEL A car is traveling a t a speed of 5 5 m iles per hour on tires
fu nctio ns of 9. (Lesson 4-1)
th a t m e asu re 2 .6 fe e t in diam eter. Find th e a p p ro xim ate angular
speed of the tires in radians per m inute. (Lesson 4-2)
1.
S k e tc h e a c h a n g le . T h e n fin d its re fe re n c e an g le . (Lesson 4-3)
14. 2 1 t t
13. 1 7 5 °
13
Find th e e x a c t v a lu e o f e a c h e x p re s s io n . If u n d e fin e d , w rite
Find th e v alu e of x. Round to th e n e a re s t te n th if necessary.
(Lesson. 4-1)
3.
^
4.
undefined. Lesson 4-3)
15. cos 3 1 5 °
16.
sec-y^
sin
18.
tan ^ r -
17.
0
Find th e e x a c t v a lu e s of th e fiv e re m a in in g trig o n o m e tric fu n c tio n s of
9.
5. SHADOWS A pine tree casts a shadow th a t is 7 .9 fe e t long w hen
th e Sun is 8 0 ° above the horizon. (Lesson 4-1)
a.
19.
(Lesson 4-3)
cos 9 = —
20. cot 9 =
5
w h e re sin 9 < 0 and tan 9 > 0
w h e re cos 9 > 0 and sin 9 > 0
Find the height of the tree.
b. Later th a t s am e day, a person 6 fe e t tall casts a s hadow 6 .7 fe e t
long. At w h a t angle is the Sun above the horizon?
Find th e m e a s u re o f a n g le 9. R ound to th e n e a re s t d e g re e
S ta te th e a m p litu d e , period, frequency, phase shift, and v ertical sh ift of
e ach functio n. Th en graph tw o full periods of th e functio n. (Lesson 4-4)
21.
if n ecessary. (Lesson 4-1)
23.
6.
y = - 3 sin ( * - f
)
22. y = 5 c o s 2 x - 2
MULTIPLE CHOICE W hich of th e functions has th e s am e graph as
y = 3 sin (x — it)? (Lesson 4-4)
F y = 3 sin (x +
tt)
G y = 3 cos (x - y j
H
J
y =
— 3 sin (x —-tv)
y = -3 c o s (x + y j
24. SPRING The m otion of an object attached to a spring oscillating
across its original position of rest can be m odeled by x(t) = A cos
u t, w h e re A is th e initial dis p la c e m en t of th e o bject from its resting
Id e n tify all an g le s th a t a re c o te rm in a l w ith th e given an g le . Th e n find
and d ra w one positive and one n e g a tiv e a n g le c o te rm in a l w ith th e
given angle. (Lesson 4-2)
position, a ; is a constant dep en d en t on the spring and the m ass of
the o bject a ttach ed to the spring, and t is tim e m easured in
seconds. [Lesson 4-4)
a.
10.
-
22°
D ra w a graph for the m otion of an object attached to
a spring and displaced 4 centim e te rs w h e re u = 3.
b. H ow long w ill it ta k e fo r the o bject to return to its initial position
for the first tim e?
11. MULTIPLE CHOICE Find the approxim ate a re a of the shaded
c. The constant uj is equal to
region. (Lesson 4-2)
w h e re k is the spring constant,
and m is the m ass of the object. H ow w ould increasing th e m ass
of an o bject a ffe c t the period of its oscillations? Explain your
reasoning.
25. BUOY The height above sea level in fe e t of a signal buoy’s
tra n s m itte r is m odeled by h = a sin bt +
268
In rough w a te rs , th e height cycles b e tw ee n 1 and
A 1 2 .2 in 2
C 85.5 in2
1 0 fe e t, w ith 4 seconds betw een cycles. Find the
B 4 2 .8 in 2
D 111.2 in2
values of a and b.
C h a p te r 4
M id -C ha pte r Quiz
■ p i
{■Graphing Other
Trigonometric Functions
Then
Now
Why?
You analyzed graphs
of trigonometric
functions.
1
There are two types of radio transmissions known as
amplitude modulation (AM) and frequency
modulation (FM). When sound is transmitted by an
AM radio station, the amplitude of a sinusoidal wave
called the carrier wave is changed to produce sound.
The transmission of an FM signal results in a change
in the frequency of the carrier wave. You will learn
more about the graphs of these waves, known as
damped waves, in this lesson.
(Lesson 4-4)
Graph tangent
and reciprocal
trigonometric
functions.
| Graph damped
trigonometric
functions.
AM signal
FM signal
A A A ir "
V V VI
3 NewVocabulary
* * damped trigonometric
function
damping factor
damped oscillation
damped wave
damped harmonic motion
Tangent and Reciprocal Functions
In Lesson 4-4, you graphed the sine and cosine
functions on the coordinate plane. You can use the same techniques to graph the tangent
function and the reciprocal trigonom etric functions— cotangent, secant, and cosecant.
1
Since tan x =
the tangent function is undefined when cos x = 0. Therefore, the tangent
function has a vertical asym ptote whenever cos x = 0. Similarly, the tangent ahd sine functions each
have zeros at integer m ultiples of t t because tan x = 0 w hen sin x = 0.
41
,1\yj/ / 4T \
I T \ |y = sin
M 1\
/
1 \ l/ /
1
/
/
1
!
\ /
h
V
/
1
!
I
/j 1I
A\ 1
\
i
/
y = t a n x
I
V
1
i
" iJ ,
---------------1-------------- f -------------- r ...............r t
ts
CM
1
1
i
1
A a
/ A
/
1
t
/
/
\
1
i
1
\
L
1 J
t
rx
/
l
/
The properties of the tangent function are sum marized below.
KeyConcept Properties of the Tangent Function
D o m a in :
xe
Range:
(—oo, oo)
x -in te rc e p ts :
m t ,n e Z
y -in te r c e p t:
0
C o n tin u ity :
infinite discontinuity at
x = y + me, n e Z
A s y m p to te s :
x=
S y m m e try :
origin (odd function)
E x tre m a :
none
E n d B eh a vio r:
R,
Y
y +
m r, n e
+ m r, n e Z
lim tan x and lim tan x
X — >— o o
X — >oo
do not exist. The function
oscillates between — oo
and oo.
period:
it
connectED.mcgraw-hili.com| 269
StudyTip
Amplitude The term amplitude
does not apply to the tangent or
cotangent functions because the
heights of these functions are
infinite.
The general form of the tangent function, w hich is sim ilar to that of the sinusoidal functions, is
y = a tan (bx + c) + d, where a produces a vertical stretch or com pression, b affects the period,
c produces a phase shift, d produces a vertical shift and neither a or b are 0.
KeyConcept Period of the Tangent Function
Words
The
period of a tangent function is the distance
between any two consecutive vertical asymptotes.
Symbols
For y = a tan (bx + c), where b ± 0, period =
1*1
Two consecutive vertical asym ptotes for y = tan x are .r = —y and x = y You can find two
consecutive vertical asym ptotes for any tangent function of the form y = a tan (bx + c) + d by
solving the equations bx + c — —y and bx + c =
You can sketch the graph of a tangent function by plotting the vertical asym ptotes, x-intercepts, and
points betw een the asym ptotes and x-intercepts.
Graph Horizontal Dilations of the Tangent Function
Locate the vertical asymptotes, and sketch the graph of y — tan Zx.
The graph of y = tan 2x is the graph of y = tan x com pressed horizontally. The period is
-jYj-
or
Find two consecutive vertical asymptotes.
y.
bx + C =
Tangent asymptote equations
—y
b = 2 , c —0
2x + 0 = - f
x = -JL
4
bx + c =
2x + 0 :
Simplify.
TT
2
x = *
4
Create a table listing key points, including the x-intercept, that are located betw een the two
vertical asym ptotes at x = —y and x = y .
Vertical
Asymptote
/-in te rc e p t
* = - f
y = tan 2x
HM
Interm ediate
Point
(0 .0 )
(?■')
(0 .0 )
If'1)
Sketch the curve through the indicated key points
for the function. Then sketch one cycle to the left
on the interval (~ "y v ~ y j ar>d one cycle to the
right on the interval |y, - y j .
p GuidedPractice
Locate the vertical asymptotes, and sketch the graph of each function.
1A. y = tan 4x
270
| L esson 4 -5
G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s
Vertical
Asymptote
II
*
y = tan x
Interm ediate
Point
*
II
Function
1B. y = tan y
IE H 2 B
Graph Reflections and Translations of the Tangent Function
Locate the vertical asymptotes, and sketch the graph of each function,
a.
2/ = - t a n |
The graph of y = —tan y is the graph of y = tan x expanded horizontally and then reflected in
the x-axis. The period is j y - or 2tt. Find two consecutive vertical asymptotes.
12 1
b = j .c =
f + 0 = - f
x = 2
f + ° = f
0
Simplify.
y j or —t t
x = 2 ^yj or
tt
Create a table listing key points, including the x-intercept, that are located betw een the two
vertical asym ptotes at x = - t t and x = t t .
y = tan x
—
y = -ta n |
Interm ediate
Point
f
( - 7 ’ - 1)
Interm ediate
Point
x-intercept
(0 ,0 )
(t
(0 ,0 )
X = — TT
>)
( f ■- 1)
Vertical
Asymptote
>5
II
ro|S
Vertical
Asymptote
Function
X = Tt
Sketch the curve through the indicated key
points for the function. Then repeat the
pattern for one cycle to the left and right
of the first curve.
b. y =
tan (x -
The graph of y = tan |x — y ^ j is the graph of y = tan x shifted ~
units to the right.
The period is jyj- or 7T. Find two consecutive vertical asymptotes.
X
StudyTip
Alternate Method When graphing
a function with only a horizontal
translation c, you can find the key
points by adding cto each of the
/-coordinates of the key points of
the parent function.
6=1, c= =
- y
+
y y
Function
y = ta n x
y= tan ( x -
or
Simplify.
TT
Vertical
Asymptote
Interm ediate
Point
x - - -
( - ? • - ')
X = TT
( f - 'l
*
2
3 tt
7r
2
x =
x-intercept
(0,0)
( f ,° )
or
2 tt
Interm ediate
Point
Vertical
Asymptote
( f - 1)
II
37T
(t -'I
x = 2 -k
Sketch the curve through the indicated key
points for the function. Then sketch one
cycle to the left and right of the first curve.
f GuidedPractice
2A. y = tan ^2x + y j
2B. y = - t a n (x -
271 m
The cotangent function is the reciprocal of the tangent function, and is defined as cot x = co s* .
Like the tangent function, the period of a cotangent function of the form y = a cot (bx + c) + d can
be found by calculating y^y. Two consecutive vertical asymptotes can be found by solving the
equations bx + c = 0 and bx + c
= 7T. The properties of the cotangent function are summarized below.
KeyConcept Properties of the Cotangent Function
z
x e R , x ± m t, n e
R an g e :
( - 00, oo)
J1
x -in te rc e p ts :
^ + n -K ,n € l
1
!
|
y -in te rc e p t:
none
C o n tin u ity :
infinite discontinuity at x = a tt, n e Z
S y m m e try :
origin (odd function)
E nd B eh a vio r:
X—
0
2 A1~
\
—4 “
i
^
none
\
\ i
11
l 1
—3 r
i
i
i
i
i
! *
' x
\
3A
2 tt
2 \ i
\ i
\ i
\ I
\ I
11
j
lim cot x and lim cot x do not exist.
►
—OO
X — >00
The function oscillates between
I
i
i
I
i
x = n fK ,n e Z
\'K\
1
—7T
iy
■1
i
11
2 - " l \y = cot x | \
\ //
1 \
- \V
1 V
1
\
1 \
V
A s y m p to te s :
E x tre m a :
3-
_ ts
D o m a in :
period: 7\ 1
and <*>.
J
You can sketch the graph of a cotangent function using the same techniques that you used to
sketch the graph of a tangent function.
Sketch the Graph of a Cotangent Function
TechnologyTip
Locate the vertical asymptotes, and sketch the graph of y = cot j .
y=
Graphing calculators
may produce solid lines where the
asymptotes occur. Setting the
mode to DOT will eliminate the line.
The graph of y = cot j is the graph of y = cot x expanded horizontally. The period is — or
Find two consecutive vertical asym ptotes by solving bx + c = 0 and bx + c — i t .
M
4 + 0 = 0
■0 =
b= ± c= 0
Simplify.
x = 3(0) or 0
x =
TT
3 ( tt)
or
3
tt
Create a table listing key points, including the x-intercept, that are located betw een the two
vertical asym ptotes at x = 0 and x = 3 t t .
Vertical
Asymptote
x=0
y = c o t|
x=0
Interm ediate
Point
/-in te rc e p t
( f - ’l
(? •» )
/ = it
(¥ ■ ’ )
(f.o )
£
CO
II
Function
II
o
o
Graphing a Cotangent
Function When using a calculator
to graph a cotangent function,
enter the reciprocal of tangent,
Interm ediate
Point
Following the same guidelines that you used for the tangent
function, sketch the curve through the indicated key points
that you found. Then sketch one cycle to the left and right
of the first curve.
GuidedPractice
Locate the vertical asymptotes, and sketch the graph of each function.
3A. y = —cot 3x
272
L esson 4 -5
G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s
3B. y = 3 cot |-
Vertical
Asymptote
3 tt.
The reciprocals of the sme and cosine functions are defined as csc x = —— and sec x =
>as
, ,
sm x
cosx
shown below.
The cosecant function has asym ptotes when sin x
=
0, w hich occurs at integer m ultiples of t t .
Likewise, the secant function has asymptotes when cos x = 0, located at odd multiples of y .
Notice also that the graph of y = csc x has a relative m inim um at each m axim um point on the
sine curve, and a relative m axim um at each m inim um point on the sine curve. The same is true
for the graphs of 1/ = sec x and y = cos x.
The properties of the cosecant and secant functions are sum m arized below.
TechnologyTip
Graphing Graphing the cosecant
and secant functions on a
calculator is similar to graphing
the cotangent function. Enter the
reciprocals of the sine and cosine
functions.
KeyConcept Properties of the Cosecant and Secant Functions
Secant Function
Cosecant Function
D o m a in :
x e R,
D o m a in :
x e R , x ± y + me, n e Z
R ange:
(— 00, — 1] and [1 , 00)
R ange:
(— 00, —1] and [1,00)
x -in te rc e p ts :
none
x -in te rc e p ts :
none
y -in te r c e p t:
none
y -in te r c e p t:
1
C o n tin u ity :
infinite discontinuity at
x = mt, n e z
C o n tin u ity :
A s y m p to te s :
x = n it, / ) € Z
A s y m p to te s :
x= y +
S y m m e try :
origin (odd function)
S y m m e try :
y-axis (even function)
E nd B eh a vio r:
ne Z
lim csc x and lim csc x do not exist.
X — ►— 0 0
X— >0 0
infinite discontinuity at
x = y + /?TV, n e l
B e h a v io r:
n-K, n e Z
lim sec x and lim secxdo not exist.
X— * — 00
X — >00
The function oscillates between --00 and 00
period: 2 tt
period: 2 it
Like the sinusoidal functions, the period of a secant function of the form y = a sec (bx + c) + d
or cosecant function of the form y = a csc (bx + c) + d can be found by calculating — . Two vertical
asym ptotes for the secant function can be found by solving the equations bx + c = —y and
bx + c = — and two vertical asym ptotes for the cosecant function can be found by solving
bx + c = —t t and bx + c =
tt.
M
connectED.m cgraw-hill.com |
273
To sketch the graph of a cosecant or secant function, locate the asym ptotes of the function and find
the corresponding relative m axim um and m inim um s points.
B T E T n ffB E l Sketch Graphs of Cosecant and Secant Functions
Locate the vertical asymptotes, and sketch the graph of each function,
a. i / = c s c ( x + y )
The graph of y = csc^x + y j is the graph of y = csc x shifted y units to the left. The period
is
or 2 tx. Two vertical asym ptotes occur w hen bx + c
3tv
Tt
asymptotes are x + — = —tt or x ■
2
2
=
— tv
and bx + c =
7V.
Therefore, two
TV
2'
Create a table listing key points, including the relative m axim um and minim um , that are
located betw een the two vertical asym ptotes at x = —y ^ and x = y .
Function
StudyTip
Finding Asymptotes and Key
Points You can use the periodic
nature of trigonometric graphs to
help find asymptotes and key
points. In Example 4a, notice that
the vertical asymptote x = ~
y = cscx
Vertical
Asymptote
Relative
M inim um
X= — TT
(-f'-’l
x= 0
I?'1)
3 It
T
(-Tt, -1 )
—
f
Vertical
Asymptote
X =
(0,1)
* =
TT
f
Sketch the curve through the indicated key points for the function. Then sketch one cycle to
the left and right. The graph is show n in Figure 4.5.1 below.
is
and x = y .
Relative
M axim um
y = esc ( * + f )
equidistant from the calculated
asymptotes, x = — y
Vertical
Asymptote
b.
y = sec j
The graph of y = sec j is the graph of y = sec x expanded horizontally. The period is yyor 8 tv . Two vertical asym ptotes occur when bx + c = —y and bx + c = y k Therefore, 4
two asym ptotes are j + 0 :
■y or x = —2 tt and ^ + 0 = y ^ or x = 6 tt.
Create a table listing key points that are located betw een the asymptotes at x = —2 tt
and x = 6tt.
f
Relative
Minim um
(0,1)
*= 7
x = 2tt
Relative
M axim um
(tt, -1 )
(4ir, —1)
Vertical
Asymptote
„
3it
2
II
(0,1)
Vertical
Asymptote
£
CO
y= sec j
—
II
I
3
y = sec x
Vertical
Asymptote
ro
Function
Sketch the curve through the indicated key points for the function. Then sketch one cycle to
the left and right. The graph is shown in Figure 4.5.2 below.
^ GuidedPractice
4A. y = csc 2x
274
| Lesson 4 -5 | G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s
4B. y = sec (x + 7r)
Damped Trigonometric Functions W hen a sinusoidal function is multiplied by
another function /(x), the graph of their product oscillates betw een the graphs of y = f( x )
and y = —f{ x ). W hen this product reduces the am plitude of the wave of the original sinusoid,
it is called damped oscillation, and the product of the two functions is know n as a damped
trigonom etric function. This change in oscillation can be seen in Figures 4.5.3 and 4.5.4 for the
graphs of y = sin x and y = 2x sin x.
2
StudyTip
y = sin x
Damped Functions Trigonometric
functions that are multiplied by
constants do not experience
damping. The constant affects the
amplitude of the function.
V
—9tt
9/
I
Figure 4.5.3
A damped trigonometric function is of the form y = f( x ) sin bx or y = f( x ) cos bx, where f( x ) is the
damping factor.
Damped oscillation occurs as x approaches
± 0 0
or as x approaches 0 from both directions.
■ I H S H i j S k e t c h Damped Trigonometric Functions
><
ii
i
CO
Identify the damping factor f(x ) of each function. Then use a graphing calculator to sketch
the graphs of fix ) , —f(x ), and the given function in the same view ing window. Describe the
behavior of the graph.
a. y = —3x cos x
The function y = —3x cos x is the product of the
functions y = —3x and y = cos x, so f( x ) — —3x.
x l \
w
y = -3 x c o s x
!/
The am plitude of the function is decreasing
as x approaches 0 from both directions.
- 3x
[—4 ir, 4 ir] scl:
tt
by [ - 4 0 , 40] scl: 5
b. y — x2 sin x
Math HistoryLink
The function y = x 2 sin x is the product of the
functions y = x 2 and y = sin x. Therefore, the
dam ping factor is/(x) = x 2.
Cathleen Synge M oraw etz
(1923—)
A Canadian, Morawetz studied the
scattering of sound and magnetic
waves and later proved results
relating to the nonlinear wave
equation.
The am plitude of the function is decreasing as
x approaches 0 from both directions.
-4 ic , 4 ir] scl:
it
by [ - 1 0 0 ,1 0 0 ] scl: 10
C. y — 2Xcos 3x
The function y = 2 T cos 3x is the product of the
functions y = 2* and y = cos 3x, so /(x) = 2 X.
The am plitude of the function is decreasing
as x approaches —0 0 .
y
GuidedPractice
5A. y = 5x sin x
5B. y = \ cos x
5C. y = 3 Xsin x
connectED.m cgraw-hill.com
275
W hen the amplitude of the m otion of an object decreases w ith time due to friction, the m otion is
called damped harmonic motion.
KeyConcept Damped Harmonic Motion
Words
An object is in damped harmonic motion
when the amplitude is determined by the
function a (f) = ke~ct.
Symbols
F o r y = /fe - c f sin w fan d y = k e ~ a cos
ojt, where c > 0, k is the displacement,
c is the damping constant, f is time, and
w is the period.
The greater the dam ping constant c, the faster the am plitude approaches 0. The m agnitude of c
depends on the size of the object and the material of w hich it is com posed.
S
M
lk
S
S
M
il
DamPecl Harmonic Motion
MUSIC A guitar string is plucked at a distance of 0.8 centim eter above its rest position and
then released, causing a vibration. The dam ping constant for the string is 2.1, and the note
produced has a frequency of 175 cycles per second.
a.
Write a trigonometric function that m odels the m otion of the string.
The m axim um displacem ent of the string occurs w hen t = 0, so y = k e~ctcos u>t can be
used to model the motion of the string because the graph of y = cos t has a y-in te rc e p t other
than 0.
The maxim um displacement occurs w hen the string is plucked 0.8 centimeter. The total
displacem ent is the m axim um displacem ent M m inus the m inim um displacem ent m, so
k = M — m = 0.8 — 0 or 0.8 cm.
You can use the value of the frequency to find u).
IcjI
y = 175
M = 350tt
Real-WorldLink
Each string on a guitar is
stretched to a particular
length and tautness. These
aspects, along with the weight
and type of string, cause it to
vibrate with a characteristic
frequency or pitch called its
fundamental frequency,
producing the note we hear.
Source:
M
2ir
: frequency
Multiply each side by 2-k .
Write a function using the values of k, u;, and c.
y = 0.8e_21,cos 350 tt/is one m odel that describes the m otion of the string.
b.
Determine the amount of time t that it takes the string to be damped so that
- 0 .2 8 < y < 0.28.
Use a graphing calculator to determ ine the value of t
when the graph of y = 0 .8 e "2'1,cos 3507rf is oscillating
betw een y = —0.28 and y = 0.28.
HowStuffWorks
Y1=0.B4''<-2.iK)*C05(3E(HT-
From the graph, you can see that it takes approxim ately
0.5 second for the graph of y = 0.8e~2,lfcos 350-Trt to
oscillate w ithin the interval —0.28 < y < 0.28.
X=.£
[ 0, 1
Y = -.2 7 3 5 5 0 2
scl: 0.5 by [ - 0 .7 5 , 0.75] scl: 0.25
» GuidedPractice
6.
MUSIC Suppose another string on the guitar w as plucked 0.5 centim eter above its rest
position with a frequency of 98 cycles per second and a dam ping constant of 1.7.
A. Write a trigonometric function that models the m otion of the string y as a function
of time t.
B. Determine the time t that it takes the string to be damped so that —0.15 < y < 0.15.
276
| Lesson 4 -5
G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s
Exercises
= Step-by-Step Solutions begin on page R29.
Locate the vertical asymptotes, and sketch the graph of each
function. (Examples 1-4)
1
. y =
2
tan x
2. y =
ta n
(x +
3. y = cot (x -
4. y = —3
s. y = - \ c o tx
6. y =
—ta n
7. y
8. y =
c o t|
=
—2 tan ( 6x — t t )
9. y = \ esc 2x
11. y = sec (x 4- 7r)
ta n
28. DIVING The end of a diving board is 20.3 centimeters
above its resting position at the m om ent a diver leaves
the board. Two seconds later, the board has moved down
and up 12 times. The dam ping constant for the board is
0.901. (Example 6)
-|
3x
~ r
20.3
resting
' position
I
10. y = csc ^4x +
12. y = —2 csc 3x
a. W rite a trigonom etric function that models the motion
of the diving board y as a function of time f.
13. y = 4 sec (x —
14. y = sec ( f + f )
15. y = | c s c ( x - ^ )
16. y -
b. Determ ine the am ount of time f that it takes the diving
board to be damped so that —0.5 < y < 0.5.
-se c:
Locate the vertical asym ptotes, and sketch the graph of each
function.
Identify the damping factor fix ) of each function. Then use a
graphing calculator to sketch the graphs of fix ), —fix ) , and
the given function in the same viewing window. Describe
the behavior of the graph. (Example 5)
17. y = t * sin x
18. y = 4x cos x
19. y = 2x2 cos x
20. y = — sin x
21. y = i-x sin 2x
22. y = (x — 2 )2 sin x
23. y = e°'5x cos x
24. y = 3 Xsin x
25. y = |x| cos 3x
26. y = In x cos x
x3 •
29. y = sec x + 3
30. y = sec
t
32. y = csc
31. y = csc
3
- 2
33. y = cot {2x +
tt)
—3
34. y = cot
+ 4
( * +?
)+ 3
( ! + ?)
-
1
35) PHOTOGRAPHY Jeff is taking pictures of a haw k that is
flying 150 feet above him. The hawk will eventually fly
directly over Jeff. Let d be the distance Jeff is from the
haw k and 9 be the angle of elevation to the haw k from
Jeff's camera.
27. MECHANICS W hen the car shown below hit a bum p in
150 ft
the road, the shock absorber was com pressed 8 inches,
released, and then began to vibrate in damped harm onic
motion with a frequency of 2.5 cycles per second. The
damping constant for the shock absorber is 3. (Example 6)
Rest Position
X
(*-? )
a. Write d as a function of 9.
b. Graph the function on the interval 0 < 9 <
tt .
C. Approxim ately how far away is the haw k from Jeff
w hen the angle of elevation is 45°?
36. DISTANCE A spider is slowly clim bing up a wall. Brianna
is standing 6 feet away from the w all watching the spider.
Let d be the distance Brianna is from the spider and 9 be
the angle of elevation to the spider from Brianna.
a. Write a trigonometric function that models the
displacement of the shock absorber y as a function of
time f. Let t = 0 be the instant the shock absorber is
released.
a. W rite d as a function of 9.
b. Determine the am ount of tim e t that it takes for the
C.
amplitude of the vibration to decrease to 4 inches.
b. Graph the function on the interval 0 < 9 < y .
A pproxim ately how far away is the spider from
Brianna w hen the angle of elevation is 32°?
to—-...... *............^
fllconnectED.mcgraw-hill.com |
277
GRAPHING CALCULATOR Find the values of 6 on the interval
— t v < 0 < 7 T that make each equation true.
GRAPHING CALCULATOR Graph each pair of functions on the
same screen and m ake a conjecture as to w hether they are
equivalent for all real num bers. Then use the properties of
the functions to verify each conjecture.
37. cot 9 = 2 sec 9
38. sin 9 = cot i
39. 4 cos 9 = csc 9
40. tan — = sin I
48. f i x ) = sec x cos x; g(x) = 1
41. csc 9 = sec 9
42. tan 9 = sec
49. f{ x ) = sec2 x; gix ) = tan2 x + 1
43. TENSION A helicopter is delivering a large mural that is to
be displayed in the center of town. Two ropes are used to
attach the mural to the helicopter, as shown. The tension
T on each rope is equal to half the downward force times
50. fi x ) = cos x csc x; gix) = cot x
51'
= ------z 1 of\' g(x) = sin x
sec Ix - —j
Write an equation for the given function given the period,
phase shift (ps), and vertical shift (vs).
sec -•
52.
function: sec; period: 3ir; ps: 0; vs: 2
(53) function: tan; period: y ps:
vs: —1
54. function: csc; period: y ps: —7r; vs: 0
55. function: cot; period: 3tt; ps: y vs: 4
a. The downward force in newtons equals the m ass of
the mural times gravity, which is 9.8 newtons per
kilogram. If the mass of the mural is 544 kilograms,
find the downward force.
b. Write an equation that represents the tension T on
each rope.
c. Graph the equation from part b on the interval
[0,180°].
56. function: csc; period: y ps: —y vs: —3
H.O.T. Problems
Use Higher-Order Thinking Skills
57. PROOF Verify that the y-intercept for the graph of any
function of the form y = ke~c> cos u)t is k.
REASONING D eterm ine w hether each statement is true or
d. Suppose the mural is 9.14 meters long and the ideal
angle 9 for tension purposes is a right angle.
Determine how much rope is needed to transport the
mural and the tension that is being applied to each
rope.
e. Suppose you have 12.2 meters of rope to use to
transport the mural. Find 9 and the tension that is
being applied to each rope.
fa ls e . Explain your reasoning.
58. If b ^ 0, then y = a + b sec x has extrema of ± {a + b).
59. If x = 9 is an asym ptote of y = csc x, then x = 9 is also an
asym ptote of y = cot x.
60. ERROR ANALYSIS Mira and Arturo are studying the graph
shown. M ira thinks that it is the graph of y = ~
tan 2x,
and Arturo thinks that it is the graph of y = —cot 2x. Is
Match each function with its graph.
either of them correct? Explain your reasoning.
r
U
1
lI
1»
n
44.
y = c s c (| + j ) - 2
45.
y = se c(f + f ) - 2
46.
y = cot (lx - j j - 2
47.
y = ta n (2 x —! ) —2
278
Lesson 4 -5
G ra p h in g O th e r T r ig o n o m e tric F u n c tio n s
61. CHALLENGE Write a cosecant function and a cotangent
function that have the same graphs as y = sec x and
y = tan x respectively. Check your answers by graphing.
62. WRITING IN MATH A damped trigonom etric function
oscillates betw een the positive and negative graphs of the
dam ping factor. Explain why a damped trigonometric
function oscillates betw een the positive and negative
graphs of the dam ping factor and why the am plitude of
the function depends on the dam ping factor.
Spiral Review
State the amplitude, period, frequency, phase shift, and vertical shift of each function. Then
graph two periods of the function. (Lesson 4-4)
63.
y = 3 sin ( i x — y j + 10
64. y = 2 cos (3x +
— 665• y — \ cos (4x — tt) 4-1
Find the exact values of the five rem aining trigonom etric functions of 9. (Lesson 4-3)
66.
sin 9 =
D
cos 9 > 0
67. cos 9 =
0
sin 9 > 0
68. tan 9 =
/
7
sin 9 > 0
69. POPULATION The population of a city 10 years ago was 45,600. Since then, the population has
increased at a steady rate each year. If the population is currently 64,800, find the annual
rate of growth for this city. (Lesson 3-5)
70. MEDICINE The half-life of a radioactive substance is the am ount of time it takes for
half of the atoms of the substance to disintegrate. N uclear m edicine technologists use the
iodine isotope 1-131, w ith a half-life of 8 days, to check a patient's thyroid function. After
ingesting a tablet containing the iodine, the isotopes collect in the patient's thyroid, and a
special camera is used to view its function. Suppose a patient ingests a tablet containing
9 microcuries of 1-131. To the nearest hour, how long will it be until there are only
2 .8 microcuries in the patient's thyroid? (Lesson 3-2)
Factor each polynomial completely using the given factor and long division. (Lesson 2-3)
71. x 3 + 2x2 — x — 2; x — 1
74.
72. x 3 + x 2 — 16x — 16; x + 4
73.
x 3 — x 2 — lOx — 8; x + 1
EXERCISE The Am erican College of Sports M edicine recom m ends that healthy adults
exercise at a target level of 60% to 90% of their m axim um heart rates. You can estimate your
maximum heart rate by subtracting your age from 220. Write a com pound inequality that
models age a and target heart rate r. (Lesson 0-5)
Skills Review fo r Standardized Tests
75.
76.
SAT/ACT In the figure, A and D are the centers of
the two circles, which intersect at points C and E. CE
is a diameter of circle D. If AB = CE = 10, w hat is AD?
A 5
C 5V 3
B 5V 2
D 10\/2
77.
W hich equation is represented by the graph?
E 10V 3
B y = cot (0 —
C y = ta n ( 9 + j j
REVIEW Refer to the figure below. If c = 14, find the
value of b.
D y = ta n (< 9 -^ )
78.
REVIEW If sin 9 =
F
V3
F T
G 14\/3
H 7
G
and tt < 9 <
then
1=
?
H f
13n
12
7it
r
4-k
J 7V3
connectlD .m cgfaw -hill.com
&
| 279
Inverse Trigonometric Functions
Then
You found and
graphed the inverses
of relations
and functions.
Now
•1
(Lesson 1-7)
S
NewVocabulary
® arcsine
arnsins fnnntinn
function
arccosine function
arctangent function
2
: Why?
Evaluate and graph
inverse trigonometric
functions.
Find compositions
of trigonometric
functions.
Inverse trigonometric functions can
be used to model the changing
horizontal angle of rotation needed
for a television camera to follow the
motion of a drag-racing vehicle.
| Inverse Trigonometric Functions
In Lesson 1-7, you
■ learned that a function has an inverse function if and
only if it is one-to-one, meaning that each y-value of the
function can be matched with no more than one x-value.
Because the sine function fails the horizontal line test, it
is not one-to-one.
If, however, we restrict the domain of the sine function to the interval
7C TT
■2 , 2 , the restricted
function is one-to-one and takes on all possible range values [—1 ,1 ] of the unrestricted function.
It is on this restricted dom ain that y = sin x has an inverse function called the inverse sine function
y = sin -1 x. The graph of y = sin -1 x is found by reflecting the graph of the restricted sine function
in the line y = x.
Notice that the dom ain of y = sin 1 x is [—1 ,1 ], and its range is
—y y
. Because angles and arcs
given on the unit circle have equivalent radian m easures, the inverse sine function is sometimes
referred to as the arcsine fu nction y = arcsin x.
In Lesson 4-1, you used the inverse relationship betw een the sine and inverse sine functions to find
acute angle measures. From the graphs above, you can see that in general,
y = sin -1 x or y = arcsin x iff sin y = x, when —1 < x < 1 and —y < y <
iff means if and only if.
This means that sin -1 x or arcsin x can be interpreted as the angle (or arc) between —y and y with a
sine o fx . For exam ple, sin -1 0.5 is the angle with a sine of 0.5.
280
| Lesson 4-6
Recall that sin t is the y-coordinate of the point on the unit circle
corresponding to the angle or arc length t. Because the range of
the inverse sine function is restricted to
Inverse S ine Values
sin
TV TV
•2 , 2 , the possible
angle m easures of the inverse sine function are located on the
right half of the unit circle, as shown.
You can use the unit circle to find the exact value of some expressions involving sin
x or arcsin x.
■ 2 2 I E E S Q Evaluate Inverse Sine Functions
Find the exact value of each expression, if it exists.
Technology Tip
Evaluate sin-1 You can also use
a graphing calculator to find the
a. sin -l 1
-1
J
.52359 87756
.5235987756
2
Therefore, sin 1 \
2
6
2
V
___________
h.
D
2
6
CHECK If s in " 1 ± = Z then sin £ = i
Make sure you select RADIAN on
the MODE feature of your
graphing calculator.
■2 , 2
T
TT
w ith a u-coordinate of —. W hen t = —, sin t = —.
s i n _1< 0 . 5 )
Jt/6
TV TV
Find a point on the unit circle on the interval
angle that has a sine of 1 .
6
6
2
✓
• ( I— — I
arcsin
Find a point on the unit circle on the interval
with a v-coordinate of
j
Therefore, arcsin
CHECK If arcsin
C.
2
W hen t =
4
TV TV
-2, 2
sin t =
-1
2
= —- j.
then sin
✓
sin 1 3
Because the dom ain of the inverse sine function is [—1 ,1 ] and 3 > 1, there is no angle w ith a
sine of 3. Therefore, the value of sin -1 3 does not exist.
► GuidedPractice
*■#)
1A. arcsin
1B. sin 1 (—2tv)
Notice in Exam ple la that while sin ^jr- is also
not in the interval
TV TV
' 2’ 2
2'
6
1C. arcsin (—1)
is
Therefore, sin -1 \ =f=
L
6
co n n e ctE D~m cgraw -hill ~com H
281
StudyTip
Principal Values Trigonometric
functions with restricted domains
are sometimes indicated with
capital letters. For example,
y = Sin x represents the function
W hen restricted to a dom ain of [0, tv ], the cosine function is one-to-one and takes on all of its
possible range values on [—1 ,1 ], It is on this restricted dom ain that the cosine function has an
inverse function, called the inverse cosine fu n ction y = c o s-1 x or arccosine function y = arccos x.
The graph of y = co s-1 x is found by reflecting the graph of the restricted cosine function in the
line y = x.
In verse C osine Function
R estricted C o sin e Function
y = sin x, where — y < x < y .
1.5- y
The values in these restricted
domains are often called principal
I y = cos x
y
values.
«
0.5/
■ \
I
\
\
/ 0
-0 .5 -
\
!
1
»
/
/
tv
I
'
2 tv
\
\
I
’
3 tt
■
A
V
-1 .5 -
Recall that cos t is the x-coordinate of the point on the unit circle
corresponding to the angle or arc length t. Because the range
of y = co s-1 x is restricted to [0, tv ], the values of an inverse
cosine function are located on the upper half of the unit circle.
Evaluate Inverse Cosine Functions
Find the exact value of each expression, if it exists.
a. co s'
(-#)
Find a point on the unit circle in the interval [0,
an x-coordinate of
2
W hen t —
4
tv]
cos t = —
w ith
2
Therefore, cos 1 j —y p j =
CHECK
b.
If co s” 1
(-#)
-
4
then cos ~
V2
4
arccos (—2)
Since the domain of the cosine function is [—1 ,1 ] and —2 < —1, there is no angle w ith a
cosine of —2. Therefore, the value of arccos (—2) does not exist.
c.
cos 1 0
Find a point on the unit circle in the interval [0,
an x-coordinate of 0. W hen t = y , cos t = 0.
tv]
w ith
(0,1)
y
^
t= P \
Therefore, co s-1 0 = y .
-il
O
CHECK If c o s " 1 0 = £ , then cos ^ = 0. ✓
-1,
y
GuidedPractice
2A. cos
282
| Lesson 4 -6
(-#)
In v e rs e T r ig o n o m e tric F u n c tio n s
2B. arccos 2.5
2C. cos
- B )
l
/
■
*
W hen restricted to a dom ain of ( - y , y ) , the tangent function is one-to-one. It is on this restricted
StudyTip
End Behavior of Inverse
Tangent Notice that when the
graph of the restricted tangent
function is reflected in the line
y = x, the vertical asymptotes at
>domain that the tangent function has an inverse function called the inverse tangent function
y = tan -1 x or arctangent function y = arctan x. The graph of y = ta n -1 x is found by reflecting the
graph of the restricted tangent function in the line y = x. N otice that unlike the sine and cosine
functions, the dom ain of the inverse tangent function is (—0 0 , 0 0 ).
x = ± y become the horizontal
asymptotes y = ±~- of the
inverse tangent function.
Therefore,
lim tan_ 1 x = - ^
2
X-»-oo
and lim tan-1 * = ■?•
2
X -*o o
You can also use the unit circle to find the value of an inverse
tangent expression. On the unit circle, tan t =
or —. The values
of y = ta n -1 x will be located on the right half of the unit circle, not
including —y and y , because the tangent function is undefined at
those points.
TechnologyTip
Evaluate tan ~ 1 You can also use
a graphing calculator to find the
angle that has a tangent of V 3 .
[U ir T O X lJ)
m
>
i
Find the exact value of each expression, if it exists.
a. tan _1V J
Find a point (x, y) on the unit circle in the interval
V3
1.047197551
ji/3
Evaluate Inverse Tangent Functions
1.047197551
( —y , y j such that — = V 3 . W hen t — y , tan t = - ~
~2
or \/3. Therefore, tan -1 \J?> = y .
Make sure you select RADIAN on
the MODE feature of your
graphing calculator.
CHECK If tan -1 V 3 = y , then tan y = \[3. </
b.
arctan 0
Find a point (x, y) on the unit circle in the interval
( —y , y ) such that y = 0. W hen t = 0, tan t = y or 0.
Therefore, arctan 0 = 0.
CHECK If arctan 0 = 0, then tan 0 = 0. ✓"
p GuidedPractice
3A. arctan
(-#)
3B. ta n ” 1 (—1)
W hile inverse functions for secant, cosecant, and cotangent do exist, these functions are rarely used
in com putations because the inverse functions for their reciprocals exist. Also, deciding how to
restrict the domains of secant, cosecant, and cotangent to obtain arcsecant, arccosecant, and
arccotangent is not as apparent. You will explore these functions in Exercise 66.
283
The three most com m on inverse trigonom etric functions are sum m arized below.
KeyConcept Inverse Trigonometric Functions
Words
The angle (or arc) between - y
Words
and y with a sine of x.
Symbols
Inverse Tangent of x
Inverse Cosine of x
Inverse Sine of x
y = sin-1 X if and only if sin y = x,
for — 1 < x < 1 and - y < y < y .
Symbols
The angle (or arc) between 0 and tv
with a cosine of x.
Words
y = cos-1 x if and only if cos y = x,
for —1 < x < 1 and 0 < y < tv.
Symbols
The angle (or arc) between - y
and y
with a tangent of x.
y = t a n _1 x if and only if tan y = x , for
— oo < x < oo and - y < y < y .
Domain: [-1,1 ]
Domain: [ - 1 , 1 ]
Domain:
Range: [ - f , f ]
Range: [0 , tv]
Range: ( - y f )
1-
1
< - * ----------- >- -TV
y
------------------ .►
i
1
1
1
y
( - o o , oo)
y=cos~1x
y = sin—1 x
y = tan-1 x
L
I
-
-J
1x
0
1
-1
^
—TV-
TT
2
You can sketch the graph of one o f the inverse trigonom etric fu nctions show n above by
rew riting the function in the form sin y = x, cos y = x, or tan y = x, assigning values to y and
m aking a table of values, and then plotting the poin ts and connecting the points w ith a sm ooth
curve.
K 5 3 j j 3 3 I a D Sketch Graphs of Inverse Trigonometric Functions
Sketch the graph of y = arccos 2x.
By definition, y = arccos 2x and cos y = 2x are equivalent on 0 < y <
7V,
so their graphs are the
same. Rewrite cos y = 2x as x = j cos y and assign values to y on the interval [0,
table of values.
WatchOut!
Remember that tv = 3.14 radians
or 180°.
o
TT
4
IV
6
7T
2
1
V2
V3
2
o
4
4
57T
6
3 TV
4
7V
V3
V2
1
4
4
2
Then plot the points (x, y) and connect them
with a sm ooth curve. Notice that this curve
has endpoints at
j , 7vj and |-i,
oj, indicating
that the entire graph of y — arccos 2x is shown.
p GuidedPractice
Sketch the graph of each function.
4A. y = arcsin 3x
284
| L esson 4 -6 j In v e rs e T r ig o n o m e tric F u n c tio n s
4B. y = tan 1 2x
tv ]
to make a
Real-World Example 5 Use an Inverse Trigonometric Function
MOVIES In a movie theater, a person's view ing angle for watching a movie changes depending
on where he or she sits in the theater.
a. Write a function modeling the view ing angle 9 for a person in the theater whose
eye-level w hen sitting is 4 feet above ground.
Draw a diagram to find the m easure of the
viewing angle. Let 9 1 represent the angle
form ed from eye-level to the bottom of the
screen, and let 02 represent the angle
formed from eye-level to the top of the
screen.
In the late 19th century, Thomas
Edison began work on a device to
record moving images, called the
kinetoscope, which would later
become the film projector. The
earliest copyrighted motion
picture is a film of one of Edison’s
employees sneezing.
32 ft
So, the view ing angle is 9 = 02 — 9V You can use the tangent function to find 9 Xand 92.
Because the eye-level of the person when seated is 4 feet above the floor, the distance
opposite 9l is 8 — 4 feet or 4 feet long.
Source: The Library of Congress
tan 0, = —
a
opp = 4 and adj = d
= tan -1 4
Inverse tangent function
The distance opposite 92 is (32 + 8) — 4 feet or 36 feet.
tan 9? = ^
a
= tan
opp = 36 and adj = d
-l 36
Inverse tangent function
So, the viewing angle can be m odeled by 9 = tan 1 ~
b.
— tan 1 j .
D eterm ine the distance that corresponds to the m axim um view ing angle.
The distance at w hich the m axim um view ing angle occurs is
the m axim um point on the graph. You can use a graphing
calculator to find this point.
From the graph, you can see that the m axim um view ing angle
occurs approxim ately 12 feet from the screen.
-.Y = S 3 .i3 0 1 0 £ „
[0 ,1 0 0 ] scl: 10 by [0, 60] scl: 5
p GuidedPractice
5.
TELEVISION Tucas has purchased a new flat-screen television. So that his fam ily will be able
to see, he has decided to hang the television on the w all as shown.
■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■HI
□
7.5 ft
I
A. Write a function m odeling the distance d of the m axim um view ing angle 9 for Lucas if his
eye level w hen sitting is 3 feet above ground.
B. Determ ine the distance that corresponds to the m axim um view ing angle.
V
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2
Compositions of Trigonometric Functions
In Lesson 1-7, you learned that if x
is in the domain of f( x ) and f ~ \ x ), then
/ [/ “ '(*)] = x
and
f ~ l lf(x )] = x.
Because the domains of the trigonom etric functions are restricted to obtain the inverse
trigonometric functions, the properties do not apply for all values of x.
For example, while sin x is defined for all x, the dom ain of sin -1 x is [—1 ,1 ]. Therefore,
sin (sin-1 x) = x is only true w hen —1 < x < 1. A different restriction applies for the com position
sin 1 (sin x). Because the dom ain of sin x is restricted to the interval
TT TT
■2 , 2
, sin
(sin x) = x is
only true w hen - ~ < x < ^ - .
These dom ain restrictions are sum m arized below.
KeyConcept Domain of Compositions of Trigonometric Functions
f[f-Hx)) = x
f- '[ f(x ) ] = x
If - 1 < x < 1, then sin (sin-1 x) = x.
If - y
< x < y , then sin-1 (sin x) = x.
If - 1 < x < 1, then cos (cos-1 x) = x.
If 0 < x < tt, then cos-1 (cos x) = x.
If - o o < x < oo, then tan (tan-1 x) = x.
If - y
< x < y , then tan-1 (tan x) = x.
V
-j
Use Inverse Trigonometric Properties
Find the exact value of each expression, if it exists.
H)
a. sin
The inverse property applies because —i lies on the interval [—1 ,1 ].
--H )
Therefore, sin
b.
i
~4'
arctan |tan y j
Because tan x is not defined w hen x = y , arctan |tan y j does not exist.
c. arcsin
7 77
WatchOut!
Compositions and
Inverses When computing
M [ f ( x ) ] with trigonometric
functions, the domain appears to
be (— oo, oo). However, because
the ranges of the inverse
functions are restricted,
coterminal angles must
sometimes be found.
(sinZr )
Notice that the angle — does not lie on the interval
w ith
4
—2
tt
or —
4
whi ch is on the interval
sin ^sin -y -j = arcsin j^sin ( ~ y j j
TT
>
Since -
' 4
Therefore,, arcsin |sin
77T
. However, ——is coterminal
4
TT TT
•2 , 2
• 7-ir
sm
^ = sin
4
TT TT
'2 ' 2
(-7)
■< —^
arcsin (sin x) = x.
JL
'4 '
G u id ed P ra ctice
6A. tan ^tan-1 y j
286
| L esson 4 -6 i In v e rs e T r ig o n o m e tric F u n c tio n s
6B. cos
6C. arcsin
in (sin f )
You can also evaluate the com position of two different inverse trigonom etric functions.
s .^ ii
Evaluate Compositions of Trigonometric Functions
Find the exact value of cos
To sim plify the expression, let u = ta n -1
-|J, so tan u =
Because the tangent function is negative in Quadrants II and
IV, and the dom ain of the inverse tangent function is restricted
to Quadrants I and IV, u m ust lie in Q uadrant IV.
Using the Pythagorean Theorem , you can find that the length
of the hypotenuse is 5. Now, solve for cos u.
cos u
adj
hyp
Cosine function
4
5
adj = 4 and hyp = 5
So, cos tan
4
5'
- H )
►GuidedPractice
Find the exact value of each expression.
7A, co s-1 (sin y j
7B. sin |arctan
Som etimes the com position of two trigonom etric functions reduces to an algebraic expression that
does not involve any trigonom etric expressions.
Evaluate Compositions of Trigonometric Functions
StudyTip
Decomposing Algebraic
Functions The technique used
to convert a trigonometric
expression into an algebraic
expression can be reversed.
Decomposing an algebraic
function as the composition
of two trigonometric functions
is a technique used frequently
in calculus.
Write tan (arcsin a) as an algebraic expression of a that does not involve trigonometric
functions.
>
Let u = arcsin a, so sin u = a.
Because the domain of the inverse sine function is
restricted to Quadrants I and IV, u m ust lie in
Quadrant I or IV. The solution is sim ilar for each
quadrant, so we will solve for Q uadrant I.
J
y
/
y
a
From the Pythagorean Theorem , you can find that
the length of the side adjacent to u is V l — a 2. Now,
solve for tan u.
opp
ta n u = — 77"
adj
=
n
V l-fl2
----- A
X
Tangent function
opp = a and adj = V i — a2
or a —
1-«
So, tan (arcsin a) = - —1 ■ a .
1 —u
p GuidedPractice
Write each expression as an algebraic expression of x that does not involve trigonometric
functions.
8A. sin (arccos x)
-
8B.cot
[sin-1 x]
V________________________________________________________
c o n n e c tE D jn c ^
i
287
Exercises
= Step-by-Step Solutions begin on page R29.
Find the exact value of each expression, if it exists.
DRAG RACE A television camera is film ing a drag race.
The camera rotates as the vehicles m ove past it. The
cam era is 30 m eters aw ay from the track. Consider 9
and x as show n in the figure. (Example 5)
(Examples 1 -3 )
1. sin-1 0
.
V3
2. arcsm - r -
. V2
3. arcsin - y
4. sin -1 j
5.
6. arccos 0
- 1 V2
7. cos 1 —
8. arccos (—1)
'A
9. arccos
V3
x
10. c o s-1 j
11. arctan 1
12. arctan (—V 3 )
V3
13. tan -1J —
14. tan-1 0
a. Write 9 as a function of x.
b. Find 9 when x = 6 meters and x = 14 meters.
15. ARCHITECTURE The support for a roof is shaped like two
right triangles, as show n below. Find 9. (Example 3)
28. SPORTS Steve and Ravi w ant to project a pro soccer
game on the side of their apartment building. They have
placed a projector on a table that stands 5 feet above the
ground and have hung a 12-foot-tall screen that is 10 feet
above the ground. (Example 5)
16. RESCUE A cruise ship sailed due west 24 miles before
turning south. W hen the cruise ship becam e disabled
and the crew radioed for help, the rescue boat found that
the fastest route covered a distance of 48 miles. Find the
angle 9 at w hich the rescue boat should travel to aid the
cruise ship. (Example 3)
a. Write a function expressing 9 in terms of distance d.
b. Use a graphing calculator to determ ine the distance
for the m axim um projecting angle.
Find the exact value of each expression, if it exists.
(Examples 6 and l)r
29.
sin |sin_1
30. sin 1 |sin y j
31.
cos ^cos-1
32.
33.
tan |tan-1
34. tan _ 1 | ta n -jj
35.
cos (tan-1 1)
36. sin -1 |cos y j
CO S
1 (cos 7v)
Sketch the graph of each function. (Example 4)
17. y = arcsin x
18. y = sin -1 2x
19. y = sin -1 (x + 3)
20. y = arcsin x — 3
21. y = arccos x
22. y = cos-1 3x
23. y = arctan x
24. y = tan -1 3x
25. y = tan -1 (x + 1)
26. y = arctan x — 1
288
Lesson 4 -6 ] In v e rs e T r ig o n o m e tric F u n c tio n s
sin (tan-1 1 — sin -1 1)
37. sin ^2 cos 1
38.
39. cos (tan- 1 1 — sin -1 1)
40. cos |cos-1 0 + sin -1
Write each trigonometric expression as an algebraic
expression of x. (Example 8)
Write each algebraic expression as a trigonometric function
of an inverse trigonom etric function of x.
(4 l) tan (arccos x)
42. csc (cos 1 x)
63.
43. sin (cos 1 x)
44. cos (arcsin x)
45.
csc (sin 1 x)
46. sec (arcsin x)
47.
cot (arccos x)
48. cot (arcsin x)
64.
Vl -X 2
65. Ǥl MULTIPLE REPRESENTATIONS In this problem , you will
explore the graphs of com positions of trigonometric
functions.
a. ANALYTICAL Consider/(x) = sin x and / -1 (x) = arcsin x.
Describe the dom ain and range o f / 0 / - 1 and/-1 o f .
Describe how the graphs of g(x) a n d /(x ) are related.
b. GRAPHICAL Create a table of several values for each
com posite function on the interval [—2, 2], Then use
the table to sketch the graphs o f / 0 / - 1 and/-1 o f .
Use a graphing calculator to check your graphs.
49. /(x) = sin -1 x and g (x ) = sin -1 (x — 1) — 2
50. /(x) = arctan x and g(x ) = arctan 0.5x — 3
51. f( x ) = cos-1 x a n d g (x ) = 3 (cos-1 x — 2)
c. ANALYTICAL Consider g(x) = cos x and g 1 (x) =
arccos x. Describe the dom ain and range of g o g -1
and g ~1 o g and m ake a conjecture as to what the
graphs of g o g _1 and g -1 o g will look like. Explain
your reasoning.
52. /(x) = arcsin x and g (x) = ^ arcsin (x + 2)
53. /(x) = arccos x and g (x) = 5 + arccos 2x
54. /(x) = tan-1 x and g (x ) = tan-1 3x — 4
d. GRAPHICAL Sketch the graphs of g o g -1 and g -1 o g.
Use a graphing calculator to check your graphs.
55. SAND W hen piling sand, the angle form ed betw een the
pile and the ground remains fairly consistent and is called
the angle o f repose. Suppose Jade creates a pile of sand at
the beach that is 3 feet in diam eter and 1.1 feet high.
e. VERBAL M ake a conjecture as to what the graphs of
the two possible com positions of the tangent and
arctangent functions will look like. Explain your
reasoning. Then check your conjecture using a
graphing calculator.
H.O.T. Problems
Use Higher-Order Thinking Skills
66. ERROR ANALYSIS Alisa and Trey are discussing inverse
a. W hat is the angle of repose?
trigonom etric functions. Because tan x =
b. If the angle of repose remains constant, how m any feet
in diameter would a pile need to be to reach a height
of 4 feet?
conjectures that ta n -1 x = sm
cos
of them correct? Explain.
Give the domain and range of each composite function.
Then use your graphing calculator to sketch its graph.
56.
y = cos (tan-1 x)
57. y = sin (cos-1 x)
58.
y = arctan (sin x)
59. y = sin -1 (cos x)
60.
y = cos (arcsin x)
61. y = tan (arccos x)
62.
67.
68.
, Alisa
x . Trey disagrees. Is either
x
CHALLENGE Use the graphs of y = sin - x and y — cos 1 .
to find the value of sin -1 x + co s-1 x on the interval
[—1 ,1 ]. Explain your reasoning.
REASONING Determ ine w hether the follow ing statem ent is
true o r false: If cos ■— = - y - , then co s-1 -^y- = -y-. Explain
your reasoning.
INVERSES The arcsecant function is graphed by restricting
the domain of the secant function to the intervals [o, y j
and ( y , i r j, and the arccosecant function is graphed by
REASONING D eterm ine w hether each function is odd, even, or
neither. Justify your answer.
y = sin -1 x
restricting the domain of the cosecant function to the
69.
intervals [—y ,
70. y — cos 1 x
0
) and ( 0 , y j .
a. State the domain and range of each function.
71. y = ta n -1 x
b. Sketch the graph of each function.
C.
Explain why a restriction on the dom ain of the secant
and cosecant functions is necessary in order to graph
the inverse functions.
72.
WRITING IN MATH Explain how the restrictions on the sine,
cosine, and tangent functions dictate the domain and
range of their inverse functions.
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289
Spiral Review
Locate the vertical asymptotes, and sketch the graph of each function. Lesson 4-5)
75. y = 3 csc -
73.
y = 3 tan 9
74. y = cot 5 9
76.
WAVES A leaf floats on the water bobbing up and down. The distance betw een its highest
and lowest points is 4 centimeters. It m oves from its highest point dow n to its low est point
and back to its highest point every 10 seconds. Write a cosine function that models the
movement of the leaf in relationship to the equilibrium point. (Lesson 4-4)
Find the value of x. Round to the nearest tenth, if necessary. (Lesson 4-1)
79.
77.
17.8
For each pair of functions, find [ f o g](x), [g o f](x ), and [ / o gl(4). (Lesson 1-6)
80.
81. f( x ) = 6 - 5 x
f( x ) = x 2 + 3x - 6
g(x) = 4x + l
82. f( x ) = \]x + 3
g {x ) = x 2 + 1
gW = j
83. EDUCATION Todd has answered 11 of his last 20 daily quiz questions correctly. His baseball
coach told him that he must raise his average to at least 70% if he w ants to play in the
season opener. Todd vows to study diligently and answer all of the daily quiz questions
correctly in the future. How m any consecutive daily quiz questions m ust he answer
correctly to raise his average to 70%? (Lesson 0-8)
Skills Review for Standardized Tests
84. SAT/ACT To the nearest degree, what is the angle of
depression 9 betw een the shallow end and the deep
end of the swimming pool?
86. REVIEW The hypotenuse of a right triangle is 67
inches. If one of the angles has a m easure of 47°, what
is the length of the shortest leg of the triangle?
A 45.7 in.
C 62.5 in.
B 49.0 in.
D 71.8 in.
87. REVIEW Two trucks, A and B, start from the
intersection C of two straight roads at the same time.
Truck A is traveling twice as fast as truck B and after 4
hours, the two trucks are 350 m iles apart. Find the
approxim ate speed of truck B in miles per hour.
A 25°
C 41°
B 37°
D 53°
E 73°
AK
:1 \
85. Which of the following represents the exact value of
sin |tan~1-^-)'?
\
:
\
90“
iff1
F --
2V5
H
J
290
\
1
Vf
350mi
\
„
5
2V5
5
| Lesson 4 -6 | In v e rs e T r ig o n o m e tric F u n c tio n s
F 39
H 51
G 44
J 78
LrfWi
The Law of Sines and
the Law of Cosines
:Then
•
You solved right
triangles using
trigonometric
functions.
: Why?
•
(Lesson 4-1)
®
NewVocabulary
oblique triangles
Law of Sines
ambiguous case
Law of Cosines
Heron’s Formula
1
Solve oblique
•
triangles by using the
Law of Sines or the
Law of Cosines.
I Find areas of oblique
■triangles.
Triangulation is the process of finding the
coordinates of a point and the distance to that point
by calculating the length of one side of a triangle,
given the measurements of the angles and sides of
the triangle formed by that point and two other
known reference points. Weather spotters can use
triangulation to determine the location of a tornado.
Solve Oblique Triangles In Lesson 4-1, you used trigonom etric functions to solve right
triangles. In this lesson, you will solve oblique triangles— triangles that are not right triangles.
1
You can apply the Law of Sines to solve an oblique triangle if you know the measures of two
angles and a nonincluded side (AAS), two angles and the included side (ASA), or two sides and
a nonincluded angle (SSA).
KeyConcept Law of Sines
If A ABC has side lengths a, b, and c representing the lengths of the
sides opposite the angles with measures
A fl,an d C ,th en ^ = ^ ® = ^ .
a
b
c
You will derive the Law of Sines in Exercise 69.
H S S E E D 3 E I Apply the Law of Sines (AAS)
Solve A A B C . Round side lengths to the nearest tenth
and angle m easures to the nearest degree.
Because two angles are given, C = 180° — (103° 4- 35°) or 42°.
Use the Law of Sines to find a and c.
sin B
sin A
sin 103° _ sin 35°
20
~
a
! sin 103° = 20 sin 35°
20 sin 35°
sin 103°
a = 11.8
sin B
Law of Sines
Substitution
Multiply.
sin 103°
20
sin C
sin 42°
c
c sin 103° = 20 sin 42°
20 sin 42°
sin 103°
Divide.
Use a calculator.
c ~ 13.7
Therefore, a = 11.8, c = 13.7, and ZC = 42°.
|fe GuidedPractice
Solve each triangle. Round side lengths to the nearest tenth and angle
measures to the nearest degree.
1A.
1B.
I
............. ........ .............. ........,.,,.1.1,...-........
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291
J j g E S E E i a S S ® Apply the Law of Sines (ASA)
SATELLITES An Earth-orbiting satellite is passing betw een the Oak Ridge Laboratory in
Tennessee and the Langley Research Center in Virginia, which are 446 miles apart. If the
angles of elevation to the satellite from the Oak Ridge and Langley facilities are 58° and 72°,
respectively, how far is the satellite from each station?
To gain an improved
understanding of the atmosphere,
land surface changes, and
ecosystem processes, NASA uses
a series of satellites as part of its
Earth Observing System (EOS) to
study the air, land, and water
on Earth.
Oak Ridge -•
446 m i
► Langley
Because two angles are given, C = 180° — (58° + 72°) or 50°. Use the Law of Sines to find the
distance to the satellite from each station.
Source: NASA
sin C _ shiB
Law of Sines
sin 50°
sin 72°
—7 7 7 — =
;—446
b
.
.
Substitution
b sin 50° = 446 sin 72°
u
Multiply.
446 sin 72°
sin 50°
b = 553.72
n. .;J„
sin C _ sin A
.
si n 50°
sin 58°
—7T7— —---------446
a
a sin 50° = 446 sin 58°
446 sin 58°
a _= ■
“
sin50'
Use a calculator.
« ~ 493.74
So, the satellite is about 554 miles from O ak Ridge and about 494 m iles from Langley.
p GuidedPractice
2.
SHIPPING Two ships are 250 feet apart and traveling to the same port as shown. Find the
distance from the port to each ship.
_ 2 5 0 ft_ _
Ship A N '~ 39°
StudyTip
4 1 V ' S h ip !
From geometry, you know that the measures of two sides and a nonincluded angle (SSA) do not
necessarily define a unique triangle. Consider the angle and side measures given in the figures
below.
Alternative Representations
The Law of Sines can also be
written in reciprocal form as
a _ b _ c
sin /I
sin 6
sin C ‘
In general, given the measures of two sides and a nonincluded angle, one of the follow ing will be
true: (1) no triangle exists, (2) exactly one triangle exists, or (3) two triangles exist. In other words,
when solving an oblique triangle for this am biguous case, there m ay be no solution, one solution,
or two solutions.
292
Lesson 4 -7 | T h e L a w o f Sines a n d th e L a w o f C osines
KeyConcept The Ambiguous Case (SSA)
Consider a triangle in which a, b, and Aare given. For the acute case, sin A = | , so h = b sin A.
A is Acute.
(A < 90°)
a < b and a < h
no solution
a < b and a = h
one solution
a < b and a > h
two solutions
one solution
To solve an ambiguous case oblique triangle, first determ ine the num ber of possible solutions. If the
triangle has one or two solutions, use the Law of Sines to find them.
E S S E E S * The Ambiguous Case— One or No Solution
StudyTip
Make a Reasonable Sketch
When solving triangles, a
reasonably accurate sketch can
help you determine whether your
answer is feasible. In your sketch,
check to see that the longest side
is opposite the largest angle and
that the shortest side is opposite
the smallest angle.
Find all solutions for the given triangle, if possible. If no solution exists, write no solution.
Round side lengths to the nearest tenth and angle m easures to the nearest degree.
a. a = 15, c = 12, A = 94°
Notice that A is obtuse and a > c because 15 > 12. Therefore,
one solution exists. A pply the Law of Sines to find C.
sin C _ sin 94°
12
15
sin C ;
Law of Sines
12 sin 94°
15
Multiply each side by 12.
c = s in -i (12sm 941) or about 53°
Definition of sin-1
Because two angles are now known, B ~ 180° — (94° + 53°) or about 33°. Apply the Law of
Sines to find b. Choose the ratios w ith the few est calculated values to ensure greater accuracy.
sin 94°
15
sin 33°
15 sin 33°
or about 8.2
sin 94°
Law of Sines
Solve for b.
Therefore, the rem aining m easures of A A BC are B ~ 33°, C ~ 53°, and b ~ 8.2.
b.
a = 9, b = 11, A — 61°
(
Notice that A is acute and a < b because 9 < 11. Find h.
sin 61° =
h = 11 sin 61° or about 9.6
Definition of sine
h = b sin A
Because a < h, no triangle can be formed with sides a = 9, b = 11,
and A = 61°. Therefore, this problem has no solution.
^ GuidedPractice
3A. a = 12, b = 8, B = 61°
3B. a = 13, c = 26, A = 30°
1 293
E 2 J J 3 H 3 3 The Ambiguous Case-Two Solutions
Find two triangles for which A = 43°, a = 25, and b = 28. Round side lengths to the nearest
tenth and angle measures to the nearest degree.
C
A is acute, and h = 28 sin 43° or about 19.1. N otice that
a < b because 25 < 28, and a > h because 25 > 19.1.
Therefore, two different triangles are possible w ith the
given angle and side measures. Angle B will be acute,
while angle B' will be obtuse.
A
Make a reasonable sketch of each triangle and apply the Law of Sines to find each solution. Start
with the case in which B is acute.
Solution 1
ZB is acute.
Find B.
B
s in
s in 4 3 °
28
Law of Sines
25
sin B =
2 8 s in 4 3 °
Solve for sin S.
25
Use a calculator.
sin B = 0.7638
TechnologyTip
B = s in -1 0.7638 or about 50°
Using sin-1 Notice that when
calculating sin-1 of a ratio, your
calculator will never return two
possible angle measures because
sin-1 is a function. Also, your
calculator will never return an
obtuse angle measure for sin- '
because the inverse sine function
has a range o f - y to y
or - 9 0 ° to 90°.
Definition of s i n - 1
Find C.
C = 180° - (43° + 50°) ~ 87°
Apply the Law of Sines to find c.
s in 8 7 °
s in 4 3 °
c
Law of Sines
25
„
2 5 s in
87°
or about 36.6
s in 4 3 °
Solution 2
Solve for c.
Z B' is obtuse.
Note that mZCB'B = mZCBB'. To find B', you need to find an
obtuse angle w ith a sine that is also 0.7638. To do this, subtract
the measure given by your calculator to the nearest degree, 50°,
from 180°. Therefore, B' is approxim ately 180° — 50° or 130°.
Find C.
C ~ 180° - (43° + 130°) or 7°
Apply the Law of Sines to find c.
s in 7 °
c
s in 4 3 °
Law of Sines
25
„
2 5 s in 7 °
s in 4 3 °
or about 4.5
Solve for c.
Therefore, the missing measures for acute A A BC are B ~ 50°, C = 87°, and c = 36.6, while the
missing measures for obtuse A AB'C are B' ~ 130°, C ~ 7°, and c ~ 4.5.
f
GuidedPractice
Find two triangles with the given angle measure and side lengths. Round side lengths to the
nearest tenth and angle measures to the nearest degree.
4A. A = 38°, a = 8, b = 10
294
L esson 4 -7 | T h e L a w o f Sines a n d th e L a w o f C osines
4B.
A
=
65°, a
= 55, b = 57
StudyTip
Law of Cosines Notice that the
You can use the Law of Cosines to solve an oblique triangle for the rem aining two cases: when
>you are given the m easures of three sides (SSS) or the m easures of two sides and their included
angle (SAS).
angle referenced in each equation
of the Law of Cosines corresponds
to the side length on the other
side of the equation.
KeyConcept Law of Cosines
In A A B C, if sides with lengths a, 6, and c are opposite angles with
measures A, B, and C, respectively, then the following are true.
a 2 = b2 + c 2 - 2bc cos A
b 2 = a2 + c2 - 2accos B
c 2 = a 2 + b 2 - 2aficos C
B
a 2 = b 2 + c 2 - 2bc cos A
y
b 2 = a 2 + c 2 - 2accos B
A
c 2 = a 2 + b 2 ~ 2a6cos C
\a
-------------bH----------
J
You will derive the first formula for the Law of Cosines in Exercise 70.
Real-World Example 5 Apply the Law of Cosines (SSS)
HOCKEY W hen a hockey player attempts a shot, he is 20 feet from the left post of the goal and
24 feet from the right post, as shown. If a regulation hockey goal is 6 feet wide, what is the
player's shot angle to the nearest degree?
20 ft /
shot
angle
StudyTip
24 ft
Since three side lenghts are given, you can use the Law of Cosines to find the player's shot
angle, A.
Check for Reasonableness
Because a triangle can have at
most one obtuse angle, it is wise
to find the measure of the largest
angle in a triangle first, which will
be the angle opposite the longest
side. If the largest angle is obtuse,
then you know that the other two
angles must be acute. If the largest
angle is acute, the remaining two
angles must still be acute.
a 2 — b 2 + c 2 — 2 b e cos A
Law of Cosines
62 = 242 + 202 - 2(24) (20) cos A
a = 6, b = 24, and c = 20
36 = 576 + 400 - 960 cos A
Simplify.
36 = 976 - 960 cos A
Add.
Subtract 976 from each side.
—940 = —960 cos A
940
960
cos
Divide each side by —960.
■cos
-1 / 9 4 0 \
= A
\9 6 0 /
Use the cos-1 function.
Use a calculator.
11.7° as A
So, the player's shot angle is about 12°.
y
GuidedPractice
5. HIKING A group of friends who are on a cam ping trip decide to go on a hike. According to the
map show n, what is the angle that is form ed by the two trails that lead to the camp?
Trail 1
I Checkpoint
Camp
2 mi
Trail 2
Checkpoint
Apply the Law of Cosines (SAS)
Solve A A BC . Round side lengths to the nearest tenth
and angle measures to the nearest degree.
EfljflTl Use the Law of Cosines to find the m issing side measure.
c2 = a 2 + b 2 — l a b cos C
Law of Cosines
c2 = 5 2 + 8 2 - 2(5)(8) cos 65°
a = 5 , b = 8, and C = 65°
c2 « 55.19
Use a calculator.
Take the positive square root of each side.
c ~ 7.4
ETTSffW Use the Law of Sines to find a m issing angle measure.
sin A
5
sin 65°
7.4
sin ^ _ sin C
a ~ c
sin A =
5 sin 65°
7.4
Multiply each side by 5.
Definition of sin-1
A = 38°
ETTflin Find the measure of the remaining angle.
B ~ 180° - (65° + 38°) or 77°
Therefore, c = 7.4, A « 38°, and B ~ 77°.
y
GuidedPractice
6. Solve A HJK if H = 34 ° ,j = 7, and k = 10.
l Find Areas of Oblique Triangles
W hen the measures of all three sides of a triangle are
i known, the Law of Cosines can be used to prove Heron's Form ula for the area of the triangle.
KeyConcept Heron’s Formula
StudyTip
If the measures of the sides of AABCare a, fi, and c, then the area of the
triangle is
Semiperimeter The measure s
used in Heron’s Formula is called
the semiperimeter oi the triangle.
B
where s = ^ ( a + b + c).
A
V
,
V
/
Area = y /s ( s - a )( s - b)(s — c),
b
C
J
............. ..... ... .............,...........
You will prove this formula in Lesson 5-1,
51 in.
Find the area of A XYZ.
The value of s is 2-(45 + 51 + 38) or 67.
Area = yjs(s ~ *)(s — 2/)(s — z)
p
Lesson 4 -7
= V 67(67 - 45)(67 - 51)(67 - 38)
s = 67, x = 45, y = 51, and z —38
= V 683,936
Simplify.
« 827 in2
Use a calculator.
GuidedPractice
7A.
296
Heron’s Formula
x = 24 cm , y = 53 cm, z = 39 cm
T h e L a w o f Sines a n d th e L a w o f C osines
7B. x = 61 ft, y = 70 ft, 2 = 88 ft
In the am biguous case of the Law of Sines, you com pared
the length of a to the value h = b sin A. In the triangle
shown, h represents the length of the altitude to side c
in A ABC. You can use this expression for the height of
the triangle to develop a formula for the area of the triangle.
Area = —c/z
Formula for area of a triangle
: — c(b sin A)
Replace h with b sin A.
Simplify,
= —be sin A
By a sim ilar argument, you can develop the form ulas
Area = —ab sin C
and
Area = —ac sin \
Notice that in each of these form ulas, the inform ation needed to find the area of the triangle is the
measures of two sides and the included angle.
KeyConcept Area of a Triangle Given SAS
StudyTip
Area of an Obtuse Triangle
This formula works for any type
of triangle, including obtuse
triangles. You will prove this in
Lesson 5-3.
W o rd s
The area of a triangle is one half the product of the lengths
of two sides and the sine of their included angle.
S y m b o ls
Area = — 6c sin A
C
Area = j a c sin B
A
Area = ^ab sin C
c
B
J
V
Because the area of a triangle is constant, the form ulas above can be w ritten as one formula.
I
l
l
Area = —be sin A = —ab sin C = —ac sin B
If the included angle m easures 90°, notice that each formula sim plifies to the formula for the area of
a right triangle, i(b a se )(h eig h t), because sin 90° = 1.
B S I i n F i n c l the Area of a Triangle Given SAS
Find the area of A G H J to the nearest tenth.
H
In A GHJ, g = 7 , h = 10, and J = 108°.
Area = ^ g h sin J
= j ( 7 ) ( l 0 ) sin 108°
= 3 3 .3
Area
of a triangle using SAS
Substitution
Simplify.
So, the area is about 33.3 square centimeters.
(►GuidedPractice
Find the area of each triangle to the nearest tenth.
8A.
25 yd
8B.
297
Exercises
= Step-by-Step Solutions begin on page R29.
Solve each triangle. Round to the nearest tenth, if necessary.
(Examples 1 and 2)
18. SKIING A ski lift rises at a 28° angle during the first 20 feet
up a m ountain to achieve a height of 25 feet, w hich is the
height m aintained during the rem ainder of the ride up
the m ountain. Determ ine the length of cable needed for
this initial rise. (Example 3)
1.
~~ ^
25 ft
3.
X
o j
28'
Ar
J O ft
Find two triangles with the given angle measure and side
lengths. Round side lengths to the nearest tenth and angle
measures to the nearest degree. (Example 4)
5. T
7. GOLF A golfer m isses a 12-foot p u tt b y p u tting 3° off
course. The hole now lies at a 129° angle b etw een the
b all and its spot before the putt. W hat d istance does
the golfer need to p u tt in order to m ake the shot?
(Examples 1 and 2)
8 . ARCHITECTURE An architect's client wants to build a home
based on the architect Jon Lautner's Sheats-Goldstein
House. The length of the patio will be 60 feet. The left side
of the roof will be at a 49° angle of elevation, and the right
side will be at an 18° angle of elevation. Determine the
lengths of the left and right sides of the roof and the angle
at which they will meet. (Examples 1 and 2)
19.
A = 39°, a = 12, b = 17
20. A = 26° , a = 5 ,b = 9
21.
A — 61°, a = 14, b = 15
22. A = 47°, a = 25, b = 34
23.
A = 54°, a = 31, b = 36
24. A = 18°, a = 8, b = 13
25. BROADCASTING A radio tow er located 38 m iles along
Industrial Parkw ay transm its radio broadcasts over a
30-m ile radius. Industrial Parkw ay intersects the
interstate at a 41° angle. How far along the interstate
can vehicles pick up the broadcasting signal? (Example 4)
Industrial
Parkway
\ Tower
\
41°
Y __
Interstate
: 9 j TRAVEL For the initial 90 miles of a flight, the pilot heads
8° off course in order to avoid a storm. The pilot then
changes direction to head toward the destination for the
remainder of the flight, making a 157° angle to the first
flight course. (Examples 1 and 2)
a. Determine the total distance of the flight.
b.
Determine the distance of a direct flight to the
destination.
10.
a = 9, b = 7, A = 108°
12.
a = 1 8 ,b = 12,A = 27°
14.
a = 14, b = 6, ,4 = 145°
15.
a=
19,b = 38, A
16.
a = 5 ,b = 6 ,A = 63°
17.
a=
10,b = V 2 0 0 , A = 45°
298
Lesson 4 -7
a=
14,b = 15, A
<
Solve each triangle. Round side lengths to the nearest tenth
and angle measures to the nearest degree. (Examples 5 and 6)
27. A A BC, if A = 42°, b = 12, and c = 19
Find all solutions for the given triangle, if possible. If no
solution exists, write no so lu tio n . Round side lengths to the
nearest tenth and angle measures to the nearest degree.
(Example 3)
11.
26. BOATING The light from a lighthouse can be seen from
an 18-mile radius. A boat is anchored so that it can just
see the light from the lighthouse. A second boat is located
25 miles from the lighthouse and is headed straight
toward it, making a 44° angle with the lighthouse and the
first boat. Find the distance betw een the two boats when
the second boat enters the radius of the lighthouse
light. (Example 4)
= 117°
. 13. a = 35, b = 24, A = 92°
= 30°
T h e L a w o f Sines a n d th e L a w o f C osines
28. A X Y Z , if x = 5, y = 18, and z = 14
29. A PQR, if P = 73°, q = 7, and r = 15
30. A JKL, if / = 125°, k = 24, and I = 33
31. A RST, if r = 35, s = 22, and t = 25
32. A FGH, if/ = 39, g = 50, and h = 64
33. A BCD, if B = 16°, c = 27, and d = 3
34. A LM N, i f £ = 12, m = 4, and n = 9
35. AIRPLANES D u rin g h er shift, a p ilo t flies from C o lu m bu s
51. DESIGN A free-stan d in g a rt p ro ject requ ires a triang u lar
to A tlan ta, a d istan ce o f 448 m iles, and th e n on to
su p p o rt p iece fo r stability. Tw o sid es o f the trian g le m u st
Phoenix, a d istan ce o f 1583 m iles. From P h o en ix, she
m easu re 18 and 15 feet in len g th and a n o n in clu d ed angle
returns h o m e to C o lu m b u s, a d istan ce o f 1667 m iles.
m u st m easu re 42°. If su p p o rt p u rp o ses requ ire the
D eterm ine the ang les o f the trian g le created b y h er flig h t
path. (Examples 5 and 6)
triang le to h av e an area o f at le ast 75 squ are feet, w h at
is the m easu re o f the third sid e? (Example 8)
36. CATCH L ola rolls a b all on the grou nd at an an g le o f 23°
to the righ t o f h e r d o g B u ttons. If the b a ll ro lls a total
d istance o f 48 feet, and sh e is stan d in g 30 feet aw ay,
ho w far w ill B u tto n s h av e to ru n to retriev e the
Use Heron's Form ula to find the area of each figure. Round
answers to the nearest tenth.
52.
(5 3
s
17 mm
b all? (Examples 5 and 6)
34 ml
79.7 m
Use Heron's Formula to find the area of each triangle. Round
to the nearest tenth. (Example 7)
QI
U
2 4 .3 m 4 3 .2 m /
37. x = 9 cm , y = 11 cm , 2 = 16 cm
1.1 m
21 mm
' 5 8 .9 m
mm
27 mm
7
H
20 mm
\4 3 m \ / 3 9 . 6 m
38. x = 29 in., y = 25 in., 2 = 2 7 in.
39. x = 58 ft, y = 40 ft, 2 = 63 ft
40. x = 37 m m , y = 10 m m , 2 = 34 m m
41 . x = 8 yd,
y=
15 yd , 2 = 8 yd
54.
8 cm
42. x = 133 m i, y = 82 m i, 2 = 77 m i
43. LANDSCAPING T h e S teele fam ily w an ts to exp an d their
backyard b y p u rch asin g a v acan t lo t ad jacen t to their
property. To g et a ro u gh m easu rem en t o f the area o f the
lot, Mr. Steele cou n ted the step s n eed ed to w alk arou nd
the b o rd er and d iag o n al o f the lot. (Example 7)
56. ZIP LINES A to u rist attractio n cu rren tly h as its base
co n n ected to a tree p latfo rm 150 m eters aw ay by a zip
lin e. T h e o w n ers n o w w an t to co n n ect the b ase to a
seco n d p latfo rm lo cated acro ss a can y o n and then
co n n e ct the p latfo rm s to e ach other. T h e bearings from
the b a se to e ach p latfo rm and from platform 1 to
p latfo rm 2 are giv en. F in d the d istan ces from the base to
p latfo rm 2 and from p latfo rm 1 to p latfo rm 2 .
a. E stim ate the area o f the lo t in steps.
b. If Mr. S teele m easu red his step to b e 1.8 feet, d eterm in e
the area o f the lo t in squ are feet.
44. DANCE D u rin g a p erfo rm an ce, a d an cer rem ain ed w ith in a
triang u lar area o f the stage. (Example 7)
57. LIGHTHOUSES T h e b e a rin g fro m the S o u th B ay lig hthou se
♦
a. Find the area o f stage u sed in the p erform an ce.
b.
If the stage is 250 squ are feet, d eterm in e the p ercen tag e
o f the stage u sed in the p erform an ce.
F in d the area o f each tria n g le to the n e a re st te n th . (Example 8)
45. A ABC, if A = 98°,
b = 13 m m ,
and c = 8 m m
46. A/XL, if L = 67°, j = 11 yd , and k = 24 yd
to the S teep R o ck lig h th o u se 25 m iles aw ay is N 28° E.
A sm all b o a t in d istress sp o tted o ff the coast b y each
lig h th o u se h as a b e a rin g o f N 50° W from S o u th Bay
and S 80° W from S teep R o ck. H o w far is each to w er from
the b o at?
■
Steep m
Rock^
- — r ' ^ ,'j
47. A RST, if R = 35°, s = 42 ft, and t = 26 ft
48. A X Y Z , if Y = 124°, x = 16 m, and 2 = 18 m
49. A FGH, if F = 41°, g = 22 in., and h = 36 in.
50. A PQR, if Q = 153°, p = 27 cm , an d r = 21 cm
50-1
7 25
sn n /
X i/S o u t h
^ Bay
mi |
■
299
Find the area of each figure. Round answers to the
nearest tenth.
H.O.T. Problem s
Use Higher-Order Thinking Skills
65. ERROR ANALYSIS M o n iq u e and R o g elio are solv in g an
acu te trian g le in w h ich Z A = 34°, a = 16, and b = 21.
M o n iq u e th in k s th at the trian g le has one solu tio n , w h ile
R o g elio th in k s th at the trian g le h as n o solu tion. Is eith er
o f th em correct? E xp lain you r reasoning.
66. WRITING IN MATH E xp lain the d ifferent circu m stan ces in
w h ich yo u w o u ld u se the L aw o f C o sin es, the L aw o f
Sin es, the P y th ag o rean T heo rem , and the trig on o m etric
ratios to so lv e a triang le.
67. REASONING W h y d oes an o b tu se m easu rem en t ap p ear on
the g rap h in g calcu lato r fo r in v erse cosin e w h ile n eg ativ e
m easu res ap p ear fo r in v erse sine?
62. BRIDGE DESIGN In the figure below , Z F D E = 45°,
Z.CED = 55°, ZFDE = ZFGE, B is the m id p o in t o f A C, and
DE = EG. If AD = 4 feet, DE = 12 feet, and CE = 14 feet,
find BF.
68. PROOF S h o w for a g iv en rh om b u s w ith a sid e len g th o f s
and an in clu d ed an g le o f 6 th at the area can b e foun d
w ith the fo rm u la A = s 2 sin 6.
69. PROOF D eriv e the L aw o f Sines.
70. PROOF C o n sid er the figu re below .
C
F
63. BUILDINGS B arbara w an ts to kn ow the d istan ce b etw een
the top s o f tw o bu ild in g s R and S. O n the top o f h er
B
a. U se the fig u re and h in ts b elo w to d eriv e the first
b u ild in g , she m easu res the d istance b etw een the p oints
T and U and find s the giv en an g le m easu res. Find the
fo rm u la a 2 = b 2 + c 2 — 2 be cos A in the L aw o f C osines.
d istan ce b etw een the tw o build ings.
• U se the P y th ag o rean T h eo rem for A D B C .
• In A A D B, c 2 — x 2 + h 2.
• cos A = jr
b. E xp lain h o w y o u w o u ld go ab o u t d eriv in g the
o th er tw o fo rm u las in the L aw o f C osin es.
71) CHALLENGE A satellite is o rb itin g 850 m iles ab ov e M ars
and is n o w p o sitio n e d d irectly ab ov e on e o f the poles.
The rad iu s o f M ars is 2110 m iles. If the satellite w as
p o sitio n ed at p o in t X 14 m in u tes ago, ap p ro x im ately how
m an y h o u rs d o es it take for the satellite to com p lete a full
orbit, assu m in g th at it trav els at a con stan t rate arou nd a
circu lar orbit?
64. DRIVING A fter a h ig h school football gam e, D ella left the
park in g lo t trav elin g at 35 m iles p er h ou r in the d irection
N 55° E. If D ev on left 20 m inu tes after D ella at 45 m iles
p er h o u r in the d irection S 10° W, how far ap art are
D ev on and D ella an h ou r and a h a lf after D ella left?
72. WRITING IN MATH D escrib e w h y solv in g a trian g le in w h ich
h < a < b u sin g the L aw o f S in es resu lts in tw o solu tions.
Is this also tru e w h e n u sin g the L aw o f C osin es? E xp lain
yo u r reasoning.
300
Lesson 4-7
The Law o f Sines and the Law o f Cosines
Spiral Review
Find the exact value of each expression, if it exists. (Lesson 4-6)
73. c o s ^ - i
74. s in ” 1 ^
-jC
■ _i V3
76. sin 1 —
75. arctan 1
Identify the damping factor/(.v) of each function. Then use a graphing calculator to sketch
the graphs of f(x ), —fix ), and the given function in the same view ing window. D escribe the
behavior of the graph. (Lesson 4-5)
77.
78. y = -|x cos x
y = —2x sin x
79. y — (x — l ) 2 sin x
80. y = —4 x 2 cos x
81. CARTOGRAPHY T h e d istan ce arou nd E arth along a g iv en latitu d e can b e fou n d u sing
C = 27Tr cos L, w here r is the rad iu s o f E arth and L is the latitu d e. T h e rad iu s o f E arth is
app roxim ately 3960 m iles. M ak e a table o f v alu es for the latitu d e and co rresp on d in g
d istance arou nd E arth th at in clu d es L = 0°, 30°, 45°, 60°, and 90°. U se the table to d escrib e
the d istances alo ng the latitu d es as yo u go fro m 0° at the e q u ato r to 90° at a pole. (Lesson 4-3)
82. RADIOACTIVITY A scien tist starts w ith a 1-gram sam p le o f lead-211. T h e am o u n t o f the
sam p le rem ain in g after v ariou s tim es is sh o w n in the tab le below . (Lesson 3-5)
Time (min)
Pb-211 present(g)
10
20
30
40
0.83
0 .6 8
0.56
0.46
a. Find an exp o n en tial reg ression eq u atio n for the am o u n t y o f lead as a fu n ctio n o f tim e x.
b. W rite the reg ressio n eq u atio n in term s o f b ase e.
C.
U se the eq u atio n from p art
b
to e stim ate w h en there w ill b e 0.01 g ram o f lead-211 present.
Write a polynomial function of least degree with real coefficients in standard form that has
the given zeros. (Lesson 2-4)
83.
- 1 ,1 ,5
84. - 2 , - 0 . 5 , 4
85. - 3 , - 2 i , 2 i
86. - 5 z , - i , i, 5i
Skills Review for Standardized Tests
f;. IT
/
/
89. FREE RESPONSE T h e p en d u lu m
87. SAT/ACT W h ich o f the fo llo w in g is the
at the rig h t m o v es acco rd in g to
p erim eter o f the trian g le show n?
9 = — cos 12f, w h e re 9 is the
A 49.0 cm
a n g u la r d is p la c e m e n t in
B 66.0 cm
rad ian s and t is the tim e in
second s.
C 71.2 cm
D 91.4 cm
22 cm
E 93.2 cm
f
6
—
4
a. S et the m od e to rad ian s and g rap h the fu n ctio n for 0
< t< 2 .
88. In A DEF, w h at is the v alu e o f 9 to the n earest d egree?
b. W h a t are the p erio d , am p litu d e, and freq u en cy of
the fu n ctio n ? W h a t do
th ey m ean in the con te x t o f this situ ation?
C. W h a t is the m ax im u m an g u lar d isp lacem en t of the
p e n d u lu m in d eg rees?
VTT5
F 26°
G
74°
d. W h a t d o es the m id lin e o f the g rap h represent?
H 80°
e. A t w h at tim es is the p e n d u lu m d isp laced 5 d egrees?
J 141°
tor--1 .. -.......
■j
connectED.m cgraw-hill.com |
301
Study Guide and Review
Chapter Summary
KeyVocabulary
Right Triangle Trigonometry
sin0 = ^P
hyp
'Lesson 4-1)
adj
hyp
csc 6 ■
opp
sec 9 =
Degrees and Radians
(Lesson 4 2)
•
•
•
tane = ^
cos 9 = — L
hyp
adj
adj
cot 9 =
opp
adj
7t radians
180° '
180°
To convert from radians to degrees, multiply by
■k radians'
To convert from degrees to radians, multiply by
angular speed (p. 236)
phase shift (p. 261)
circular function (p. 248)
quadrantal angle (p. 243)
cosecant p. 220)
radian (p. 232)
cosine (p. 220)
reciprocal function (p. 220)
cotangent (p. 220)
reference angle (p. 244)
coterminal angles p. 234)
secant ip. 220)
sector (p. 237)
sine (p. 220)
w = j , where 9 is the angle of rotation
damped wave (p. 275)
sinusoid (p. 256)
(in radians) moved during time t
damping factor (p. 275)
standard position (p. 231)
frequency (p. 260)
tangent (p. 220)
initial side (p. 231)
terminal side (p. 231)
inverse trigonometric
function (p. 223)
trigonometric functions (p. 220)
(Lesson 4-3)
For an angle 9 in radians containing (x, y), cos 9 = j , sin 9 = j ,
For an angle t containing (x, y) on the unit circle, cos 9 = x,
Law of Cosines (p. 295)
Law of Sines (p. 291)
linear speed (p. 236)
(Lesson 4-4)
trigonometric ratios (p. 220)
unit circle (p. 247)
vertical shift (p. 262)
midline (p. 262)
A sinusoidal function is of the form y = a sin (bx + c) + d or
y = a cos (b x- 1- c) + d, where amplitude = \a\, period =
Ih\
frequency = V - , phase shift = —
and vertical shift = d.
27T
ID]
Graphing Other Trigonometric Functions
(Lesson 4-5)
A damped trigonometric function is of the form y = f(x) sin bx or
y = f(x) cos bx, where f(x) is the damping factor.
Inverse Trigonometric Functions
(Lesson 4-6)
•
y = s i n -1 x iff sin y = x, for —1 < x < 1 and - y < y < y .
•
y = cos-1 x iff cos y = x, for - 1 < x < 1 and 0 < y < tt.
•
y = tan~1 x iff tan y = x, for
-o o <
x<
oo
The Law of Sines and the Law of Cosines
and
(Lesson 4-7)
Let A A B C be any triangle.
•
periodic function (p. 250)
Angular speed:
Graphing Sine and Cosine Functions
•
angle of elevation p 224)
damped trigonometric
function (p. 275)
sin 9 = y, and tan 0 = | -
•
period (p. 250)
v = ~t , where s is the arc length traveled
during time t
and tan Q = \ , where r = \J x 2 + y 2.
•
oblique triangles (p. 291)
Linear speed:
Trigonometric Functions on the Unit Circle
•
amplitude (p. 257)
angle of depression (p. 224)
The Law of Sines:
The Law of Cosines:
sin C
c
2be cos A
a2 + c 2 ■ la c cos B
c 2 = a2 + b2 - la b cos C
C h a p te r 4
State whether each sentence is true or false. If false, replace
the underlined term to make a true sentence.
1. The sine of an acute angle in a right triangle is the ratio of the lengths
of its opposite leg to the hypotenuse.
2.
The
secant ratio is the reciprocal of the sine
ratio.
3. An angle of elevation is the angle formed by
a horizontal lineand
an observer’s line of sight to an object below the line.
4. The radian measure of an angle is equal to the ratio of the length
of its intercepted arc to the radius.
5. The rate at which an object moves along a circular path is called
its linear speed.
6. 0°,
tt ,
and - y are examples of reference angles.
7. The period of the graph of y = 4 sin 3x is 4.
sin A _ sin B
a
b
a2 = b2 + c2
b2
302
VbcabularyGheck
S tu d y G u id e a n d R e v ie w
8. For f(x) = cos bx, as b increases, the freouencv decreases.
9. The
10.
range of the arcsine function is [0, -tt],
The Law of Sines can be used to determine
or angle measures of some triangles.
unknown sidelengths
Lesson-by-Lesson Review
c ^
^ "f (j/V
Right Triangle Trigonometry (pp. 220 - 230 )
Find the exact values of the six trigonometric functions of 0.
12 .
11.
'
Example 1
Find the value of x. Round to the
nearest tenth, if necessary.
41
Tangent function
tane = ^
ad]
tan 38° =
Find the value of x. Round to the nearest tenth, if necessary.
10
0 = 38°, opp = 10,
and adj = x
13.
Multiply each side by x.
X
tan 38°
^
Find the measure of angle 9. Round to the nearest degree,
if necessary.
(
Use a calculator.
X s: 12.8
K
Divide each side by tap 38°
/
15.
Degrees and Radians (pp. 231 - 241 )
Write each degree measure in radians as a multiple of
radian measure in degrees.
17. 135°
18. 450°
19' -74t
20.
tt
and each
Exam ple 2
Identify all angles coterminal w ith
. Then find and draw one
positive and one negative coterminal angle.
1 3 tt
All angles measuring
10
5 tt
12
2n-rr are coterminal with a ^
angle.
Let n = 1 and - 1 .
Identify all angles coterminal w ith the given angle. Then find and
draw one positive and one negative angle coterminal w ith the given
angle.
21. 342°
22.
6
^
+ 2tt(1) = ^
- 2 tt(—1) =
2-7l_ IT T t
Find the area of each sector.
23.
z' ' —
24.
/ _____ \
[3 -1
V i4
\
1
in.
/
/
\
/
\ 10 m A
\
47tt
30
*r
j
/
/
303
' . ■
.
"
Study Guide and Review
' .
.
.
:
'
'
. •
.
.
... :: " . .
'Vs'*-
. ..
Continued
Trigonometric Functions on the Unit Circle (pp. 242-253)
Sketch each angle. Then find its reference angle.
25.
240°
26. 75°
28.
27.
Example 3
Let cos 9 =
11 TT
where sin 9 < 0. Find the exact values of the
five remaining trigonom etric functions of 9.
18
Find the exact values of the five remaining trigonometric
functions of 0 .
Since cos 9 is positive and sin 9 is negative, 9 lies in Quadrant IV.
This means that the x-coordinate of a point on the terminal side of 9
is positive and the y-coordinate is negative.
29.
Since cos 9 = j =
cos 0 = • where sin 9 > 0 and tan 9 > 0
5
( 30. tan 0 = ~ | , where sin 9 > 0 and cos 9 < 0
31. sin 9 =
where cos 9 > 0 and cot 9 < 0
32> cot 9 = |, where sin 9 < 0 and tan 9 > 0
i=
use x = 5 and r = 13 to find y.
Pythagorean Theorem
V r2- x2
= V169 - 25 or 12
r — 13 and
sin 6» = 7 o r t a n 0 = | o r ^ | '1 3
*
CSC 6 - L or™
y
Find the exact value of each expression. If undefined, write
undefined.
12
o
x —
5
sec0 = ^ o r^ |A
o
cot0 = | o r A
11 TT
33. sin 180°
34. cot-
35. sec 450°
36. cos -
Hr)
Graphing Sine and Cosine Functions (pp. 256-266)
Describe how the graphs of f(x ) and g(x) are related. Then find the
amplitude and period of g(x), and sketch at least one period of both
functions on the same coordinate axes.
37.
38. f(x) = cos x
f(x) = sin x
g(x) =.cos 2x
g(x) = 5 sin x
39.
40.
f(x) = sin x
9 (x)
= Tj-sin x
f ( x ) = cos x
Example 4
State the amplitude, period, frequency, phase shift, and vertical
shift of y = 4 sin
— y j — 4. Then graph two periods of the
function.
In this function, a = 4, b = 1, c =
and d = - 4 .
Amplitude: \a\ = |4| or 4
Period: t t =
g[x) = -c o s
\b\
\b\
|1 |
1
F re q u e n c y :- = - o r State the amplitude, period, frequency, phase shift, and vertical shift
of each function. Then graph two periods of the function.
41.
y = 2 cos (x - tv)
42.
y = - s in 2x + 1
43.
y = lc o s ( x + ^ )
44.
y = 3 s in ( x + ^ )
304
Chapter 4
Study G uide and Review
Phase shift: —
t t
|
1|
o r 2 tv
Vertical shift: rfo r - 4
2
\b\
|1 | ° r 2
First, graph the midline y = - 4 . Then graph y = 4 sin xshifted •
units to the right and 4 units down.
Other Trigonometric Functions (pp. 269-279)
Locate the vertical asymptotes, and sketch the graph of each
function.
Exam ple 5
45. y = 3 tan x
46. y = ^ta n ( * - y )
y = 2 sec ( x + j ) .
47. y = c o t ( x + y j
48. y = - c o t (x - -it)
Because the graph of y = 2 sec I x + y j is the graph of
49. y = 2 sec |- |j
50. y = - c s c ( 2 x )
y = 2 sec xshifted to the left y units, the vertical asymptotes for one
51 . y = sec (x -
-tt)
52. y = | c s c ( x + | )
Locate the vertical asymptotes, and sketch the graph of
period are located at - ¥ . T . a n d ^ .
4
the interval
inverse irigonometric Functions
4
4
Graph two cycles on
3 tt
1 3 tt
4
4
(pp. 280-290)
Find the extict value of each expression, if it exists.
53. s in "1 ( - 1 )
54. c o s "1 ~ -
55. ta n " 1
56. arcsin 0
57. arctan ( - 1 )
58. arccos - y -
59. sin-1
sin ( - y )
V2
Find the exact val ue of arctan - V 3 .
Find a point on the unit circle in the interval ( - y . y ) w ith a tangent
of -\/3 .W h e n t == - f . t a n f = - V 3 .
Therefore, arctan -V 3 = - f .
60. c o s -1 [cos ( - 3 * ) ]
V
,
-J
The Law of Sines and the Law of Cosines (pp. 291-301)
Find all solutions for the given triangle, if possible. If no solution
exists, write no solution. Round side lengths to the nearest tenth and
angle measurements to the nearest degree.
61.
a = 11, b = 6, AY 22°
62.
a = 9 , b = 1 0 , A = 42°
63.
a = 20, 6 = 1 0 , 4 = 78°
64.
a = 2 , 6 = 9 ,4 = 88°
Solve each triangle. Round side lengths to the nearest tenth and
angle measures to the nearest degree.
65. a = 1 3 , 6 = 1 2 , c = I
66.
Exam ple 7
Solve the triangle if a = 3, b = 4,
and A = 71°.
In the figure, h = 4 sin 71 ° or about 3.I
Because a < h, there is no triangle
that can be formed with sides a = 3,
b = 4, and A = 71 °. Therefore, this
problem has no solution.
a = 4, 6 = 5 , C = 9 6 °
connectED.m cgraw-hill.com
305
Applications and Problem Solving
67. CONSTRUCTION A construction company is installing a three-foot-
high wheelchair ramp onto a landing outside of an office. The angle
of the ramp must be 4°. (Lesson 4-1)
a. What is the length of the ramp?
b. What is the slope of the ramp?
68.
NATURE For a photography project, Maria is photographing deer
from a tree stand. From her sight 30 feet above the ground, she
spots two deer in a straight line, as shown below. How much farther
away is the second deer than the first? (Lesson 4-1)
72. AIR CONDITIONING An air-conditioning unit turns on and off to
maintain the desired temperature. On one summer day, the air
conditioner turns on at 8:30 a .m . when the temperature is 80°
Fahrenheit and turns off at 8:55 a . m . when the temperature
is 74°. (Lesson 4-4)
a. Find the amplitude and period if you were going to use a
trigonometric function to model this change in temperature,
assuming that the temperature cycle will continue.
b. Is it appropriate to model this situation with a trigonometric
function? Explain your reasoning.
73. TIDES In Lewis Bay, the low tide is recorded as 2 feet at
4:30 a . m ., and the high tide is recorded as 5.5 feet at
10:45 a . m . (Lesson 4-4)
a. Find the period for the trigonometric model.
b. At what time will the next high tide occur?
j Ll
69.
FIGURE SKATING An Olympic ice skater performs a routine in
which she jumps in the air for 2.4 seconds while spinning 3 full
revolutions. (Lesson 4-2)
a. Find the angular speed of the figure skater.
b. Express the angular speed of the figure skater in degrees per
minute.
74. MUSIC When plucked, a bass string is displaced
1.5 inches, and its damping factor is 1.9. It produces a note
with a frequency of 90 cycles per second. Determine the
amount of time it takes the string’s motion to be dampened so
that —0.1 < y < 0.1. (Lesson 4-5)
75. PAINTING A painter is using a 15-foot ladder to paint the side
of a house. If the angle the ladder makes with the ground is less
than 65°, it will slide out from under him. What is the greatest
distance that the bottom of the ladder can be from the side of the
house and still be safe for the painter? Lesson 4-6)
70. TIMEPIECES The length of the minute hand of a pocket watch
is 1.5 inches. What is the area swept by the minute hand
in 40 minutes? (Lesson 4-2)
76. NAVIGATION A boat is 20 nautical miles from a port at a bearing
30° north of east. The captain sees a second boat and reports to
the port that his boat is 15 nautical miles from the second boat,
which is located due east of the port. Can port personnel be
sure of the second boat’s position? Justify your answer. (Lesson 4-7)
71. WORLD’S FAIR The first Ferris wheel had a diameter of 250 feet
and took 10 minutes to complete one full revolution. (Lesson 4-3)
a. How many degrees would the Ferris wheel rotate in 100
seconds?
a. Find C.
b. How far has a person traveled if he or she has been on the Ferris
wheel for 7 minutes?
c. How long would it take for a person to travel 200 feet?
306
C h a p te r 4
77. GEOMETRY Consider
quadrilateral ABCD. (Lesson 4-7)
Study G uide and Review
b. Find the area of ABCD.
Find the value of x. Round to the nearest tenth, if necessary.
16. TIDES The table gives the approximate times that the high and low
tides occurred in San Azalea Bay over a 2-day period.
1-
2X
High 1
Low 1
High 2
2:35 a . m .
8:51 a . m .
3:04 p . m .
9:19 p . m .
9:48 a . m .
3:55 p . m .
1 0 :2 0
Tide
Day 1
Day 2
Find the measure of angle 9. Round to the nearest degree,
if necessary.
3:30 a . m .
Low 2
p .m .
a. The tides can be modeled with a trigonometric function.
4.
Approximately what is the period of this function?
b.
The difference in height between the high and low tides is 7 feet.
What is the amplitude of this function?
c. Write a function that models the tides where f is measured in
hours. Assume the function has no phase shift or vertical shift.
5. MULTIPLE CHOICE What is the linear speed of a point rotating at an
angular speed of 36 radians per second at a distance of 12 inches
from the center of the rotation?
A 420 in./s
C 439 in./s
B 432 in./s
D 444 in./s
Write each degree measure in radians as a multiple of ir and each
radian measure in degrees.
6.
200 °
8.
Find the area of the sector of the circle shown.
Locate the vertical asymptotes, and sketch the graph of
each function.
17. y = ta fr(x + | )
18. y = l s e c 4 x
Find all solutions for the given triangle, if possible. If no
solution exists, w rite no solution. Round side lengths to
the nearest tenth and angle measurements to the nearest
degree.
19. a = 8, * = 1 6 , A = 22°
21.
a = 3, b = 5, c = 7
20. a = 9 , b = 7, A = 84°
22. a = 8, 6 = 1 0 , C = 4 6 °
Find the exact value of each expression, if it exists.
23.
Sketch each angle. Then find its reference angle.
9.
165°
10.
21 TV
13
cos” 1
24. s in - 1 ( - 1 )
25. NAVIGATION A boat leaves a dock and travels 45° north of west
averaging 30 knots for 2 hours. The boat then travels directly west
averaging 40 knots for 3 hours.
Find the exact value of each expression.
11. s e c ^
6
13.
12.
cos (-2 4 0 ° )
MULTIPLE CHOICE An angle 9 satisfies the following inequalities:
csc 9 < 0, cot 9 > 0, and sec 9 < 0. In which quadrant does 9 lie?
F I
H III
G II
J IV
State the amplitude, period, frequency, phase shift, and vertical
shift of each function. Then graph tw o periods of the function.
14. y = 4 cos-| - 5
15. y = - s i n ( x + y )
a. How many nautical miles is the boat from the dock after
5 hours?
b.
How many degrees south of east is the dock from the boat’s
present position?
^connectEDjncgraw^^^^ 307
If air is being pumped into a balloon at a given rate, can we find the rate at which the volume of the balloon is
expanding? How does the rate a company spends money on advertising affect the rate of its sales? Related rates
problems occur when the rate of change for one variable can be found by relating that to rates of change for other
variables.
# Model and solve related
rates problems.
Suppose two cars leave a point at the same time. One car is
traveling 40 miles per hour due north, while the second car is
traveling 30 miles per hour due east. How far apart are the two
cars after 1 hour? 2 hours? 3 hours? We can use the formula
d = r t and the Pythagorean Theorem to solve for these values.
In this situation, we know the rates of change for each car. What
if we want to know the rate at which the distance between the
two cars is changing?
Activity 1 Rate of Change
Two cars leave a house at the same time. One car travels due north at 35 miles per hour, while
the second car travels due east at 55 miles per hour. Approxim ate the rate at which the
distance betw een the two cars is changing.
ETTTT1
M ak e a sk etch o f the situ atio n .
k W '.H
W rite eq u ation s fo r the d istan ce traveled
b y e ach car after t hou rs.
Find the d istan ce trav eled b y each car
after 1, 2, 3, and 4 h ou rs.
E T7m
U se the P y th ag orean T h eo rem to find the
N
105» n
70
3 5*.
‘X
X
f= 3
t=2
\
d istan ce b etw een the tw o cars at each p o in t
in tim e.
EVM I1
Find the av erage rate o f ch an g e o f the
d istan ce b etw een the tw o cars for
1 < t < 2 , 2 < f < 3, and 3 < t < 4.
55
11C
pAnalyze the Results
1. M ak e a scatter p lo t d isp lay in g the total d istan ce b etw een the tw o cars. L et tim e t b e the
in d ep en d en t v ariab le and total d istan ce d b e the d ep en d en t v ariab le. D raw a lin e throu gh
the points.
2. W h at typ e o f fu n ctio n does the grap h seem to m o d el? H o w is y o u r co n jectu re su p p o rted by
the v alu es foun d in Step 5?
3. W h at w o u ld h ap p en to the av erag e rate o f ch an g e o f the d istan ce b e tw e e n the tw o cars if one
o f the cars slow ed d ow n? sp ed up ? E xp lain y o u r reasoning.
^
V ............................................................................................................................................................................................
The rate that the distance between the two cars is changing is related to the rates of the two cars. In calculus,
problems involving related rates can be solved using im plicit differentiation. However, before we can use advanced
techniques of differentiation, we need to understand how the rates involved relate to one another. Therefore, the first
step to solving any related rates problem should always be to model the situation with a sketch or graph and to write
equations using the relevant values and variables.
308
C h a p te r 4
Activity 2 Model Related Rates
A rock tossed into a still body of water creates a circular ripple that grows at a rate of
5 centim eters per second. Find the area of the circle after 3 seconds if the radius of the circle is
5 centim eters at t = 1.
M ake a sk etch o f the situ ation.
. t= 3
W rite an eq u atio n fo r the rad iu s r o f the circle after f second s.
/
r= 5 t
. t= 2
/ ' t= 1
i / /
Find the rad iu s for f = 3, and then fin d the area.
'\ '\ H
5 cm
^ Analyze the Results
\ \
)
~~
/
/
'
’i
1
1
/
4. Find an eq u atio n for th e area A o f the circle in term s o f f.
5. Find the area o f the circle fo r t = 1, 2 , 3 , 4 , and 5 second s.
6 . M ak e a g rap h o f the v alu es. W h at typ e o f fu n ctio n d o es the g rap h seem to m od el?
You can use the difference quotient to calculate the rate of change for the area of the circle at a certain point in time.
Activity 3 Approximate Related Rate
Approxim ate the rate of change for the area of the circle in Activity 2.
StudyTip
Difference Quotient Recall that
the difference quotient for
calculating the slope of the line
tangent to the graph of f(x) at the
point (x, f(x)) is
m
S u b stitu te the exp ressio n for the area o f the
m■
circle in to the d ifferen ce qu otient.
it[5 ( f + h)]2 —TT(5f)2
Ti
A p p ro x im ate the rate o f ch an g e o f the circle at 2 secon d s.
L et h = 0.1, 0.01, and 0.001.
f(x+tl)~f(x)
R ep eat Step s 1 and 2 for t = 3 seco n d s and t = 4 second s.
y Analyze the Results
7. W h at do the rates o f ch an g e ap p ear to ap p roach for e ach v a lu e o f t ?
8 . W h at h ap p en s to the rate o f ch an g e o f the area o f the circle as the rad iu s in creases? Explain.
9. H o w d o es this ap p roach d iffer from the ap p roach y o u u sed in A ctiv ity 1 to fin d the rate of
ch an g e fo r the d istan ce b e tw e e n the tw o cars? E xp lain w h y th is w as necessary.
Model and Apply
10.
A 13-fo ot lad d er is lean in g ag ain st a w all so th at the b a se o f th e lad d er is exactly 5 feet from
the b a se o f the w all. If the b o tto m o f the lad d er starts to slid e aw ay fro m the w all at a rate of
2 feet p er secon d , h o w fast is the top o f the lad d er slid in g d o w n the w all?
a. S k etch a m o d el o f th e situ atio n . L et rf b e the d istan ce from the top o f the lad d er to the
g ro u n d and m b e the rate at w h ich the top o f the lad d er is slid in g d o w n the w all.
b. W rite an e xp ressio n fo r the d istan ce from the b a se o f the lad d er to the w all after
t second s.
c. F in d an eq u atio n for the d istan ce d from the top o f the lad d er to th e grou nd in term s o f t b y
su b stitu tin g the e xp ressio n fou n d in p art b in to the P y th ag o rean T heorem .
d. U se the P y th ag o rean T h eo rem to fin d the d istan ce d from the top o f the lad d er to the
grou nd fo r t = 0 ,1 , 2, 3, 3.5, and 3.75.
e. M ak e a g rap h o f the v alu es. W h a t typ e o f fu n ctio n d o es the g rap h seem to m od el?
f. U se the d ifferen ce q u o tien t to ap p rox im ate the rate o f ch an g e m fo r the d istan ce from the
top o f the lad d er to the grou nd at t = 2. L et h = 0 .1 ,0 .0 1 , and 0.001. A s h ap p roaches 0, w hat
d o the v alu es for m ap p ear to ap p roach ?
hp=
§
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309
Get Ready for the Chapter
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trig o n o m e tric ide n tity
p. 312
identidad trigonom etrica
reciprocal ide n tity
p. 312
identidad reciproca
qu o tie n t ide n tity
p. 312
cociente de identidad
Pythagorean ide n tity
p. 313
P itagoras identidad
o d d -e ve n -id e n tity
p. 314
im p a r-in clu so -d e identidad
cofu nction
p. 314
co fun cion
v e rify an ide n tity
p. 320
v e rific a r una identidad
sum ide n tity
p. 337
sum a de identidad
redu ctio n ide n tity
p. 340
identidad de reduccion
d ou ble -a ng le ide n tity
p. 346
d ob le-a n gu lo de la
identidad
pow e r-re d u cin g ide n tity
p. 347
p od er-re du cir la identidad
h a lf-a n g le ide n tity
p. 348
m edio angulo identidad
Solve each equation by factoring. [Lesson 0-3!
1.
x2 + 5 x - 2 4 = 0
2. x2 — 11 x + 28 = 0
3. 2x2 — 9x — 5 = 0
4. 15x2 + 26x + 8 = 0
5. 2x3 — 2x2 — 12x = 0
6.
12x3 + 78x2 — 42x = 0
7. ROCKETS A rocket is projected vertically into the air. Its distance in
feet after t seconds is represented by s(t) = - 1 6 f2 + 192f. Find the
amount of time that the rocket is in the air. Lesson 0-3}
Espanol
Find the missing side lengths and angle measures of each
triangle. Lessons 4-1 and 4-7)
8.
ReviewVocabulary
e xtraneous solu tio n p. 91 solucion extraha a solution that does not
satisfy the original equation
q uadrantal angle p. 243 angulo cua dran ta an angle 0 in standard
position that has a terminal side that lies on one of the coordinate axes
u n it circle p. 247 c irc u lo u nitario a circle of radius 1 centered at the
origin
p eriodic fu n c tio n p. 250 fu n cio n perio dica a function with range
values that repeat at regular intervals
trig o n o m e tric fu n c tio n s p. 220 fu n cio n e s trig o n o m e trica s Let e
be any angle in standard position and point P(x, y) be a point on the
terminal side of 0. Let /represent the nonzero distance from Pto the
Find the exact value of each expression. (Les:
12. cot 420°
14. sec
10tv
16. csc
2n
2
3
13. cos ~~4
15. tan 480°
origin or |r| = V * 2 + y 2 #
as follows.
Then the trigonometric functions of 9 are
sin 0 =
cos 6 = j
tan 0 = ^ x ^ 0
csc 0 = -jj, y =£ 0
sec 0 = -j, x ^ 0
co t0 = ^ O
17. sin 510°
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311
• You found
trigonom etric values
using the unit circle.
(Lesson 4-3)
NewVocabulary
identity
trigonometric identity
cofunction
1
Identify and use basic
trigonom etric identities to
find trigonom etric values.
Use basic trigonom etric
■identities to sim plify and
rewrite trigonometric
expressions.
•
Many physics and engineering applications,
such as determining the path of an aircraft,
involve trigonometric functions. These functions
are made more flexible if you can change the
trigonometric expressions involved from one
form to an equivalent but more convenient form.
You can do this by using trigonometric identities.
Basic Trigonometric Identities
1
A n eq u atio n is an identity if the left sid e is equ al to the
rig h t sid e for all v alu es o f the v ariab le fo r w h ic h b o th sid es are d efined . C o n sid er the
equ atio n s below .
x —9
= x + 3
x —3
This is an identity since both sides of the equation are defined and equal for all x such that x
3.
This is not an identity. Both sides of this equation are defined and equal for certain values, such as
sin x = 1 — cos x
when x —0, but not for other values for which both sides are defined, such as when x =
Trigonometric identities are id en tities th at in v o lv e trig o n o m etric fu n ction s. You alread y k n o w a
few b asic trig o n o m etric identities. T h e recip ro cal an d q u o tie n t id e n tities b e lo w fo llow d irectly fro m
the d efin itio n s o f the six trig o n o m etric fu n ctio n s in tro d u ced in L esso n 4-1.
KeyConcept Reciprocal and Quotient Identities
Reciprocal Identities
sin 9 = —
CSC 0
CSC
V
6
=
-4 -r
sin 6
Quotient Identities
cos e = —
sec 6
tan 0 = 4 cot 9
tan 0 = ^ 4
sec 6 = —
cos 9
cot 9 = —
tan 9
cot 9 = ^
sin 9
COS 9
You can u se these basic trig o n o m etric id en tities to find trig o n o m etric v alu es. A s w ith an y fraction ,
the d en o m in ator can n o t equ al zero.
Use Reciprocal and Quotient Identities
a. If csc 6 =
4
b.
find sin 8.
1
csc 9
1
2
V5
If cot x = — — and sin x = — , find cos x.
5 V5
Reciprocal Identity
csc 0 =
c o tx =
t
5V5
COS X
s in
x
cos
x
V5
Quotient Identity
cot X= —
2
5 V5"
•
V5
si n x = - y -
3
Divide.
2
V5
5V 5
3
Vs
= COS X
— = cos x
f
Lesson 5-1
Simplify.
GuidedPractice
1A. If sec x = —, find cos x.
312
Multiply each side by - y - .
1B. If csc /3 = -y- and sec fi =
find tan /3.
R ecall from L esso n 4-3 th at trig on o m etric fu n ctio n s can b e d efined
on a u n it circle as show n. N o tice th at for an y an g le 0, sin e and cosin e
are the d irected len g th s o f the legs o f a rig h t trian g le w ith
h y p o ten u se 1. W e can ap p ly the P y th ag o rean T h eo rem to th is rig h t
trian g le to estab lish an o th er b asic trig o n o m etric identity.
(sin 0 )2 + (cos 0 )2 = l 2
Pythagorean Theorem
s in 2 0 + c o s2 0 = 1
Simplify,
W h ile the sig n s o f th ese d irected len g th s m a y ch an g e d ep en d in g on the q u ad ran t in w h ich the
trian g le lies, n o tice th at b e cau se th ese len g th s are sq u ared , the eq u atio n abo v e h o ld s true for any
v alu e o f 0. T h is eq u atio n is one o f th ree Pythagorean identities.
ReadingMath
m
a
KeyConcept Pythagorean Identities
Powers of Trigonometric
Functions sin2 9 is read as sine
squared theta and interpreted as
the square of the quantity sin 9.
sin2 9 + cos2 9 = 1
cot2 0 +
tan2 9 + 1 = sec2 9
1=
t
i
csc2 6
V.
J
You will prove the remaining two Pythagorean Identities in Exercises 69 and 70.
N otice the shorth an d n o tatio n u sed to rep resen t p o w ers o f trig o n om etric fu n ctions: s in 2 0 = (sin 0 )2,
co s2 0 = (cos 0)2, ta n 2 0 = (tan 0) 2, and so on.
Use Pythagorean Identities
If tan 0 = —8 and sin 0 > 0, find sin 0 and cos 0.
U se the P y th ag o rean Id en tity th at in v o lv es tan 0.
tan2 0 + 1 = sec2 0
Pythagorean Identity
tan 9 = —8
(—8)2 + 1 = sec2 0
65 = sec2 0
Simplify.
± V 6 5 = sec 0
Take the square root of each side.
±V&5 = —i —
Reciprocal Identity
COS 0
±-
V65
Solve for cos 9.
: CO S I
65
S in ce tan 9 =
cos
9
is n e g ativ e and sin 0 is p o sitiv e, cos 0 m u st b e rie
You can th en use this q u o tien t id en tity ag ain to find sin 0.
tan 0 =
-8
9
s in
Quotient Identity
cos 9
9
V65
s in
V65
tan 9 = —8 and cos 9 =
65
65
8 \ /6 5
65
StudyTip
Checking Answers It is beneficial
to confirm your answers using a
different identity than the ones
you used to solve the problem, as
in Example 2, so that you do not
make the same mistake twice.
/
CHECK
= sin I
Multiply each side by ■
s in 2 0 + c o s 2 i
= 1
V 65 )2 ± 1
65 I
ff
65 = 1
V65
65 ‘
Pythagorean Identity
sin 0 =
8V65
65
and cos 0 -
V65
65
Simpiify.
p GuidedPractice
Find the value of each expression using the given information.
2A. csc 0 and tan 0; co t 0 = —3, cos 0 < 0
2B. cot x and sec x; sin x = —, cos x > 0
6
i
......... ...................
nl.connectED .m cgraw -hill.com |
313
A n o th er set o f b asic trig o n om etric id en tities in v olv e cofu n ctio n s. A trig o n o m etric fu n ctio n / is a
cofunction o f an o th er trig o n o m etric fu n ctio n g if f ( a ) = g(J3) w h e n a and (3 are com p lem en tary
angles. In the rig h t trian g le sh ow n , an g les a and (3 are co m p le m e n tary angles. U sin g the righ t
trian g le ratio s, yo u can show th at the fo llo w in g statem en ts are true.
sin a = co s /3 = cos (90° — a ) = j
tan a = cot (3 = co t (90° — a ) = —
sec a = csc /3 = csc (90° — a ) = —
From these statem en ts, w e can w rite the fo llo w in g co fu n ctio n id en tities, w h ich are v alid for all real
nu m bers, n o t ju st acu te angle m easures.
StudyTip
Writing Cofunction Identities
Each of the cofunction identities
can also be written in terms of
degrees. For example,
sin 9 = cos (90° - 9).
KeyConcept Cofunction Identities
sin 9 = cos
- flj
sec 6 = csc ( j - 0 j
cot 9 = tan | y - 0 j
csc 9 = sec ( y - e j
tan 9 = cot
- ej
cos 9 = sin ( y - 0 j
J
I
You w ill prove these identities for any angle in Lesson 5-3,
You h av e also seen th at each o f the b asic trig o n o m etric fu n ctio n s—
sin e, cosine, tan g en t, cosecan t, secan t, and cotan g en t— is eith er odd
or even. U sin g the u n it circle, y o u can sh o w th at the fo llo w in g
statem ents are true.
sin (—a ) = —y
cos (—a ) = x
sin a = i/
cos a = x
R ecall from L esso n 1-2 th at a fu n ctio n / is e v en if for e v ery x in the
d o m ain o f /,/(—* ) = f ( x ) and od d if fo r ev ery x in the d o m ain of/,
f ( —x) = —f(x ). T h ese relatio n sh ip s lead to the fo llo w in g o d d -ev en
id entities.
KeyConcept Odd-Even Identities
sin (-9 ) = - s in 9
cos (-9 ) = cos 9
tan ( - 0 ) = - t a n 0
csc (—9) = —csc 9
Sec (— 0) = sec 0
cot ( - 0 ) = - c o t 0
J
You can use co fu n ctio n and o d d -ev en id en tities to fin d trig o n o m etric valu es.
Use Cofunction and Odd-Even Identities
If tan 0 = 1.28, find cot
— y j.
co t ^0 - y j = cot [ ~ ( y ~ ^ ) ]
Factor.
Odd-Even Identity
f
= —tan 0
Cofunction Identity
= - 1 .2 8
tan 0 = 1 .2 8
GuidedPractice
3. If sin x = —0.37, fin d cos
314
Lesson 5-1 | Trigo no m etric Identities
— y j.
2
Simplify and Rewrite Trigonometric Expressions
To sim p lify a trigo n o m etric
exp ressio n , sta rt b y rew ritin g it in term s o f o n e trig o n o m etric fu n ctio n or in term s o f sine and
cosin e only.
Simplify by Rewriting Using Only Sine and Cosine
Simplify csc 0 sec 9 — cot 9.
Solve Algebraically
csc 9 sec 9 — co t 9 =
1
s in
cos
1
8
cos
9
s in
cos
1
8
s in
8c o s 8
s in
s in 0 c o s
9
9c o s (
Rewrite fractions using a common denominator.
Subtract.
s in
TechnologyTip
Reciprocal and Quotient Identities.
Multiply.
1 —cos2 0
8c o s 9
sin2 9
s in 9 c o s 9
s in 8
o r tan
cos 8
Rewrite in term s of sine and cosine using
9
9
cos2
1
s in
8
8
Pythagorean Identity
Divide the numerator and denominator by sin 9.
9
Support Graphically T h e graphs o f y = csc 9 sec 9 — cot 9 and y = tan 9 appear to be identical.
Graphing Reciprocal
Functions When using a
calculator to graph a reciprocal
function, such as y = csc x, you
can enter the reciprocal of the
function.
P lo ti M o t£ Plots
\Y iB l/s in < X >
\V i=
\V j =
\V h=
[ —2tv, 2tv] scl: f
\V e=
Guided Practice
sVb=
nV?=
4.
by [ - 2 , 2] scl: 0.5
S im p lify sec x — tan x sin x.
S om e trig o n o m etric exp ressio n s can b e sim p lified b y ap p ly in g id en tities an d factorin g.
W
W
H
simolifv by Factoring
Simplify sin 2 x cos x — sin ( y — *)•
Solve Algebraically
Cofunction Identity
s in 2 x cos x — s in ( Y ~ x ) = s*n2 x cos * ~ cos x
WatchOut!
Graphing While the graphical
approach shown in Examples 4
and 5 can lend support to the
equality of two expressions, it
cannot be used to prove that
two expressions are equal. It is
impossible to show that the
graphs are identical over their
entire domain using only the
portion of the graph shown on
your calculator.
Support Graphically
= —co s x ( —s in 2 x + 1)
Factor —cos / from each term.
= —cos x (1 — sin2 x)
Commutative Property
= —cos x (cos2 jc) or —cos3 x
Pythagorean Identity
T h e grap h s b elo w ap p ear to b e id en tical.
y = sin2 x c o s x - s i n | y - x j
zz:
A
.
. A
■
/
A
,
s
----------------h |
y = - c o s 3x
[—2 ir, 2 tt] scl: f by [ - 2 , 2] scl: 0.5
[—2 ir, 2 tt] scl: y by [ - 2 , 2] scl: 0.5
^ Guided Practice
5.
Sim p lify —c sc ^ y — x) — ta n 2 x sec ;
lconnectED .m cgraw -hill.com I
315
You can simplify some trigonom etric expressions by com bining fractions.
S
M
3
W
Simplify
s in
1
x cos
— s in
Simplify by Combining Fractions
I
sin x cos x
1 —sin x
x
+ s in
1
x
_
1
x
sin a
cos x
+
x cos x
s in
x
cos
(1
(c o s
x)
(1
— s in x ) ( c o s x )
s in
cos
x
+ s in x ) ( l — s in x )
(c o s
1 —sin2 x
x —s in x c o s x
x cos x
x cos x
—s i n
x cos2 x
x — s in x c o s x
cos
x — s in
x cos
Multiply.
cos
s in
cos
Common denominator
x)(\ —s i n x)
cos
x
cos2 x
x cos x
Pythagorean Identity
— s in
Subtract.
x
(c o s 2 x )( s in x — 1 )
Factor the numerator and denominator.
( — c o s x )(s in x — 1 )
Divide out common factors.
= —cos x
f
GuidedPractice
Simplify each expression.
cos X
6A.
1 + s in x
+
1
+ s in x
6B.
cos x
CSC X
1
+ sec x
■4 -
CSC X
1
-
sec x
In calcu lu s, you w ill som etim es n eed to rew rite a trig o n o m etric exp ressio n so it does n o t in v o lv e
a fraction . W h e n the d en o m in ato r is o f the fo rm 1 ± u or u ± 1, y o u can som etim es do so b y
m u ltip lyin g the n u m erato r and d en o m in ato r b y the con ju g ate o f the d en o m in ato r and ap p lyin g
a P yth ag o rean identity.
Review Vocabulary
BT#
conjugate a binomial factor
which when multiplied by the
original binomial factor has a
product that is the difference of
two squares (Lesson 0-3)
R ew rite
1 + cos x
1
Rewrite to Eliminate
as an expression that does not involve a fraction.
1
1 + cos X
1 —
1 + cos X
_
Fractions
cosX
1 — cos X
1 — COS X
Multiply numerator and denominator by the conjugate
of 1 + cos x, which is 1 — cos x.
Multiply.
1 — COS2 X
_
1 — COS X
Pythagorean Identity
s in 2 x
1
COS X
1
s in
=
x
Write as the difference of two fractions.
COS X
1
s in x
s in x
C SC2 X — c o t
XCSC X
Factor.
Reciprocal and Quotient Identities
► GuidedPractice
Rewrite as an expression that does not involve a fraction.
7A.
316
Lesson 5-1
c o s '- x
1
— s in x
Trigo n o m etric Identities
70
7B.
4
sec x + tan x
m
Exercises
= Step-by-Step Solutions begin on page R29.
Find the value of each expression using the given
information. (Example 1)
1. If cot I
5
Simplify each expression. (Example 6)
32.
cos X
sec x + l
gg
1 — cos x
tan x
34.
1
■+ ■ 1
sec x + l
sec x — 1
35.
cos x cot x ,
sin x
sec x + tan x
sec x —tan x
-, find tan 0.
2. If cos x = —, find sec x.
3. If tan a =
5
find cot a .
4. If sin /3 = —§> find csc /3.
6
5. If cos x = -r and sin x =
6
6. If sec ip = 2 and tan
6
30
fin d cot x.
7. If csc Q; = — and co t a =
fin d sec
s in
csc
= V 3 , find sin <p.
x
x+
.
COS X
sec x — 1
sin x
1 + cos x
_ j_
l
s in
csc
x
x —1
( 3 7 | SUNGLASSES M an y su n g lasses are m ad e w ith polarized
len ses, w h ich red u ce the in te n sity o f light. T he intensity
o f lig h t em erg in g fro m a sy ste m o f tw o p o larizin g lenses
a.
8. If sec 0 = 8 and tan 9 = 3 V 7 , find csc 9.
I can b e calcu lated b y I = /0 ------ —— , w here I0 is the
csc2
Axis 1
9. sec 0 and cos 0; tan 0 == —5, cos 0 > 0
10. cot 0 and sec 0; sin 0 ==j , tan 0 < 0
Si*
Unpolarized
11. tan 0 and sin 0; sec 0 == 4, sin 0 > 0
li®ht
from a sy ste m o f tw o p o larized lenses.
b. If a p air o f su n g lasses co n tain s a system o f tw o
p o larizin g len ses w ith ax e s at 30° to one another,
w h a t p ro p o rtio n o f th e in ten sity o f lig h t enterin g
the su n g lasses em erg es?
15. cot 0 and sin 0; sec 0 == - j , sin 0 > 0
16. tan 0 and csc 0; cos 0 == - j , sin 0 < 0
Find the value of each expression using the given
information. (Example 3)
—y j .
19. If tan 9 = - 1 .5 2 , find co t ( f l - f ) .
20.
21 .
If sin 6= 0.18, find cos (0 - f ).
If cot x = 1.35, find tan |x —y j .
22. csc x sec x — tan x
23. csc x — co s x co t x
24. sec x co t x — sin x
25.
1 — sin2 x
CSC
28.
x —1
sec x csc x —tan x
sec x csc x
30. cot x — csc2 x co t x
27.
29.
sin x
cot X
39.
CSC X —
40.
cot X
sec x — tanx
42.
3 tan x
1 - cos x
44.
sin x
1 — sec x
46.
Simplify each expression. (Examples 4 and 5)
26.
Rewrite as an expression that does not involve a fraction.
(Example 7)
38.
If csc 9 =—1.24, find sec ^0 —
If cos x = 0.61, find sin
Lens 2
a. S im p lify th e fo rm u la fo r th e in te n sity o f lig ht em ergin g
14. sin 0 and cos 0; cot 0 == 8, csc 0 < 0
18.
Lens1
sin 0 < 0
13. cos 0 and tan 0; csc 0 == -|, tan 0 >•0
17.
9
in ten sity o f lig h t en terin g the sy ste m o f lenses and 0 is
the an g le o f the axis o f th e seco n d lens in relatio n to that
o f the first len s. (Example 6)
Find the value of each expression using the given
information. (Example 2)
12. sin 0 and cot 0; cos 0 ==
-v
41.
43.
45.
5
47.
sec x + l
CSC X
1
— s in
cot
1
+ s in
2
cot
x
x
x
s in
x+
x
csc
c o t " x cos
csc
x
x
x —1
s in
x ta n x
cos x + l
Determine w hether each parent trigonom etric function
shown is odd or even. Explain your reasoning.
tan x + sin x sec x
csc x tan x
H i
J
csc x cos x + cot x
I
sec x cot x
n
cot2 x + l
31 . co t
x
— cos3 X
n
CSC X
connectED.mcgraw-hill.com 1
&
317 y n
50. SOCCER W h en a soccer b all is kick ed from the grou nd , its
61.
zEU
y-
2v02 cos2 0
LIGHT WAVES W h e n lig h t sh in es thro u gh tw o n arro w slits,
a series o f lig h t an d d ark frin g es appear. T h e an g le 9, in
h eigh t y and h o rizo n tal d isp lacem en t x are related b y
rad ian s, lo catin g the m th frin g e can b e calcu lated b y
+ T s'n , w here i>0 is the in itial v elo city of
cos 6
sin 9 =
the b all, 9 is the angle at w h ich it w as k ick ed , and g is the
acceleration d u e to gravity. R ew rite this eq u atio n so th at
a
w h ere d is the d istan ce b etw een the tw o slits,
and A is the w av e le n g th o f light.
tan 9 is the o n ly trigo n o m etric fu n ctio n th at ap p ears in
the equation.
Write each expression in terms of a single trigonometric
function.
51 . tan x — csc x sec x
52. cos x + tan x sin x
53. csc
x
tan 2 x — s e c 2 x csc x
a. R ew rite the fo rm u la in term s o f csc 9.
54. sec x csc x — cos x csc x
55.
b. D eterm in e the an g le lo catin g the 100th fring e w h en
lig h t h a v in g a w av e le n g th o f 550 n an om eters is shined
th rou g h d o u b le slits sp aced 0.5 m illim eters apart.
* MULTIPLE REPRESENTATIONS In this p ro blem , y o u w ill
in v estig ate the v erificatio n o f trig o n om etric id entities.
C o n sid er the fu n ctions show n.
i. i/j = tan x + 1
H.O.T. Problems
y 2 = sec x cos x — sin x sec x
ii. y3 = tan x sec x — sin x
y4 = sin x ta n 2 x
62.
g rap h in g the functions.
—2tt
—IT
PROOF P rove th at the area o f the trian g le is
A =
a. TABULAR C o p y and com p lete the table below , w ith ou t
Use Higher-Order Thinking Skills
s(s — a)(s — b)(s — c) w h ere s = j ( a + b + c).
(Hint: T h e area o f an o bliqu e trian g le is A = j be sin A.)
0
7T
27T
B
63. ERROR ANALYSIS Je n e lle and C h lo e are sim p lify in g
b. GRAPHICAL G rap h e ach fu n ctio n on a grap h in g
— \ — sin x —. Je n e lle th in k s th at the exp ression
sin2 x — cos x
calculator.
c. VERBAL M ake a con jectu re ab ou t the relatio n sh ip
betw een y 1 and y 2. R ep eat fo r y 3 and y 4.
sim p lifies to c sc 2 x — ta n 2 x. Is eith er o f th em correct?
E xp lain y o u r reasoning.
d. ANALYTICAL A re the con jectu res th at you m ad e in
p art c valid fo r the entire d o m ain o f e ach fun ctio n ?
E xplain y ou r reasoning.
Rewrite each expression as a single logarithm and simplify
the answer.
56. In |sin x\ — In |cos x\
2
sim p lifies to — cos
, and C h lo e th in k s th at it
1 —2 cos x
CHALLENGE Write each of the basic trigonometric functions in
terms of the following functions.
64. sin x
65. cos x
66. tan x
REASONING Determ ine w hether each statem ent is true or
57. In |sec x\ — In |cos x\
fa ls e . Explain your reasoning.
58. In (co t2 x + 1) + In |sec x\
67. c sc 2 x tan x = csc x sec x is tru e fo r all real nu m bers.
59. In (sec2 x — ta n 2 x ) — In (1 — c o s2 x )
68. T h e o d d -ev en id en tities can b e u sed to p ro v e th at the
g rap h s o f y = co s x and y = sec x are sy m m etric w ith
resp ect to th e y-axis.
60. ELECTRICITY A cu rren t in a w ire in a m ag n etic field cau ses
a force to act o n the w ire. T h e stren gth o f the m ag n etic
field can be d eterm in ed u sin g the fo rm u la B = — —— ,
w here F is the force on the w ire, I is the cu rren t in the
w ire, £ is the len gth o f the w ire, and 9 is the angle the w ire
m akes w ith the m ag n etic field. S o m e p h y sics b o o k s give
the form u la as F = I£B sin 9. S h ow th at the tw o fo rm u las
are equivalent.
318
Lesson 5-1
Trigo no m etric Identities
PROOF Prove each Pythagorean identity.
(69| ta n 2 9 + 1 = s e c 2 9
71.
70. c o t2 9 + 1 = c sc 2 6
PREWRITE U se a ch art or a tab le to help y o u org an ize the
m ajo r trig o n o m etric id en tities fo u n d in L esso n 5-1.
Spiral Review
Solve each triangle. Round to the nearest tenth, if necessary. (Lesson 4-7)
73.
75.
76.
Find the exact value of each expression, if it exists. (Lesson 4-6)
H)
78. cot (s in -1
79. tan (arctan 3)
80. cos
81 . cos — — cos
82. cos 1 (s in " 1 f )
83. sin (c o s -1
84. ANTHROPOLOGY Allometry is the stu d y o f the relation sh ip b e tw e e n the size
Growth of the Average American M ale
(0 -3 years of age)
o f an org an ism and the size o f any o f its p arts. A research er d ecid ed to test
for an allom etry b e tw e e n the size o f the h u m an h ead com p ared to the
Head Circum ference (in.)
h u m an b o d y as a p erso n ages. T h e data in the table rep resen t the av erage
A m erican m ale. (Lesson 3-5)
a. Find a qu ad ratic m o d el relatin g th ese d ata b y lin earizin g the d ata and
find ing the lin ear reg ressio n equ ation.
b. U se the m od el fo r the lin earized d ata to find a m o d el fo r the o rigin al
data.
C. U se y o u r m o d el to p red ict the h e ig h t o f an A m e rica n m ale w h o se h ead
circu m ference is 24 inches.
Let U = { 0 ,1 , 2, 3, 4, 5 }, A = {6 , 9 }, B = {6 , 9 ,1 0 } , C = { 0 ,1 , 6, 9 ,1 1 },
D = {2 , 5 ,1 1 }. Determine whether each statem ent is true or fa l s e .
Explain your reasoning. (Lesson 0-1)
85. A C i
Height (in.)
14.1
19.5
18.0
26.4
18.3
29.7
18.7
32.3
19.1
34.4
19.4
36.2
19.6
37.7
Source: National Center for Health Statistics
86. D C U
Skills Review for Standardized Tests
87. SAT/ACT If x > 0, then
*2
_
1
(X +
x+ l
l ) 2
x+ 2
1
89. W h ich o f the fo llo w in g is e q u iv ale n t to
1 - sin2 (
(* + 2)2 x + 3
1 — cos2 t
•tan 91
A (x + l )2
A tan 0
C
B (x — l )2
B
D cos i
co t 9
sin t
C 3 x -l
90. REVIEW R efer to the figure. If cos D = 0.8, w h a t is the
D 3x
len g th o f D F?
E 3(x — l )2
88. REVIEW If sin x = m and 0 < x < 90°, then tan x :
H
1 - m2
J
F
5
G
4
«V1 - m 2
H 3.2
1 —m2
J I
F
4
m
D
1 — m2
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3
&
3
1
9
y
i
•
You simplified
trigonometric
expressions.
(Lesson 5-1)
Verify trigonometric
identities.
•
Determine whether
equations are
identities.
Two firew orks travel at the same speed v. The
fireworks technician wants to explode one
firew ork higher than another by adjusting the
angle 0 of the path each rocket makes with
the ground. To calculate the maximum height h
of each rocket, the form ula h = v tan„ e could
2 g sec2 9
be used, but would h -
v sin e give the same
2g
result?
NewVocabulary
verify an identity
Verify Trigonometric Identities
1
In L esso n 5-1, you u sed trig o n o m etric id en tities to
rew rite exp ressio n s in eq u iv alen t and so m etim es m ore u sefu l form s. O n ce v e rifie d , th ese new
id en tities can also b e u sed to solv e p ro b lem s o r to rew rite o th er trig o n o m etric expression s.
To verify an identity m ean s to prove th at b o th sid es o f the e q u atio n are equ al fo r all v alu es o f the
v ariable for w h ich b o th sid es are d efined . T h is is d on e b y tran sfo rm in g the exp ressio n o n one sid e
o f the id en tity in to the exp ressio n on the oth er sid e th ro u g h a seq u en ce o f in term ed iate exp ression s
that are each eq u iv alen t to the first. A s w ith o th er ty p es o f p ro o fs, e ach step is ju stifie d b y a reason,
u su ally an o th er v erified trig o n o m etric id e n tity o r an alg eb raic op eratio n .
You w ill find th at it is often easier to start the v erification o f a trigo nom etric identity b y b eg in n in g on
the side w ith the m ore com plicated exp ression and w ork in g tow ard the less com plicated expression.
Verify a Trigonometric Identity
Verify that csc * — - = cos2 x.
csc x
The left-h an d sid e o f this id en tity is m ore com p licated , so start w ith th at exp ressio n first.
= “ 4 ^
Pythagorean Identity
= c o t2 x s in 2 x
Reciprocal Identity
s if te r
Quotient Identity
Pythagorean Identity
N o tice th at the v erificatio n end s w ith the e xp ressio n on the o th er sid e o f the identity.
► GuidedPractice
V erify each identity.
1A. s e c 2 9 c o t2 9 — 1 = c o t2 9
1B. tan a = sec a csc a tan a — 1
Th ere is u su ally m o re th an one w ay to v e rify a n identity. F o r e xam p le, the id e n tity in E xam p le 1
can also b e v erified as follow s.
csc
x —1
Write as the difference of two fractions.
sin2 x
Simplify and apply a Reciprocal Identity,
Pythagorean Identity
320
I Lesson 5-2
W h en there are m u ltip le fractio n s w ith d ifferen t d en o m in ato rs in an exp ressio n , y ou can find a
com m on d en o m in ato r to red u ce the e xp ressio n to o n e fraction .
J j J E E E E Verify a Trigonometric Identity by Combining Fractions
Verify that 2 csc x =
—
csc x + cot x
-\--------------------------- .
csc x — cot x
The rig h t-h an d sid e o f th is id en tity is m o re co m p licated , so start th ere, rew ritin g e ach fraction
u sin g the co m m o n d en o m in ato r (csc x + cot x)(csc x — co t x).
1_______ ,_______ 1
csc x + cot x
Start with the right-hand side
of the identity.
csc x — cot x
csc x — cot x
,
(csc x + cot x)(csc x — cot x)
(csc
X
csc x + cot x
+ cot x)(csc X — cot x)
Common denominator
__________ 2 csc x__________
(csc x + cot x)(csc X — cot x)
Add.
2 csc x
csc x — c o t- x
Multiply.
Pythagorean Identity
= 2 csc x ✓
►GuidedPractice
2.
V erify th at - - ^ S.Q + J
1 + sin a
1
cos a
0
— = 2 sec a .
To elim in ate a fractio n in w h ich the d en o m in ato r is o f the form 1 ± u or u + 1, rem em ber to try
m u ltip ly in g the n u m erato r and d en o m in ato r b y the co n ju g ate o f the d en om inator. T h en yo u can
p o ten tially ap p ly a P y th ag o rean Id e n tity
■ S E H U p e r i f y a Trigonometric Identity by Multiplying
StudyTip
Alternate Method You do not
always have to start with the
more complicated side of
the equation. If you start with
the right-hand side in Example 3,
you can still prove the identity.
csc a + cot a
Verify that - - s‘n Q— = csc a + cot a .
1 —cos a
B ecau se the left-h an d sid e o f th is id e n tity in v o lv es a fractio n , it is slig h tly m ore com p licated
th an the rig h t side. So, start w ith the left side.
sin a
_
1 — cos a
1 + cos a
_
1 + cos a
sin
1 — cos2 a
a
sin a (1 + cos a)
1 - cos a
sin a
sin a
1 - cos a
1 + cos a
1 + cos a
sin a (1 + cos a)
_
1 + cos a
— sin a
sin a
_
sin a
1 — cos a
1 - cos
sin2 a
a
,
_ 1 + cos a
sin a
sin a
sin a
= csc a + co t a V
Multiply numerator and denominator by the conjugate of
1 — cos a , which is 1 + cos a .
Multiply.
Pythagorean Identity
Divide out the common factor of sin a .
Write as the sum of two fractions.
Reciprocal and Quotient identities
^ GuidedPractice
3.
tan x
V erify that
= csc x - cot x.
sec x + 1
U n til an id e n tity h as b e e n v erified , yo u can n o t assu m e th at b o th sid es o f the eq u atio n are equal,
th e re fo re , y o u can n o t u se the p ro p erties o f eq u ality to p erfo rm alg eb raic o p eratio n s on each side
o f an identity, su ch as ad d in g the sam e q u an tity to e ach sid e o f the equ ation.
c o n n e c tE tU iic g r a ^ iilU o m 1
321
W hen the more com plicated expression in an identity involves powers, try factoring.
B f T f T H T T i Verify a Trigonometric Identity by Factoring
Verify that cot 0 sec 6 csc2 6 —cot3 6 sec 6 = csc 0.
co t 9 sec 9 c sc 2 9 — c o t3 9 sec 9
Start with the left-hand side of the identity.
= co t 9 sec 9 (csc 2 9 — c o t2 9)
Factor.
= co t 9 sec 9
Pythagorean identity
cos 9
sin 9
1
cos i
Reciprocal and Quotient Identities
1
sin 8
Multiply.
Reciprocal Identity
= csc 9 y/
p GuidedPractice
4.
V erify th at s in 2 x ta n 2 x c sc 2 x + c o s 2 x ta n 2 x c sc2 x = s e c 2 x.
It is som etim es h elp fu l to w o rk e ach sid e o f an id e n tity sep arately to o b tain a co m m o n in term ed iate
expression .
StudyTip
Additional Steps When verifying
an identity, the number of steps
that are needed to justify the
verification may be obvious.
However, if it is unclear, it is
usually safer to include too many
steps, rather than too few.
Verify an Identity by Working Each Side Separately
>
Verify that —
+ sec x
B o th sid es lo ok com p licated , b u t the left-h an d sid e is slig h tly m ore co m p licated sin ce its
d en o m in ato r in v o lv es tw o term s. So, start w ith the exp ressio n on the left.
/
ta n 2 * - sec2 * - 1
1 + sec x
1 + sec x
Pythagorean Identity
(sec x — l)(s e c x + l )
1 + sec x
Factor.
Divide out common factor of sec x + 1.
= sec x — 1
From h ere, it is u n clear h o w to tran sfo rm sec x — 1 into 1 COs°^'V>so start w ith the righ t-h an d
sid e and w o rk to tran sfo rm it into th e in term ed iate fo rm sec x — 1.
1 ~coscos
i —
x * = —
cos
x - !!§
cos!■x
Write as the difference of two fractions,
Use the Quotient Identity and simplify.
= sec x — 1
To com p lete the proof, w o rk b ack w ard to co n n e ct th e tw o p arts o f the proof.
ta n 2 x _ sec2 x — 1
1 + sec x
1 + sec x
Pythagorean Identity
(sec x — l)(s e c x + l )
1 + sec x
Factor.
= sec x — 1
Divide out common factor of sec x + 1.
= cosx “ f § f f
Usethe Quotient ^ < 1% and write 1 a s |3 § £
1 — cos X
,
COS X
Combine fractions.
p GuidedPractice
5.
322
V erify th at sec4 x — se c 2 x = ta n 4 x + ta n 2 x.
| Lesson 5-2 | V e rifyin g Trigo n o m etric Identities
ConceptSum m ary Strategies for Verifying Trigonometric Identities
• Start with the more complicated side of the identity and work to transform it into the simpler side, keeping the other side
of the identity in mind as your goal.
• Use reciprocal, quotient, Pythagorean, and other basic trigonometric identities.
• Use algebraic operations such as combining fractions, rewriting fractions as sums or differences, multiplying
expressions, or factoring expressions.
• Convert a denominator of the form 1 ± u or u + 1 to a single term using its conjugate and a Pythagorean Identity.
• Work each side separately to reach a common intermediate expression.
• If no other strategy presents itself, try converting the entire expression to one involving only sines and cosines.
v
__________ ___ __________________________ _____ __________________________ ____________________________________ ______________
Identifying Identities and Nonidentities
You can u se a g rap h in g calcu lator to
in v estigate w h eth er an eq u atio n m ay b e an id en tity b y g rap h in g the fu n ctions related to each
sid e o f the equation.
2
WatchOut!
Using a Graph You can use a
graphing calculator to help confirm
a nonidentity, but you cannot use a
graphing calculator to prove that
an equation is an identity. You
must provide algebraic verification
of an identity.
K
>
I B
E
Determine Whether an Equation is an Identity
Use a graphing calculator to test w hether each equation is an identity. If it appears to be an
identity, verify it. If not, find a value for which both sides are defined but not equal.
a.
cos 0 + 1 _
c o sff
sec 0 + 1
ta n 2 0
T h e grap h s o f th e related fu n ctio n s d o n o t coin cid e fo r all v alu es o f x fo r w h ich the b oth
fu n ctio n s are d efined . W h e n x =
Yx = 1.7 b u t Yz ~ 0.3. T h e eq u atio n is n o t an identity.
[ —2-jt, 2 it ] scl:
|j
COS 0 + 1 _
ta n 2 0
tt
by [ - 1 , 3 ] scl: 1
[ — 2 t t , 2 tt]
scl:
it
by [ - 1 , 3 ] scl:
1
COS 0
sec 0 — 1
T h e eq u atio n appears to b e an id e n tity b e cau se the g rap h s of
the related fu n ctio n s co in cid e. V erify th is algebraically.
cos 0 _
cos 0
sec 0 — 1
sec 0 + 1
sec 0 — 1
sec 0 + 1
cos 0 sec 0 + cos 0
sec2 0 —1
cos 0\(—
\cos 0 ]\
1 + cos 0
sec 0 —1
i
tan 0
conjugate of sec 0 —1.
Multiply.
+ cos 0
sec 2 0 —1
cos 0 + 1
Multiply numerator and
denominator by the
v
Reciprocal Identity
Simplify.
[—2-rr, 2-it] scl: -k by [ - 1 , 3] scl: 1
Commutative Property
and Pythagorean Identity
Guided Practice
CB
,,
cot 8 tan2 8 + cot (
6A. csc 0 = -
sec 9
6B.
cos
X +
1
sec x — 1
$
connectED.mcgraw-hill.com 1
BP
!
323
Exercises
= Step-by-Step Solutions begin on page R29.
Verify each identity. (Examples 1 -3 )
Verify each identity. (Examples 4 and 5)
1. (sec2 9 — 1) c o s2 9 = sin 2 9
\2o] (csc 9 + cot 0)(1 — cos 9) = sin 9
2 . ' s e c 2 9( 1 — c o s 2 9) = tan 2 9
. 2 < sin 2 9 ta n 2 9 = ta n 2 9 - s in 2 9
3. sin
r^2
1 — tan2 9 _ cos2 9 —1
1 —cot2 1
cos 9
^
1 + csc 6
■sin 9 c o s2 9 = s in 3 1
(ffl,
4. csc 9 — cos 9 co t 9 = sin 9
5. c o t2 9 csc 2 9 — c o t2 9 = co t4
{ 6. tan 9 csc 2 9 — tan 9 = co t 9
7." sec e
sin 9
(a
9.
lo .
11.
12.
v j
sin 9 _ CQt g
cos i
sin 9
1 — cos
, 1 — cos 9
H
:— ; —
sin 9
1
+
— cot l
1
1 — tan
1
-+
1 — tan2 9
1
oc 7 1 + tan2 9
1
la . <
=
1 — tan2 9
1
2 cos2 I
2 CSC
2
26. ta n 2 9 cos
cos2 9 = 1 —cos2
_ 2 f . , sec 9 — cos 9 = tan 9 sin 9
- = sin 9 + cos 9
-= 1
1 — cot2 o
■H-------- \— - = 2 s e c 2 9 sin 9
csc 9 + 1
2 _ 1 — cos 9
+ cos 9
24. (csc 9 - cot 9 )2 = j
r,
cos 9
+ tan 9 — sec i
1 + sin 9
,
= co s 9 + co t 9
28. 1 — ta n 4 9 = 2 se c 2 9 — sec 4 9
29. (csc 9 - co t 9 )2 = } •••cos I
j/'
30.
1 + tan 9
sin 9 + cos 9
w
2 + csc 9 sec 6
= (sin 9 + cos 9 )2
csc 9 sec
csc 9 — 1
13. (csc 9 — cot #)(csc 9 + co t 9) = 1
1 + cos 9
1
: sec 9
14. co s4 9 — sin 4 9 = c o s 2 9 — s in 2 9
15.
ljg\
1
.
r+ ■
1 — sin 9
1
1 + sin i
cos 9
,
cos 9
1 + sin 9
1 — sin l
- 2 se c 2 i
32. OPTICS If tw o p rism s o f the sam e p o w er are p laced n ex t
to each other, th eir total p o w er can be d eterm in ed u sing
z = 2 p cos 9, w h ere 2 is the com b in ed p o w er o f the
=2 sec 9
p rism s, p is the p o w er o f the in d iv id u al p rism s, and
9 is th e an g le b e tw e e n the tw o p rism s. V erify th at
2 p co s 9 = 2 p (l — s in 2 9) sec 9. (Example 4)
17. csc4 9 — co t4 9 = 2 c o t2 9 + 1
18.
csc2 9 + 2 csc 9 —3 _ csc 9 + 3
CSC2
9 -1
csc 9 + 1
33. PHOTOGRAPHY T h e a m o u n t o f lig h t p assin g th ro u g h a
level, the m axim u m h eig h t th at it reach es is giv en by
p o larizatio n filter can b e m o d eled u sin g I = I m c o s2 9,
w h ere I is the am o u n t o f lig h t p assin g th rou gh the filter,
I m is the am o u n t o f lig h t shined on the filter, and 9 is
the an g le o f rotation b etw een the lig h t sou rce and the
h = 0 ^
filter. V erify th at
(l9 ) FIREWORKS If a rocket is lau nched from ground
®, w h ere 9 is the an g le b etw een the ground
and the in itial p ath o f the rocket, v is the ro ck e t's in itial
speed , and g is the acceleration d u e to gravity,
9.8 m eters p er second squared. (Example 3)
cos2 9 = l m
tan x + l _ 1 + cot x
tan x — 1
1 —cot x
35. sec x + tan x
1
sec x —tan x
36. sec x — 2 sec x tan x + tan x 37.
b.
324
v2 tan2 9
2g sec2 9
Su p p ose a second ro cket is fired at an angle o f 80°
from the ground w ith an in itial sp eed o f 110 m eters
p er second . Find the m ax im u m h eig h t o f the rocket.
| Lesson 5-2 | V e rifyin g T rigo n o m etric Identities
(Example4)
GRAPHING CALCULATOR Test w hether each equation is an
identity by graphing. If itappears to be an identity, verify it.
If not, find a value for which both sides are defined but not
equal. (Example 6)
2^
v2 sin2 9
a. Verify that
2g
^ ------.
cor 9+ 1
38.
on
cot2 j — 1
1 + cot2 X
= 1 — 2 s in 2 x
tan x —sec x
tan x -
tan x + sec x
2
-2
39. cos x — sin x =
cot x —tan x
—
tan x + cot x
1 — cos x
1 + cos x
Verify each identity.
40.
GRAPHING CALCULATOR Graph each side of each equation. If
the equation appears to be an identity, verify it algebraically.
/sin
x tan x
I
= |sin x\
sec x
sec x —1
sec z + 1
41.
sec x
55. cos X
I sec x — 1 I
I tan x I
tan x sec x
CSC X
1
56. sec x — cos.2 x. csc x = tan x sec x
42. In |csc x + co t x\ + In |csc x — co t x\ = 0
57. (tan x + sec x )( l — sin x) = cos x
43. In |cot x\ + In |tan x cos x\ = In |cos x|
58.
Verify each identity.
44. s e c 2 9 + ta n 2 9 = sec4
45. —2 co s2 9 = sin 4
tan * 6
sec x cos x
1
cot2 x
tan2 x — sin2 x tan2 x
= -1
59. i § l MULTIPLE REPRESENTATIONS In this p ro blem , you w ill
in v estig ate m eth od s u sed to so lv e trigo n o m etric
equ atio n s. C o n sid er 1 = 2 sin x.
c o s4 9 — 1
46. sec2 9 sin 2 9 = s e c 4 — (tan 44 9, + sec 2 9)
a. NUMERICAL Iso late the trig o n o m etric fu n ctio n in the
eq u atio n so th at sin x is the o n ly exp ression o n one
47. 3 sec2 9 ta n 2 9 + 1 = s e c 6 0 — ta n 6 9
sid e o f the equ atio n .
48. s ec4 x = 1 + 2 ta n 2 x + ta n 4 x
b. GRAPHICAL G rap h the left and rig h t sid es o f the
eq u atio n y o u fo u n d in p a rt a on the sam e grap h over
49. sec2 x csc2 x = sec2 x + csc 2 x
[0, 27t). L ocate an y p o in ts o f in tersection and express
the v alu es in term s o f rad ian s.
50. ENVIRONMENT A b io lo g ist stu d y in g p o llu tio n situ ates
a net across a riv er and positio n s in stru m en ts at tw o
d ifferent statio n s on the riv er b an k to collect sam ples.
In the d iag ram sh ow n , d is the d istan ce b e tw e e n the
c. GEOMETRIC U se the u n it circle to v erify the answ ers
stations and w is w id th o f the river.
d. GRAPHICAL G rap h the le ft and rig h t sid es o f the
e q u atio n yo u fou n d in p a rt a o n the sam e grap h over
—27T < x < 277. L o cate an y p o in ts o f in tersection and
y o u fo u n d in p a rt b .
exp ress the v alu es in term s o f rad ians.
e. VERBAL M ake a co n jectu re as to the solu tio n s of
1 = 2 sin x. E xp lain y o u r reasoning.
a.
H.O.T. Problems
D eterm in e an eq u atio n in term s o f tan g en t a th at can
b e used to find the d istan ce b e tw e e n the stations.
b.
Verify th at d ■
60.
w cos (90° — a)
Use Higher-Order Thinking Skills
REASONING C an su b stitu tio n b e u sed to d eterm in e
w h eth er an eq u atio n is an id en tity ? E xp lain you r
reasoning.
C. C om p lete the table sh o w n for d = 40 feet.
w
20
40
60
80
100
120
a
d. If a > 60° or a < 20°, the in stru m en ts w ill n o t fu n ctio n
properly. U se the table from p a rt c to d eterm in e
w h eth er sites in w h ich the w id th o f the riv er is 5, 35,
61) CHALLENGE V erify th at the area A o f a trian g le is g iv en b y
a 2 sin /3 sin 7
2 sin (f3 + 7)
w h ere a, b, and c rep resen t the sid es o f the trian g le and a ,
(3, and 7 are the resp ectiv e o p p o site angles.
62. WRITING IN MATH U se the p ro p erties o f lo g arith m s to
or 140 feet cou ld b e u sed fo r the exp erim en t.
e xp lain w h y the su m o f th e n a tu ra l lo garith m s o f the six
b asic trig o n o m etric fu n ctio n s for any angle 9 is 0.
HYPERBOLIC FUNCTIONS The hyperbolic trigonom etric
functions are defined in the following ways.
sin h x = i ( e * - e - * )
csch x = —7—-— , x ± 0
s in h x
cosh x = ~ (ex + e x)
sech x =
tanh x -
s in h
co sh
x
x
coth
c o s h ,Y
1
X —
;
ta n h
x
, X ±
0
63. OPEN ENDED C reate id e n tities for sec x and csc x in term s
o f tw o o r m o re o f the oth er b asic trig o n o m etric functions.
64. REASONING If tw o an g les a and (3 are com plem entary, is
c o s 2 a + c o s 2 f3 = 1? E xp lain y o u r reasoning. Ju stify your
answ ers.
Verify each identity using the functions shown above.
51.
co sh 2 x — sin h 2 x = 1
52. s in h (—x) = —sin h x
53.
sech 2 x = 1 — ta n h 2 x
54. c o s h (—x) = cosh x
65.
WRITING IN MATH E xp lain h o w you w o u ld v erify a
trig o n o m etric id e n tity in w h ich b o th sid es o f the equ ation
are eq u ally com plex.
325
Spiral Review
Simplify each expression. (Lesson 5-1)
69.
6 8 . sin 9 co t i
67. tan 9 cot I
66. cos 9 csc i
cos 9 csc 9
tan 9
70.
sin 9 csc 9
cot 9
71.
1 - cos2 9
sin2 9
72. BALLOONING A s a h o t-air b allo o n crosses ov er a straig h t p o rtio n o f
in terstate highw ay, its p ilo t eyes tw o co n secu tiv e m ilep o sts on the
sam e sid e o f the ballo o n. W h en v ie w in g the m ilep osts, the an g les of
d ep ression are 64° and 7°. H ow h ig h is the b a llo o n to the n earest
foot? (Lesson 4-7)
Locate the vertical asymptotes, and sketch the graph of each function. (Lesson 4-5)
1
y = — tan x
73.
1
75. y = — sec 3x
74. y = csc 2x
Write each degree measure in radians as a multiple of it and each radian measure in
degrees. (Lesson 4-2)
76.
660°
79. —
4
77. 570°
78. 158°
80. iZlL
6
81. 9
Solve each inequality. (Lesson 2-6)
82. x 2 - 3x
- 18 > 0
83. x 2 + 3x - 28 <
85. x 2 + 2x
> 24
86. - x 2 -
88.
84. x 2 - 4x < 5
0
87. - x 2 - 6x + 7 < 0
x + 12 > 0
FOOD T he m an ager o f a b ak ery is ran d o m ly ch eck in g slices o f cake prep ared b y em p lo y ees
to ensure th at the correct am ou n t o f flav o r is in each slice. E ach 12-o u n ce slice should
con tain h a lf ch o colate and h alf v an illa flavo red cream . T h e a m o u n t o f ch o co late b y w h ich
each slice varies can b e rep resented b y g(x) = j| x — 12|. D escrib e the tran sfo rm atio n s in the
function. T h en grap h the fun ction. (Lesson 1-5)
Skills Review for Standardized Tests
89. SAT/ACT
a, b, a, b, b, a, b, b, b, a, b, b, b ,b , a , ...
91. REVIEW W h ich o f the fo llo w in g is n o t eq u iv alen t to
cos 9 w h e n 0 < 9 < y ?
cos 9
If the sequ ence con tin u es in this m an ner, h o w m an y
bs are there b etw een the 44th and 4 7 th ap p earan ces of
the letter a ?
A 91
C 138
B 135
D 182
B 1 - sin2
cos 9
E 230
90. W h ich exp ression can be used to form an id en tity
sec 9 + csc I
-w h e n tan 8 =/= —1?
w it h :
92.
F 2 sin 9
1
sin 9
F sin (
H cos2 6
H tan i
J csc (
326
| Lesson 5-2 | V e rifyin g Trigo n o m etric Identities
D tan 9 csc i
REVIEW W h ich o f the fo llo w in g is eq u iv alen t to
sin 9 + cot 9 cos 9 ?
1 + tan f
G cos i
C co t 9 sin i
J
sin 9 + cos (
sin2 9
■§ Solve trigonom etric
•
I equations using
algebraic techniques.
2
Solve trigonom etric
equations using basic
identities.
A baseball leaves a bat at a launch angle 0 and
returns to its initial batted height after a
distance of d meters. To find the velocity v0 of
the ball as it leaves the bat, you can solve the
2Kn2 sin 0 cos 0
trigonom etric equation d = — — — ---------- .
■4 Use Algebraic Techniques to Solve
In L esso n 5-2, y o u v erified trig o n om etric equations
1 called id en tities th at are tru e fo r all v alu es o f the v ariab le for w h ich b o th sid es are d efined . In
th is lesson w e w ill co n sid er conditional trig o n o m etric e q u atio n s, w h ich m a y be tru e fo r certain
v alu es o f the v ariab le b u t false fo r others.
C o n sid er the grap h s o f b o th sid es o f the con d itio n al trig o n o m etric eq u atio n cos x = j .
X = -52L - 27r
3
1 ^
V V - ---------------------* i \
I
/
1
\ v
\y
^
x =
y= !
r
2
1
II —
(Lesson 5-2)
•
1
You verified
trigonometric
identities.
i
1
tt
/
y
- T
- 2*
T h e grap h sh o w s th at cos
y
Sin ce
y=
cos
x h as
3
i
2 ir ,
y
^
! rS\ 2 T r ^ / '
X
x= f + 2.
^ h as tw o solu tio n s o n the in terv al [0,
a p erio d o f
i
I JT
i
2 tt
y = cos x
x=
x = - 5 f + 2iT
I o
*
•
2 tt),
x=
-j
and
x=
cos x = \ h as in fin itely m an y solu tio n s o n the in terv al (—oo, oo).
A d d itio n al solu tio n s are fo u n d b y ad d in g in teg er m u ltip les o f the p erio d , so w e exp ress all
solu tio n s b y w ritin g
x=
y + 2 « tt
and
x=
+ 2 mr,
w h ere
n is an
integer.
To solv e a trig o n om etric eq u atio n th at in v o lv es o n ly o n e trig o n o m etric exp ressio n , b e g in b y
iso latin g th is exp ression .
Solve by Isolating Trigonometric Expressions
Solve 2 tan x — V 3 = tan x.
Original equation
2 tan x — V 3 = tan x
tan x — \ / 3
= 0
Subtract tan xfrom each side to isolate the trigonometric expression,
Add V3 to each side.
tan x = \/3
The p erio d o f tan g e n t is
so you o n ly n eed to find solu tio n s on the in terv al [0,
tt,
so lu tio n on th is in terv al is
ad d in g in teg er m u ltip les
x=
tt.
T h erefore, the g en eral form o f the so lu tio n s is
x=
p GuidedPractice
1.
Solv e 4 sin
tt).
The only
-j. T h e solu tio n s on the in terv al (—oo, oo) are th en foun d b y
x = 2 sin x + \[2.
—■+ H7t,
w h e re
n is
an integer.
p i M i T l f f t Solve by Taking the Square Root of Each Side
StudyTip
Find Solutions Using the Unit
Circle Since sine corresponds to
the y-coordinate on the unit circle,
you can find the solutions of sin
Solve 4 sin2 x + 1 = 4.
• 2
3
sm x = —
using the unit circle, as shown.
1 V3
\
2’ 2
'K\
4it
3
O
\ 3
5tv
J
2
fl
and sin x = —
wh e n x = ^
and x =
S in ce sin e has a p erio d o f 27T, the solu tio n s o n the in terv al (— oo, oo) h av e the gen eral
form x
+ 2
= y
m it , x
=
■
2htt, x =
+ 2 n7r, an d x = - ^ + 2« ir, w here n is an integer.
_V 3\
1
V3
w h e n x = y an d x =
X
/
r'H
1
2’
Take the square root of each side.
O n the in terv al [0, 2ir), sin x =
\
3y
f^l
\
\
V3
\ 2’ 2 )
3
Divide each side by 4.
sin x = +■
\y(l
/2 it
/
Subtract 1 from each side.
4 sin 2 x = 3
x = ± ^ - on the interval [0 ,2it]
I
Original equation
4 sin 2 x + 1 = 4
Any angle coterminal with these
angles will also be a solution of
the equation.
►GuidedPractice
2. S olv e 3 co t2 x + 4 = 7.
W h en trig o n om etric fu n ctio n s can n o t be co m b in ed on one sid e o f an equ ation , try facto rin g and
ap p lyin g the Z ero P ro d u ct Property. If the eq u atio n h as q u ad ratic form , facto r if p ossible. If n o t
p o ssib le, ap p ly the Q u ad ratic Fo rm u la.
Solve by Factoring
Find all solutions of each equation on the interval [0, 2ir).
a.
Watch Out!
cos x sin x = 3 cos x
Original equation
cos x sin 9 = 3 co s x
Dividing by Trigonometric
Factors Do not divide out the
cos x in Example 3a. If you were
to do this, notice that you might
conclude that the equation had no
solutions, when in fact, it has two
on the interval [0, 2-k).
isolate the trigonometric expression.
cos x sin x — 3 cos x = 0
Factor,
cos x(sin x — 3) = 0
cos x = 0
or
sin x — 3 = 0
Zero Product Property
sin x = 3
Solve for x on [0 , 2it].
T h e eq u atio n sin x = 3 h as n o so lu tio n sin ce the m ax im u m v alu e the sin e fu n ctio n
can attain is 1. T herefore, o n the in terv al [0, 2 7 t ) , the solu tio n s o f the o rigin al eq u atio n
TT
j
3TV
are — and —
b. cos4 x + cos2 x — 2 = 0
Original equation
c o s4 x + c o s2 x — 2 = 0
Write in quadratic form.
(co s2 x) 2 + c o s2 x — 2 = 0
Factor.
(cos2 x + 2 )(co s2 x — 1) = 0
c o s2 x + 2 = 0
or
Zero Product Property
cos x — 1 = 0
c o s 2 x = —2
cos x = + V —2
Solve for cos2 x.
c o s2 x = 1
cos x = ± V l o r + 1
Take the square root of each side.
The eq u atio n cos x = ± V —2 h as n o real solu tio n s. O n the in terv al [0, 2ir), the eq u atio n
cos x = ± 1 has solu tio n s 0 and 7r.
i» GuidedPractice
3A. 2 sin x cos x = \ fl cos x
V______________________________
328
| Lesson 5-3
So lving Trigo n o m etric Equations
3B. 4 cos2 x + 2 cos x — 2\/2 cos x = \ fl
So m e trig o n o m etric eq u atio n s in v olv e fu n ctio n s o f m u ltip le an g les, su ch as cos 2x =
these e q u atio n s, first solv e for the m u ltip le angle.
StudyTip
R eal-W orld E x a m p le 4
Exact Versus Approximate
Solutions When solving
trigonometric equations that are
not in a real-world context, write
your answers using exact
values rather than decimal
approximations. For example, the
general solutions of the equation
tan x = 2 should be expressed as
x = t a n _1 2 + m vor
x = arctan 2 + mr.
Tri
To solve
ic Functions of Multiple Angles
BASEBALL A baseball leaves a bat with an initial speed of 30 meters per second and clears a
fence 90.5 meters away. The height of the fence is the same height as the initial height of the
v02s in 26
batted ball. If the distance the ball traveled is given by d = -------
9 .0
-, where 9.8 is in meters
per second squared, find the interval of possible launch angles of the ball.
vg2s in 29
<0i
s in
29
9 0 0 s in
29
30
9 0 .5 :
90.5 =
886.9 = 900 sin 29
8 8 6 .9
900
= sin 29
-,n - l
8 8 6 .9 _
Sln
~ 9 0 0 ~ -29
Original formula
d = 90.5 and i/0 = 30
Simplify.
Multiply each side by 9.8.
Divide each side by 900.
Definition of inverse sine
R ecall from L esso n 4-6 th at the ran ge o f the in v e rse sin e fu n ctio n is restricted to acute an g les o f I
in the in terv al [—90°, 90°]. S in ce w e are fin d in g the in v e rse sin e o f 2 9 in stead o f 9, w e n eed to
co n sid er an g les in the in terv al [ - 2 ( 9 0 ° ) , 2(90°)] o r [ - 1 8 0 ° , 180°]. U se y o u r calcu lato r to find the
acu te an g le and the referen ce an g le relatio n sh ip sin (180° — 9) = sin 9 to find the o btu se angle.
„ i n - l 8 8 6 .9 .
sin
W
Definition of inverse sine
" 20
80.2° or 99.8° = 29
s in - 1
40.1° or 49.9° = 9
Divide by 2.
a 80.2° and sin (180° - 80.2°) = sin 99.8°
T h e in terv al is [40.1°, 49.9°]. T h e b a ll w ill clear the fen ce if the an g le is b e tw e e n 40.1° and 49.9°.
CHECK S u b stitu te the an g le m easu res in to the o rigin al eq u atio n to co n firm the solu tion.
v02s i n 20
9 .8
StudyTip
Optimal Angle Ignoring wind
resistance and other factors,
a baseball will travel the greatest
distance when it is hit at a
45° angle. This is because sin
2(45) = 1, which maximizes the
distance formula in the example.
3 0 2 s in (2 • 4 0 .1 ° )
90.5
:
90.5
~ 90.497 ✓
9 .8
Original formula
0 = 40.1° or 0 = 49.9°
Use a calculator.
vn2s in 29
d = ■
9 .8
n n r
? 3 0 2 s in ( 2 - 4 9 . 9 ° )
9 0 5
= ---------------- 9£ ---------
90.5
~ 90.497 ✓
p GuidedPractice
>
4.
BASEBALL Find the in terv al o f p o ssib le lau n ch an g les requ ired to clear the fence if:
A. th e in itial sp eed w as in creased to 35 m eters p er second .
B. th e in itial sp eed rem ain ed th e sam e, b u t the d istan ce to the fen ce w as 80 m eters.
connectED.m cgraw-hill.com j
329
2
Use Trigonometric Identities to Solve
You can u se trig o n o m etric id en tities alo ng w ith
algebraic m eth o d s to solv e trig o n o m etric equ atio n s.
Example 5 Solve by Rewriting Using a Single Trigonometric Function
Find all solutions of 2 cos2 x — sin x — 1 = 0 on the interval [0, 2 ji).
2 (1
— sin2 x )
— 1 ( 2 s in
bo
and
L
x
x —
s in
x—1
s in
x—1=
s in
x —1
s in x +
— l) ( s in x +
s in x — 1 = 0
1
=
Pythagorean Identity
0
=
1
) =
Multiply.
Simplify.
0
Factor.
0
s in x +
or
1
=
Zero Product Property
0
sin x = —1
3 tt
—
x = £ o r^
6
6
on
Original equation
0
= 0
s in x = j
Alternate Method An alternate
way to check Example 5 is to
graph y = 2 cos^ x - sin x - 1,
it has zeros ■£,
—
s in 2
—2
2
—
— 2 s in 2 x —
2
StudyTip
x
cos2
2
2
Solve for sin x.
Solve for x on [ 0 ,2 ir].
CHECK T h e grap h s o f Y i = 2 c o s 2 x — sin x and Y 2 = 1 in tersect
the interval [0 ,2 it) as shown.
at
6
an d 4 ? on the in terv al [0, 2 ttJ as sh o w n v''
2
6
p Guided Practice
[0, 2-rc] scl: f
by [ - 2 , 4] scl: 1
[0, 2 ir] scl: f
by [ - 2 , 4 ] scl: 1
Find all solutions of each equation on the interval [0, 2 it).
5A. 1 — cos x = 2 s in 2 x
5B. co t2 x c sc 2 x + 2 c sc 2 x — c o t2 x = 2
S o m etim es y o u can o b tain an eq u atio n in o n e trig o n o m etric fu n ctio n b y sq u arin g each sid e, b u t this
tech n iq u e m ay p rod u ce extran eo u s solu tions.
M ' i S S S Q Solve bV Squaring
Find all solutions of csc
CSC X — c o t X =
CSC X =
(csc x) 2
-
CS C 2 X :
1
+ c o t2
x
=
=
Original equation
1 +
(1
Add cot x to each side.
cot X
+
cot
1
+ 2
1
+
cot
cot
*= ? “
CHECK
— cot x = 1 on the interval [0, 2n].
1
0 = 2
0
x
cot
2
Square each side.
x )2
x +
cot
x+
x
c o t2
x Multiply.
c o t2
x Pythagorean Identity
Subtract 1 + cot2 xfrom each side.
Divide each side by 2.
x
Solve for x on [ 0 ,2-rr].
¥
x —co t x = 1
It
. TT ? h
CSC —---------- C O t — = 1
2
2
csc
1 — 0 = 1*'''
Original equation
CSC
3 ir
Substitute.
Simplify.
3 3 0 | Lesson 5-3 | So lving Trigo n o m etric Equations
1
- 1 - 0 # IX
Guided Practice
6A. sec x + 1 = ta n x
. 3 ir ?
csc —------ co t — = 1
2
2
Therefore, the o n ly v alid so lu tio n is ^ on the in te rv al [0, 2 tt].
y
x — co t x = 1
6B. co s x = sin x — 1
Exercises
= Step-by-Step Solutions begin on page R29.
Solve each equation for all values of x. (Examples 1 and 2)
Find all solutions of each equation on the interval [0, 2 tt].
(Examples 5 and 6)
1 . 5 sin x + 2 = sin x
2. 5 = s e c 2 x + 3
3. 2 = 4 c o s 2 x + 1
4. 4 tan x — 7 = 3 ta n x — 6
5. 9 + c o t2 x = 12
6. 2 — 10 sec x = 4 — 9 sec x
22.
sec
7. 3 csc x = 2 csc x + \[2
8. 11 = 3 c sc 2 x + 7
23.
ta n 2
10. 9 + sin 2 x = 10
24.
csc
12. 7 cos x = 5 cos x + \/3
25. 2
9. 6 ta n 2 x — 2 = 4
11. 7 co t x — V 3 = 4 co t x
21. 1 =
c o t2
x=
ta n
x= 1
x+
2
—
x+
cot
cos2
x —4
x
csc
x+ 1
— sec
x
x
=
=
x
1
x+
s in
x
1
Find all solutions of each equation on [0, 2iv]. (Example 3)
26.
13. sin 4 x + 2 s in 2 x — 3 = 0
27. 3
14. —2 sin x = —sin x cos x
28.
c o t2
x csc2 x —c o t 2 x =
29.
sec2
x —1
30.
sec2 x ta n 2
15. 4 cot x = co t x s in 2 x
16. csc 2 x — csc x + 9 = 11
cos
s in
x
=
=
s in
3 —3
—
x
cos
+ ta n
x+3
4
x
9
— V 3 ta n x =
sec2
V 3
x —2 t a n 2 x
=
3
17. c os3 x + c o s 2 x — cos x = 1
18. 2 sin 2 x = sin x + 1
19) TENNIS A ten n is b all leav es a racq u et and head s tow ard a
31.
OPTOMETRY O p to m etrists so m etim es jo in tw o o bliqu e or
tilted p rism s to co rrect v isio n . T h e resu ltan t refractiv e
po w er P R o f jo in in g tw o o b liq u e p rism s can be calcu lated
net 40 feet away. T h e h eig h t o f the n e t is the sam e h eig h t
as the in itial h eig h t o f the ten n is ball. (Example 4)
b y first reso lv in g e ach p rism in to its h o rizo n tal and
v ertical com p o n en ts, P H an d P v .
P o s itio n o f B a s e
O p h th a lm ic P ris m
R e s o lv in g
P ris m P o w e r
P v = P R sin 9
a. If the b all is h it at 50 feet p er seco n d , n eg lectin g air
P H= P R cos 9
Base
resistance, use d = -^ v 02 sin 29 to find the in terv al of
po ssible an g les o f the b a ll n eed ed to clear th e net.
b. Find 9 if the in itial v elo city rem ain ed the sam e b u t the
U sin g the e q u atio n s ab o v e, d eterm in e fo r w h at v alu es of
9 P v a n d P H are equ iv alen t.
d istance to the n e t w as 50 feet.
20. SKIING In the O ly m p ic aerial skiin g com p etitio n , skiers
speed d o w n a slop e th at lau n ch es th em into the air, as
show n. T h e m ax im u m h eig h t a sk ier can reach is g iv e n by
V r? s i n 2
ft_eak = —
9
2 -------' w “ ere
o
S *s
acceleratio n d u e to
g rav ity or 9.8 m eters p er secon d squared . (Example 4)
Find all solutions of each equation on the interval [0, 2 tv].
32.
33.
34.
35.
ta n 2 x
sec x
+ co s x = 2
cos x
1 + s in x
1 + sin x
+ ■
= -4
sin x + cos x , 1 - sin x
ta n x
C O t x CO S X +
H
:
=
CO S X
1 =
sec x — 1
ta n
x
a. If a skier obtain s a h eig h t o f 5 m eters abo v e th e end o f
the ram p, w h at w as the s k ie r's in itial speed ?
b. U se you r answ er from p a rt a to d eterm in e ho w
long it to o k the skier to reach the m ax im u m heigh t
•f 1
1
peak —
vn
Us i n 9
g
'
GRAPHING CALCULATOR Solve each equation on the interval
[0, 2ir] by graphing. Round to the nearest hundredth.
36.
3 cos 2x = ex + 1
37. sin
38.
x 2 = 2 cos x + x
39. x lo g x + 5x cos x = —2
tt x
+ cos 7rx = 3x
^connectECUricgm^^^^^J 331
8
40. METEOROLOGY T h e av erage d aily tem p eratu re in
d egrees F ah ren h eit for a city can be m o d eled b y
t=
8.05 cos
(~x — Trj
56. REFRACTION W h e n lig h t trav els fro m one tran sp aren t
m ed iu m to an o th er it b e n d s o r
refracts, as show n.
+ 66.95, w here x is a fu n ctio n of
tim e, x = 1 represents Jan u ary 15, x = 2 represents
Febru ary 15, and so on.
a. U se a grap h in g calcu lator to estim ate the tem p eratu re
o n Jan u ary 31.
b. A p p roxim ate the n u m b er o f m on ths th at the av erage
d aily tem p eratu re is greater th an 70° th rou g h ou t the
entire m onth.
c. E stim ate the h ig h e st tem p eratu re o f the y ear and the
m on th in w h ich it occurs.
lig h t is e xitin g , 9 1 is the an g le o f in cid en ce, and 92 is the
an g le o f refractio n .
Find the x-intercepts of each graph on the interval [0, 2 ttI.
41.
R efractio n is d escrib ed b y n 2 sin 9 1 = n 1 sin 02, w h ere n 1
is the in d ex o f refractio n o f the m ed iu m the lig h t is
e n terin g , n2 is the in d ex o f refractio n o f the m ed iu m the
a. F in d 92 fo r e ach m aterial
sh ow n if the an g le of
y = s ir r x c o s x - c o s x
M aterial
Index of
Refraction
glass
1.52
ice
1.31
plastic
1.50
water
1.33
in cid en ce is 40° and the
in d ex o f refractio n for
air is 1 .0 0 .
b. If the an g le o f in cid en ce
is d ou bled to 80°, w ill
the resu ltin g an g les o f
refractio n b e tw ice as large
as th o se fou n d in p a rt a?
H.O.T. Problems
Use Higher-Order Thinking Skills
57. ERROR ANALYSIS V ijay and A licia are solv in g
ta n 2 x — tan x + V 3 = V 3 tan x. V ijay th in k s th at
the so lu tio n s are x = ~ + H7T, x =
Find all solutions of each equation on the interval [0, 4 tv].
45. 4 tan x = 2 sec2 x
46. 2 sin 2 x + 1 = —3 sin x
47. csc x co t2 x = csc x
48. sec x + 5 = 2 sec x + 3
47T
4
4
+ W7T, x = ^ + nix,
3
and x = — + « 7r. A licia th in k s th at the so lu tio n s are
x = y + 777T an d x = y + nix. Is eith er o f th em correct?
E xp lain y o u r reasoning.
49. GEOMETRY C o n sid er the circle below .
CHALLENGE Solve each equation on the interval [0, 2 tt].
58. 16 sin 5 x + 2 sin x = 12 sin 3 x
( 5 9 ) 4 c o s 2 x — 4 s in 2 x c o s 2 x + 3 s in 2 x = 3
a. T he len gth s o f arc AB is giv en b y s = r(29) w here
60. REASONING A re the solu tio n s o f csc x = \ fl and
c o t2 x + 1 = 2 e q u iv alen t? If so, v erify y o u r an sw er
a lg e b ra ica lly If no t, e xp lain y o u r reasoning.
0 < 9 < 77. W h en s = 18 and AB = 14, the rad iu s is
s in
-. U se
8’
a grap h in g calcu lator to fin d the
m easu re o f 29 in rad ians,
OPEN ENDED Write a trigonom etric equation that has each of
the following solutions.
b. The area o f the shad ed reg io n is g iv en by
A =
r~(8 -
s in
8)
. U se a grap h in g calcu lato r to fin d the
rad ian m easu re o f 9 if the rad iu s is 5 in ch es and the
area is 36 square inches. R o u n d to the n earest
hund redth.
Solve each inequality on the interval [0, 2 tt).
50. 1 > 2 sin
x
52. cos
51. 0 < 2
V3
(!-* )* ■ *
V|
54. cos x < —
2
332
| Lesson 5-3
COS
X—V2
53. sin ^x — y j < tan x co t x
55. \fl sin x — 1 < 0
So lving Trigo n o m etric Equations
63. WRITING IN MATH E xp lain the d ifference in the tech n iq u es
th at are u sed w h e n solv in g eq u atio n s and v erify in g
identities.
Spiral Review
Verify each identity. (Lesson 5-2)
1 + s in
s in
8_
9
9— 1
c o t2
9
csc
gc 1
+ ta n
1
+ cot
8_
8
s in
9
9
eg
'
cos
1
sec 2
,
9
1
csc2
_
8
Find the value of each expression using the given information. (Lesson 5-1)
67. tan 6} sin 6 = j , tan 8 > 0
68. csc 8, cos 6 = —
69.
csc 8 < 0
sec 8; tan 8 = —1, sin 8 < 0
70. POPULATION The p o p u latio n o f a certain sp ecies o f d eer can be m o d eled b y p = 30,000 + 20,000 cos |-^f j,
w here p is the p o p u latio n and t is the tim e in years. (Lesson 4-4)
a. W h at is the am p litu d e of the fun ctio n ? W h at d o es it rep resent?
b.
W h at is the p erio d o f the fu n ctio n ? W h a t d oes it rep resent?
c. G rap h the fun ction.
Given f ( x ) = l x 2 — 5 x + 3 and g(x) = 6 x + 4, find each. (Lesson 1-6)
71. ( f - g ) ( x )
72. (f.g ) ( x )
73. (£ )(x )
74. BUSINESS A sm all b u sin ess o w n er m u st h ire season al
Number of Additional
Employees Needed
Month
w orkers as the n eed arises. T h e fo llo w in g list sh o w s the
nu m b er o f em p loy ees hired m o n th ly fo r a 5-m o n th p eriod.
August
September
October
November
December
If the m ean o f these d ata is 9, w h at is the p o p u latio n
standard d ev iation for these d ata? R ou n d to the n earest
U
tenth. (Lesson 0-8)
Skills Review fo r Standardized Tests
75. SAT/ACT For all p o sitiv e valu es o f m and n, if
= 2 , th en x =
—
77. W h ich o f the fo llo w in g is not a so lu tio n of
0 = sin 8 + cos 6 ta n 2 91
3 tt
A
B
C
D
E
4
3
llI
4
3 + 2«
2m
2m - 3
2n
C 2 tt
D
5 tv
2
2m
3 + 2n
3
2m — 2 n
78. REVIEW T h e g rap h o f y = 2 cos 8 is sh ow n . W h ich is
a solu tio n for 2 cos 0 = 1 ?
76. If cos x = —0.45, w h at is sin
F - 0 .5 5
G
- 0 .4 5
H 0.45
J
0.55
F
G
8tt
H
3
IO tv
1 3 tv
I JJZIL
J
3
J
j
3
connectED.mcgraw-hill.com I
333
...........
Graphing Technology Lab
no o o o
^Solving Trigonometric
Inequalities
Objective
• Use a graphing calculator
to solve trigonometric
inequalities.
oooo
oooo
V
C D O O
y
You can use a graphing calculator to solve trigonometric inequalities. Graph each inequality. Then locate
the end points of each intersection in the graph to find the intervals within which the inequality is true.
Activity 1
Graph a Trigonometric Inequality
Graph and solve sin 2x > cos x.
ETEfffl R ep lace each sid e o f th is in eq u ality w ith y to fo rm the n ew inequ alities.
sin 2x > Yi and Y 2 > cos x
R E ffW G rap h e ach inequality. M ak e e ach in eq u ality sy m b o l b y scro llin g to the le ft o f the equal
sig n and selectin g Ie n t e r j u n til the sh ad ed trian g les are flashing . T h e trian g le abo v e
rep resen ts greater than, and the trian g le b e lo w rep resen ts less than (Figu re 5.3.1). In the
MODE 1m en u , select RADIAN.
G rap h the eq u atio n s in the ap p rop riate w ind ow . U se the d o m ain and ran ge o f each
trig on o m etric fu n ctio n as a g u id e (Figu re 5.3.2).
Ploti Plot2 Plot3
tY iB s in (2 X)
''Y z B c o s C X )
\Yj =
\Yh =
\Ys =
\Ys =
\Y? =
Figure 5.3.1
Figure
5.3.2
ETTSm T h e d ark ly sh ad ed area in d icates the in te rsectio n o f the grap h s and the so lu tio n o f the
sy ste m o f in eq u alities. U se CALC: in te rse c t to lo cate th ese in tersection s. M o v e the cu rso r
o v er the in tersection and select e n t e r 13 tim es.
[— 2 tt,
2 t v ] scl: j
by [ - 1 , 1 ] scl: 0.1
ETHTI3 T h e first in tersection is w h en y = 0.866 o r
in tersection is at x =
solu tio n in terv als is
O
S in ce cos -j- = -^r-, the
T h e n ex t in tersectio n is at x =
TT TT
6' 2
. A n o th er in terv al is
5 tt 3 tt
6 ' 2
Z
T h erefore, one o f the
. T h ere are an infin ite
n u m b er o f in terv als, so th e solu tio n s for all v alu es o f x are ^ +
6
and
6
+ 2«tt, ^
1
2m tt , ^
2
+
+ 2mr
Exercises
Graph and solve each inequality.
334
Lesson 5-3
1.
sin 3x < 2 co s x
2. 3 cos x > 0.5 sin 2x
3. sec x < 2 co s x
4.
csc 2x > sin 8,r
5. 2 tan 2x < 3 sin 2x
6 . ta n x > cos x
2m
r
Mid-Chapter Quiz
Lessons 5-1 through 5-3
Find the value of each expression using the given information.
Find all solutions of each equation on the interval [0 ,2 ir].
(Lesson 5-1)
(Lesson 5-3)
1.
sin 0 and cos 0, cot 0 = 4, cos 9 > 0
16. 4 sec 9 + 2 V 3 = sec 9
2. sec 9 and sin 9, tan 9 = —| , csc 9 > 0
3. tan 9 and csc 9 , cos 9 =
17. 2 tan 0 + 4 = tan 0 + 5
sin 9 > 0
18. 4 cos2 0 + 2 = 3
19. cos 9 - 1 = sin 9
Simplify each expression. (Lesson 5-1)
sin (- x )
tan (-x )
5.
sin (90° - x)
cot2 (90° - x) + 1
7.
cot2 X + 1
20.
sin x
1 + sec x
MULTIPLE CHOICE Which of the following is the solution set for cos
9 tan 9 - sin2 0 = 0 ? (Lesson 5-3)
F yn , where n is an integer
G y + nix, where n is an integer
8. ANGLE OF DEPRESSION From his apartment window, Tim can see
the top of the bank building across the street at an angle of
elevation of 9, as shown below. (Lesson 5-1)
H 7r + 2/77T, where n is an integer
J nix, where n is an integer
(employee
Solve each equation for all values of 9. (Lesson 5-3)
21. 3 sin2 0 + 6 = 2 sin2 9 + 7
a. If a bank employee looks down at Tim’s apartment from the top
of the bank, what identity could be used to conclude that sin
9 = cos 9’?
b.
If Tim looks down at a lower window of the bank with an angle
of depression of 35°, how far below his apartment is the bank
window?
22.
sin 9 + cos 9 = 0
23.
sec 9 + tan 9 = 0
24.
2 tan2 0 = 1
25. PROJECTILE MOTION The distance tfth a t a kick ball travels (in
feet) is given by d =
9. MULTIPLE CHOICE Which of the following is not equal to csc 91
vn2 sin 29
^ — , where i/0 is the object s initial speed,
0 is the angle at which the object is launched, and 32 isin feet per
second squared. If the ball is kicked with aninitial speed of82 feet
per second, and the ball travels 185 feet, what is the launch angle of
the ball? (Lesson 5-3)
(Lesson 5-1)
A sec (90° - 9)
B Vcot2 9 + 1
1
C
sin 9
26.
1
sin (90° - 9)
FERRIS WHEEL The height h of a rider in feet on a Ferris wheel
after f seconds is shown below. (Lesson 5-3)
Verify each identity. (Lesson 5-2)
10.
cos 9
1 + sin 8
cos ‘ = - 2 tan i
1 - sin 6
/7(f) = 7 5 + 70 sin ( £ t - f )
11. csc2 9 — sin2 9 — cos2 9 — cot2 9 = 0
12. sin 9 + ^ 4
tan 8
= csc i
13.
cos 8 = sec 9 - tan 9
1 + sin 8
14.
csc|
sin 8
15.
1 + sin 8
sin 8
cote = cot2 £ + csc £ + 1
cos 9
sin 8
csc (
1 - sin 8
1 - sin 8
a. If the Ferris wheel begins at f = 0, what is the initial height of a
rider?
b.
When will the rider first reach the maximum height of 145 feet?
c o n n e c t^ m c g ra w -h ilL c o m |
335
You found values of
trigonometric
functions by using
the unit circle.
* 1
(Lesson 4-3)
NewVocabulary
reduction identity
When the picture on a television screen is blurry
or a radio station w ill not tune in properly,
the problem is too often due to interference.
Interference results when waves pass through
the same space at the same tim e. You can use
trigonom etric identities to determine the type of
interference that is taking place.
Use sum and difference
identities to evaluate
trigonometric functions.
Use sum and difference
identities to solve
trigonometric equations.
Evaluate Trigonometric Functions
In L esso n 5-1, y o u used b asic id en tities inv o lv ing
o n ly one v ariable. In this lesso n , w e w ill co n sid er id en tities in v o lv in g tw o variab les. O n e o f
th ese is the cosine o f a difference identity.
1
(3 b e an g les on the in terv al [0, 27t],
a > (3 as sh ow n in Figu re 5.4.1. B ecau se e ach p o in t is lo cated o n the u n it circle, x 2 + i/,2 =
xi + Vi = 1' and x 3 2 + J/3 2 = 1- N o tice also th at the m easu re o f arc C D = a — (a — (3) or /3 and
the m easu re o f arc AB = (3.
L et p o in ts A, B, C, and D b e located o n th e u n it circle, a and
and
1,
_
0 (x2, y.)
D(x,,y.)
Figure 5.4.1
Since arcs AB and CD hav e the sam e m easure, chords AC and BD show n in Figure 5.4.2 are congruent.
Chords AC and BD are congruent.
AC = BD
Distance Formula
V (x2 - l
)2
+ (y 2 -
0)2
=
(x2 - l
)2
+ (y 2 -
0 )2
= (x3 - X j) 2 + ( 1/3 - y j ) 2
Square each side.
y 22
=
Square each binomial.
2Z -
2x2
+ 1+
\ j
( x 3 - X j ) 2 + ( 1/3 - y
x 3z - 2 x 3^
+
t )2
X jz + y 32 - 2y3y 1 + y xl
(*22 + 1/22) ~ 2x2 + 1 = (x l 2 + 3/l2) + <x 32 + J/32) - 2*3*1 ~ 23/33/l
1 — 2x2 + 1 =
1
+
1
Substitution
— 2 X3X2 — 2 y3yj
Add.
2 - 2 x 2 = 2 - 2 x 3x j - 2 y 3 y j
- 2 x 2 = - 2 x 3x j -
Group similar squared terms.
Subtract 2 from each side.
2y3y x
Divide each side by —2.
x 2 = x 3x j + y 3 y j
In Figu re 5.4.1, n o tice th at b y the u n it circle d efin itio n s fo r cosin e and sin e, x , = co s /3,
x 2 = cos ( a — /3), x 3 = cos a , y x = sin (3, and y 3 = sin a . S u b stitu tin g , x 2 = x 3x x + y 3y x beco m es
(a
co s
B y rew ritin g a — (3 as
cosin e of a sum .
cos
[a
a
— 13) = co s
a
cos
336
Lesson 5-4
sin
a
sin
(3.
Cosine Difference Identity
+ (—f3) and ap p ly in g the abo v e identity, w e o b tain the id e n tity for the
+ (—/3)] = cos
a
cos
= co s a cos
llll
(3 +
{—(3) + sin a sin
(3 — sin a sin (3
(—f3)
Cosine Difference Identity
Even-Odd Identity
T h ese tw o cosin e id en tities can be u sed to e stab lish e ach o f the oth er su m and d ifferen ce id entities
listed below .
KeyConcept Sum and Difference Identities
Sum
Difference Identities
Identities
cos ( a + /3) = cos a cos /3 - sin a sin /3
sin ( a + 0) = sin a cos (3 + cos a sin /3
,
tan a + tan 0
tan a + /3) = - — r------- r~ ^
1 - t a n a tan 0
cos [ a - 13) = cos a cos /3 + sin a sin f3
sin ( a - / 3 ) = sin a cos /3 - cos a sin f3
. ,
tan a - tan 0
tan (a - /3 = - — ;------- 7— ^
1 + tan a tan 0
r-J
\
You will prove the sine and tangent sum and difference identities in Exercises 5 7 -6 0 .
B y w ritin g an g le m easu res as the su m s or d ifferen ces o f sp ecial an g le m easu res, you can use these
su m and d ifferen ce id en tities to find e xact v alu es o f trig o n o m etric fu n ctio n s o f ang les th at are less
com m on.
J ^ ^ ^ ^ j E v a l u a t e a Trigonometric Expression
Find the exact value of each trigonom etric expression,
a.
sin 15°
W rite 15° as the su m o r d ifferen ce o f an g le m easu res w ith sin es th at you know.
StudyTip
4
4 5 ° -30° = 15°
sin 15° = sin (45° - 30°)
Check Your Answer You can
check your answers by using
a graphing calculator. In Example
1a, sin 1 5 ° « 0.259 and
= sin 45° co s 30° — sin 30° cos 45°
V2
2
~ 0.259. Make sure
’
V | _i
2
2
'
V2
2
= V6 _ V|
that the calculator is in the correct
mode.
4
sin 30° = 1 sin 45° = cos 45° =
cos 30° = -y-
Multiply.
4
V6 - V2
b.
Sine Difference Identity
Combine the fractions.
ta n £
. Z3L
ta n l f = t a n ( f + f )
ta n y
: 12
+ ta n y
i - t a n f tan ^
4
V3+ 1
tan i = V3 and tan ~ = 1
3
1 - V3(l)
V3 + 1
V3+ 1
1 + V3
1 - V3
1 + V3
V3 + 1 + 3 + V3
1 -V 3 + V 3 -3
-2
= - 2 - V3
4
Simplify.
1 - V3
4 + 2V3
Tangent Sum identity
Rationalize the denominator.
Multiply.
Simplify.
Simplify.
► GuidedPractice
1A.
cos 15°
IB . s i n f
337
Sum and difference identities are often used to solve real-world problems.
-> M
D
m
B
Use a Sum or Difference Identity
ELECTRICITY An alternating current i in amperes in a certain circuit can be found after
t seconds using i = 3 (sin 165 )t, where 165 is a degree measure.
a.
Rewrite the formula in terms of the sum of two angle measures.
Original equation
i = 3 (sin 1 65)f
= 3 sin
b.
(120 + 4 5 )f
120 + 45 = 165
Use a sine sum identity to find the exact current after
Rewritten equation
i = 3 sin (120 + 4 5 )t
= 3 sin (120 + 45)
f= 1
= 3[(sin 120)(cos 45) + (cos 120)(sin 45)]
Sine Sum Identity
1 -atm
Real-WorldCareer
Lineworker Lineworkers are
responsible for the building
and upkeep of electric power
transmission and distribution
facilities. The term also applies
to tradeworkers who install and
maintain telephone, cable TV,
and fiber optic lines.
1 second.
Substitute.
Multiply.
3V 6 - 3V 2
Simplify.
T h e e xact cu rren t after 1 secon d is
am p eres.
p GuidedPractice
2.
ELECTRICITY A n altern atin g cu rren t i in am p eres in an o th e r circu it can b e fo u n d after
t seco n d s u sin g i = 2 (sin 2 85)f, w h ere 285 is a d eg ree m easu re.
A. R ew rite the fo rm u la in term s o f the d ifferen ce o f tw o an g le m easures.
B. U se a sin e d ifference id en tity to find the e xact cu rren t after 1 second .
If a trig on o m etric e xp ressio n h as the form o f a su m o r d ifferen ce identity, y o u can use the id en tity
to fin d an e xact v alu e or to sim p lify an e xp ressio n b y rew ritin g the exp ressio n as a fu n ctio n o f a
sin gle angle.
B P IE f f f f T T i Rewrite as a Single Trigonometric Expression
a. Find the exact value of
ta n 3 2 ° + ta n 13°
1 — ta n 32° ta n 1 3 £
b.
tan 32° + tan 13°
1 - t a n 32° tan 13° ‘
= tan (32° + 13°)
Tangent Sum Identity
= tan 45° o r 1
Simplify.
Simplify sin x sin 3x — cos x cos 3x.
sin x sin 3x — cos x c o s 3x = —(cos x cos 3x — sin x sin 3x)
p GuidedPractice
8
3B. S im p lify ,tan 6x ~ tan 7* .
r
1 + tan 6x tan 7x
338
| Lesson 5-4 | Sum and D ifference Identities
cos
+ sin — ■sin
24
8
Properties
Tangent Sum Identity
= —cos (x + 3x) o r —cos 4x
3A. Find the e xact v alu e o f cos ^
Distributive and Commutative
24
Sum and difference identities can be used to rewrite trigonometric expressions as algebraic expressions.
Write as an Algebraic Expression
Write sin (arctan y/3 + arcsin x) as an algebraic expression of x that does not involve
trigonom etric functions.
A p p ly in g the S in e o f a Su m Identity, w e find that
sin (arctan V 3 + arcsin x) = sin (arctan V 3 ) cos (arcsin x) + cos (arctan V 3 ) sin (arcsin x).
If w e let a = arctan V 3 and /3 = arcsin x, th e n tan a = \f?> an d sin (3 = x. S k etch one rig h t
trian g le w ith an acu te an g le a and an o th er w ith an acu te an g le [3. L a b el the sid es su ch that
tan a = \/3 and the sin (3 = x. T h en u se the P y th ag o rean T h eo rem to exp ress the len gth o f each
third side.
U sin g th ese trian g les, w e fin d th at sin (arctan V 3 ) = sin a o r
cos (arctan \/3) = cos a or j ,
c o s (arcsin x) = co s (3 o r V l — x 2, and sin (arcsin x) = sin /3 o r x.
N ow ap p ly su b stitu tio n and sim plify.
sin (arctan \/3 + arcsin x) = sin (arctan V 3 ) cos (arcsin x) + cos (arctan V 3 ) sin (arcsin x)
V3 |
2
( V 1 - x 2 ) + jX
V3 - 3x2
p Guided Practice
Write each trigonom etric expression as an algebraic expression.
4A. cos (arcsin 2x + arccos x)
4B. sin ^arctan x — arccos - jj
S u m and d ifference id en tities can b e u sed to v e rify o th er identities.
ReadingMath
Cofunction Identities The “co”
in cofunction stands for
“complement.” Therefore, sine
and cosine, tangent and
cotangent, and secant and
cosecant are all complementary
functions or cofunctions.
B Q 2 G 2 3 3 S Ver'fy Cofunction Identities
— x j = cos x.
Verify sin
■ ITX — x\\ =
sm (y
=
•
sin
—TTc o s
l( c o s
= cos
X
x) — 0
✓
'Jt
s in ;
x — cos —
(s in
x)
Sine Difference identity
sin
2
= 1 and cos
2
=
Multiply.
Guided Practice
Verify each cofunction identity using a difference identity.
5A. co s
— x j = sin x
5B. csc ^
= sec 0
339
S u m and d ifference id en tities can b e u sed to rew rite trig o n o m etric e xp ressio n s in w h ich o n e o f the
ang les is a m u ltip le o f 90° o r y rad ian s. T h e re su ltin g id e n tity is called a reduction identity b ecau se
it reduces the co m p lexity o f the exp ression .
P E S u H S Q Verify Reduction Identities
Verify each reduction identity,
a. sin ^0 +
= —cos 9
= sin 0 cos ^
sin ^0 +
b.
Sine Sum Formula
+ cos ® sin
= sin 0 (0 ) + cos 0 (—1)
cos ~
= —cos 0 ✓
Simplify.
2
= 0 and sin
2
= -1
tan (x — 180°) = tan x
ta n
tan (x — 180)° :
1
ta n
1
x-
+ ta n
ta n 1 8 0 °
x ta n
x—
Tangent Sum Formula
180°
0
tan 180° = 0
+ ta n x (0 )
Simplify.
= tan x ✓
► GuidedPractice
Verify each cofunction identity.
6B. sin
6A. cos(360°\— 0) = cos 0
I Solve Trigonometric Equations
( f + «) =
COS X
You can solv e trig o n o m etric eq u atio n s u sin g the sum
i and d ifference id en tities and the sam e tech n iq u es th at y o u u sed in L esson 5-3.
j E a J H E H Solve a Trigonometric Equation
+ x j + cos
Find the solutions of cos
— xj = j on the interval [0, 2ir].
C0S(f+x)+C
os(f-*)=I
\
cos y cos x — sin y sin x + cos -j cos x + sin -j sin x = ~
■|(cos x) -
~ ( s i n x) + i ( c o s x) + ^ - ( s i n x) = -|
1
2
cos x =
TechnologyTip
Viewing Window When checking
your answer on a graphing
calculator, remember that one
period for y = sin x or y —cos x
is 2tt and the amplitude is 1. This
will help you define the viewing
window.
O n the in terv al [0, 2 tt], co s x = — w h e n x = — and x —
>
CHECK T h e g rap h o f y = cos
h as zeros at -j and
+ cos [~J ~ x) ~ 2
o n the in terv al [0 , 2 tv] . ✓
► GuidedPractice
7.
340
Find the solu tio n s o f cos (x + tt) — sin (x — 7r) = 0
o n the in terv al [0, 2 tt],
| Lesson 5 -4 | Sum and D ifference Identities
5 tv
Original equation
Cosine Sum Identities
Substitute.
Simplify.
Exercises
Find the exact value of each trigonometric expression.
Find the exact value of each expression. (Example 3)
(Example 1)
11.
1. cos 75°
3. sin
5.
II tt
12
tan
23tt
12
2 . sin ( - 2 1 0 °)
4.
cos
ta n 4 3 ° — ta n 13°
1 + ta n 4 3° ta n 13°
" ^ c o s 4 ? cos v + sin ~
1 7 tt
12
12
13.
4
sin ^
12
4
sin 15° cos 75° + cos 15° sin 75°
6. ta n y j
( l £ ) sin J cos
7. VOLTAGE A n aly sis o f the v o ltag e in a h aird ry er inv o lv es
term s o f the form sin (nzvt — 90°), w h ere n is a p o sitiv e
integer, w is the freq u en cy o f the v o ltag e, and t is tim e.
U se an id en tity to sim p lify th is exp ressio n . (Example 2)
- cos y sin
15. cos 40° cos 20° — sin 40° sin 20°
167)
ta n 4 8 ° + ta n 12°
1 -
ta n 4 8° ta n 12°
Simplify each expression. (Example 3)
8. BROADCASTING W h en the su m o f the am p litu d es o f tw o
17.
w av es is greater th a n th at o f th e co m p o n e n t w av es, the
resu lt is constructive interference. W h e n the co m p o n en t
8.
28 — t a n 8
29 t a n (
ta n
1 + ta n
cos y cos x + sin y sin x
w av es com bin e to h av e a sm aller am p litu d e, destructive
19. sin 3y cos y + cos 3y sin y
interference occurs.
f o ) cos 2x sin x — sin 2x co s x
21 ■ cos x co s 2x + sin x sin 2x
<2 ^
^
58 +
ta n
ta n 5 0 ta n
9
9—1
ta n
(23) SCIENCE A n electric circu it co n tain s a capacitor, an
ind uctor, and a resistor. T h e v o ltag e d rop across the
in d u cto r is g iv en b y VL = IivL cos (zvt + y j , w here I is
C on sid er tw o sig nals m o d eled b y y = 10 sin (It + 30°)
the p e a k cu rren t, w is the frequ ency, L is the ind u ctance,
and y = 10 sin (2 1 + 210°). (Example 2)
and t is tim e. U se the cosin e su m id en tity to exp ress VL
as a fu n ctio n o f sin wt. (Example 3)
a. Find the su m o f the tw o fu n ctions.
b. W h at typ e o f in terferen ce results w h e n the sig nals
m o d eled b y the tw o eq u atio n s are com b in ed ?
9. WEATHER T he m o n th ly h ig h tem p eratu res fo r M in n eap o lis
can be m od eled b y / (x ) = 31.65 sin
— 2.09 j + 52.35,
w here x rep resen ts th e m o n th s in w h ich Ja n u a ry = 1,
Febru ary = 2, and so on. T h e m o n th ly low tem p eratu res
for M in n eap o lis can b e m od eled b y
g(x) = 31.65 sin
Write each trigonom etric expression as an algebraic
expression. (Example 4)
24. sin (arcsin x + arccos x)
25. cos (sin 1 x + cos 1 2x)
26. cos |sin_1 x — ta n - 1 -^y-J
27. sin |sin_1 -^y- — ta n " 1
xj
— 2.09 j + 32.95. (Example 2)
a. W rite a n ew fu n ctio n h(x ) b y ad d in g the tw o fu n ction s
and d iv id in g th e re su lt b y 2 .
b. W h at d oes the fu n ctio n y o u w ro te in p a rt a rep resen t?
28. co s (arctan V 3 — arcco s x)
29. tan (cos 1 x + tan 1 x)
30. tan |sin-1
— c o s -1 xj
31. tan |sin-1 x + y j
10. TECHNOLOGY A blin d m ob ility aid u ses the sam e id ea as
a b a t's son ar to en ab le p eo p le w h o are v isu ally im p aired
to detect o bjects arou nd them . T h e sou n d w av e em itted
by the d ev ice for a certain p atie n t can b e m o d eled b y
b = 30 (sin 195°)f, w here t is tim e in seco n d s and b is air
pressu re in pascals. (Example 2)
a. R ew rite the form ula in term s o f the d ifferen ce o f tw o
Verify each cofunction identity using one or more difference
identities. (Example 5)
32.
ta n
(f - * )
=
cot X
33.
sec
(f - * )
=
CSC
34.
cot |
angle m easures.
b. W h at is the p ressu re after 1 seco nd ?
(? - * )
= ta n
X
x
§
connectED.m cgraw-hill.com |
341
Verify each reduction identity. (Example 6)
35.
cos
56.
— 9) = —cos 9
( tt
|§&. cos (2tt + 9) = cos 9
37. sin
( tt
ANGLE OF INCLINATION T h e angle o f inclination 9 o f a lin e is
the an g le fo rm ed b e tw e e n the p o sitiv e x -ax is and the lin e,
w h ere 0° < 9 < 180°.
a.
P rove th at the slop e m
y
o f lin e £ sh o w n at the
rig h t is g iv en b y
m = tan 9.
— 9) = sin 9
38. sin (90° + 9) = cos 9
/\
0
39. cos (270° — 9) = —sin 9
b. C o n sid er lin es
(Example 7)
41.
COS (TT
(ffij cos
43.
+ x) +
X
and i 2 b elo w w ith slo p es
and m 2,
respectively. D eriv e a form ula for the an g le 7 form ed
b y the tw o lines.
+ x\ = 0
+ x j - sin
e
/
/
Find the solution to each expression on the interval [0, 2 tt],
^ST cos
t/
(TT + x) = 1
COS
+ x j + sin
+ xj = 0
s in (f+ x ) + s in ( f - x ) = i
4?, sin
45.
+ x j + sin
tan
( tt
+ x) + tan
+ xj = -2
( tt
+ x) = 2
c. U se the fo rm u la yo u fo u n d in p art b to find the angle
fo rm ed b y y = ^ - x and y = x.
Verify each identity.
(x —y)
cos xcos y
c o s (a + /3 )
s in
46. tan x — tan y
47. cot a — tan (3
(ta n
48.
(ta n
uu+
ta n
ta n
v)
v)
s in
a
s in
s in (u- v)
(u + v)
cos
H.O.T. Problems
f3
PROOF Verify each identity.
__
,
,
ta n
(3)=
1 —t a n a
tan ( a —
f3)
57.
GRAPHING CALCULATOR Graph each function and make a
conjecture based on the graph. Verify your conjecture
algebraically.
58.
59. sin (a +
50. y
60. sin ( a — (3)=
--[sin (x
+
2
tt)
+
sin (x — 2 t t ) ]
51. y = c o s2 (x + f ) + c o s2 (x - f ' j
a + ta n B
tan (a +
49. 2 sin a cos b = sin (a + b) + sin (a — b)
=
Use Higher-Order Thinking Skills
f3)=
ta n
ta n
(3
a —t a n f3
a ta n
1 + ta n
/3
sin a cos /3 + cos a sin /3
sin a cos
[3 — cos a
sin
(3
61. REASONING U se the su m id en tity fo r sin e to d eriv e an
id e n tity fo r sin (x + y + z) in term s o f sin es and cosines.
PROOF Consider A X Y Z . Prove each identity.
(Hint: x + y + z = tt)
CHALLENGE If sin x =
and cos y =
find each of the
following if x is in Quadrant IV and y is in Quadrant I.
62. cos (x + y)
(63) sin (x — y)
64.
tan (x + y)
y
Y
65. REASONING C o n sid er sin 3x cos 2x = cos 3x sin 2x.
a. Find the solu tio n s o f the eq u atio n ov er [0, 2tt]
52. cos(x + y) — —cos z
algebraically.
53. sin z = sin x cos y + cos x sin y
b. S u p p ort y o u r an sw er graphically.
54. tan x + tan y + tan z = tan x tan y tan z
55. CALCULUS The difference quotient is given by
f{x + h) -f(x)
h
a. L et/ (x) = sin x. W rite and exp an d an exp ressio n for
PROOF Prove each difference quotient identity.
66.
the d ifference quotient.
b. Set you r answ er from p art a equ al to y. U se a grap h in g
calcu lato r to grap h the fu n ctio n for the fo llow in g
v alu es o f h : 2 , 1, 0 .1, and 0 .0 1 .
c. W h at fu n ctio n d o es the grap h in p a rt b resem b le as h
app roach es zero?
342
Lesson 5-4
Sum an d D ifference Identities
__
67.
68.
{x + h) —s i n x
=
h
c o s (x + h) — c o s x
s in
;
.
I c o s h —1 \
s in h
+ c o s x ——
\ h
I
h
c o s h—
1
s in h
sin x -—
= cos x
sm x
WRITING IN MATH C an a tan g en t su m o r d ifferen ce id en tity
b e u sed to solv e an y tan g en t red u ctio n fo rm u la? E xp lain
y o u r reaso n in g.
Spiral Review
69. PHYSICS A cco rd in g to S n e ll's law , th e an g le at w h ich lig h t en ters w ater a is related to the
angle at w h ich lig h t trav els in w ater (3 b y sin a = 1.33 sin (3. A t w h at an g le d o es a b eam of
lig ht en ter the w ater if the beam trav els at an an g le o f 23° th ro u g h the w ater? (Lesson 5-3)
Verify each identity. (Lesson 5-2)
70.
cos 9
■= sec i
1 - sin2
71.
- = co t I
Find the exact value of each expression, if it exists. (Lesson 4-6)
72.
73. ta n -1 V 3
s in -1 (—1)
74.
tan (arcsin |-j
75. MONEY S u p p ose yo u d ep o sit a p rin cip al am o u n t o f P d ollars in a b a n k accou n t th at pays
com poun d interest. If the an n u al in terest rate is
r (exp ressed
as a d ecim al) and the b an k
m akes in terest p ay m en ts n tim es e v ery year, the a m o u n t o f m o n ey A y o u w o u ld h av e after
t years is giv en b y A(t) = P ( l + ■£)” . (Lesson 3-1)
a. If the p rin cip al, in terest rate, and n u m b er o f in terest p ay m en ts are k n o w n , w h a t typ e o f
fu n ctio n is A(t) = P ( l + £ )" * ? E xp lain y o u r reasoning.
b. W rite an eq u atio n giv in g the am o u n t o f m o n ey yo u w ou ld h av e after t y ears if you
d ep o sit $1000 into an accou n t p ay in g 4% an n u al in terest com p o u n d ed q u arterly (four
tim es p er year).
C. Find the acco u n t b alan ce after 20 years.
List all possible rational zeros of each function. Then determ ine which, if any, are zeros. (Lesson 2-4)
76. p(x) = x4 + x3 — l l x — 5x + 30
77. d(x) = 2x4 — x 3 — 6x2 + 5x — 1
78.
f(x ) = x3 — 2x2 — 5x — 6
Skills Review for Standardized Tests
79. SAT/ACT T h ere are 16 green m arbles, 2 red m arbles,
81 . Find the e xact v alu e o f sin 6.
and 6 y ellow m arb les in a jar. H o w m an y y ellow
m arb les n eed to b e ad d ed to the ja r in ord er to d ou ble
the p ro b ab ility o f selectin g a y ello w m arble?
A 4
C 8
B 6
D 12
E 16
A
B
80. REVIEW R efer to the figure below . W h ich eq u ation
cou ld b e u sed to find m ZG ?
C
H
D
82.
V2 + V6
4
V 2 -V 6
4
2 + V3
4
2 - V3
REVIEW W h ich o f the fo llo w in g is eq u iv alen t to
cos 0 (cot2 6 + 1) ?
csc 9
F sin G = ■
G cos G = —
4
H co t G = -7
4
I tan G = ■
F,
tan 9
G co t 0
H sec 8
J
csc 8
E
c o n n ec tiD ~m cg ra w -h ilL co n ^
343
Graphing Technology Lab
oooo
oooo
oooo
CDOO
Reduction Identities
ftJLH ■ M
r
Another reduction identity involves the sum or difference of the measures of an angle and a quadrantal
•
Use Tl-Nspire technology
and quadrantal angles to
reduce identities.
angle. This can be illustrated by comparing graphs of functions on the unit circle with Tl-Nspire technology.
Activity 1
Use the Unit Circle
Use the unit circle to explore a reduction identity graphically.
R T f l l l A dd a Graphs page. S elect Zoom-Trig from the W indow m en u , and select Show Grid
from the View m enu. From the File m en u u n d er Tools, ch o o se Docum ent Settings, set the
Display Digits to Float 2 , and con firm th at the an g le m easu re is in rad ians.
PT7TW S elect Points & Lines an d th en Point fro m the m en u . P lace a p o in t at (1 ,0 ). N ext, select
Shapes, and th en Circle from the m en u. To co n stru ct a circle cen tered at the origin
th ro u gh (1 ,0 ) , click o n the screen and d efin e th e cen ter p o in t at the origin. M o v e the
cu rso r aw ay from the center, and the circle w ill appear. S top w h en yo u get to a rad iu s
o f 1 and (1, 0 ) lies on the circle.
E T T lT lfl
P lace a p o in t to the rig h t o f the circle on the
y
d s
x-axis, and lab el it A. C h o o se Actions and then
Coordinates and Equations fro m the m en u ,
and th e n d o u b le-click th e p o in t to d isp lay its
coord inates. From the C onstruction m en u ,
f
‘
'•"■ a
2*71*23*9 r a d
AO B
A -n
0.858407
-6
*
7B .
(4 ,0 )
. . .
t
v.
\A
>
6.2 f
A
B
ch o o se Measurement transfer. S e le ct the
x-co o rd in ate o f A , the circle, and the p o in t at
(1 ,0 ). L abel the p o in t created on the circle as B,
ESXSSSSH I
(-0.653644,-0.7 56802)
.
4
a
and d isp lay its coord inates.
W ith O as the o rig in , calcu late and lab el the m easu re o f ZAO B. S e le ct Text from the
Actions m en u to w rite the exp ressio n a — 7V. T h en sele ct Calculate from the Actions
m en u to calcu late the d ifferen ce o f the x -co o rd in ate o f A and tv.
StudyTip
M ov e A alo ng the x-axis, and o bserve the
Angle Measure The Tl-Nspire
only measures angles between 0
and tt.
effect o n the m easu re o f /.A O B.
m m
F ro m the C onstruction m en u , ch o o se
Location of A
mZAOB (radians)
(4,0)
2.7124
(3,0)
3.0708
(2,0)
2.5708
(5,0)
2.2123
(-2 ,0 )
0.5708
1.16
RAD A U T O REAL
Measurement transfer. S elect the x-ax is and
the v alu e o f a — tv. L abel th e p o in t as C and
d isp lay its coo rd in ates. U sin g Measurement
transfer again , select the x-co o rd in ate o f C,
, ‘4: n .
,
0.858407
the circle, and the p o in t at (1, 0). L abel the
p o in t as D and d isp lay its coord inates.
( "p. 653644,-p. 456802 j
^ tO B
2'.71'239rarf
* [ 0.653644,0,756802)
, D
(
~y { 0.85^407.0,)
(4>o) 6,2^
,
V___________________________________________________________________________________
Analyze the Results
1A. In Step 5, h o w are the lo catio n o f A and th e m easu re o f /.A O B related ?
IB . C o n sid er the lo catio n s o f p o in ts B and D. W h at red u ctio n id en tity or id en tities d o es this
relation ship su g g est are true?
IC . MAKE A CONJECTURE If yo u ch an g e the exp ressio n a — tt to a + 7V, w h at red u ctio n id en tities do
you think w o u ld result?
344
| Lesson 5-4
Activity 2
Use Graphs
Use graphs to identify equal trigonom etric functions.
O p e n a n ew Graphs p ag e. S e le ct Zoom-Trig fro m the W in d ow m en u.
EflSffEl G rap h / (x) = cos x ,f(x ) = cos (x — tt) , and f(x ) = sin x. U sin g the A ttributes featu re from
the Actions m en u , m ak e the lin e w e ig h t o f f(x ) = cos (x — tt) m e d iu m and the lin e style
o f/ (x) = sin x dotted.
4'
*
y=sin(x)
--------------
«(r)-cos(x
f6U’=cosU_nJ
I
B S E
U se tran slation s, reflectio n s, o r d ilatio n s to tran sfo rm / (x ) = sin x so th at the graph
coin cid es w ith / (x) = cos x. S elect the g rap h and d rag it o v er/ (x ) = cos x. A s you m ov e
the g rap h , its fu n ctio n w ill ch an g e o n the screen.
E S B
U se tran slatio n s, reflectio n s, o r d ilatio n s to tran sfo rm / (x ) = cos (x — tt) s o th at the grap h
co in cid es w ith the o th er tw o grap h s. A g ain , as you m o v e th e g rap h , its fu n ctio n w ill
ch an g e o n th e screen.
I ________________________________________________________________________
Analyze the Results
2A. W rite the id e n tity th at resu lts fro m y o u r tran sfo rm atio n o f/ (x) = sin x in Step 3. G rap h the
fu n ctio n s to con firm y o u r identity.
2B. W rite the id en tity th at resu lts fro m y o u r alteratio n o f/ (x) = co s (x -
tt )
in Step 3. G rap h the
fu n ctio n s to con firm y o u r identity.
2C. MAKE A CONJECTURE W h a t d oes the reflectio n o f a g rap h su g g est fo r th e p u rp o se o f d ev elop in g
an id en tity ? a tran slatio n ?
Exercises
Use the unit circle to write an identity relating the given expressions. Verify your identity by
graphing.
1. cos (90° — x), sin x
2. cos
— x j, sin x
Insert the trigonometric function that completes each identity.
3.
cos x = __________ (x -
4. co t x = ------------------- (x + 90°)
5.
sec x = __________ (x — 180°)
6. csc x = ------------------- |x + y j
Use transform ations to find the value of a for each expression.
7. sin ax = 2 sin x cos x
9.
a s in 2 x = 1 — cos 2x
cos 4 ax = c o s 2 x — s in 2 x
8.
10.1 + cos 6ax = 2 c o s 2 x
connectED.m cgraw-hill.com |
345
Multiple-Angle and
Product-to-Sum Identities
: Then
• You proved and used
sum and difference
identities.
(Lesson 5-4)
I
Use double-angle, power-reducing,
and half-angle identities to evaluate
trigonometric expressions and
solve trigonometric equations.
The speed at which a plane travels can be
described by a mach num ber, a ratio of
the plane’s speed to the speed of sound.
Exceeding the speed of sound produces a shock wave in the
shape of a cone behind the plane. The angle 9 at the vertex
of this cone is related to the mach number M describing the
plane’s speed by the half-angle equation sin
2
Use product-to-sum identities to
evaluate trigonometric expressions
and solve trigonometric equations.
I
Use Multiple-Angle Identities
B y letting a and (3 bo th equal 9 in each o f the angle sum
identities you learned in the previous lesson, you can d erive the follow ing double-angle identities.
KeyConcept Double-Angle Identities
sin 29 = 2 sin 6 cos 9
cos 29 = cos2 6 - sin2 9
tan 2 0 =
cos 2 0 = 2 cos2 ® “ 1
1 - tan2 0
cos 26 =
1-
2 sin2 0
P roof Double-Angle Identity for Sine
sin 20 = sin(0 + 9 )
s»
29 = 9 + 0
= sin 0 cos 0 + cos 0 sin 0
Sine Sum Identity where a = 0 = 0
= 2 sin 0 cos 0
Simplify.
J
You will prove the double-angle identities for cosine and tangent in Exercises 6 3 -6 5 .
H O S S H 3 0 I Evaluate Expressions Involving Double Angles
If sin 9 =
on the interval |tv, ^y-j, find sin 29, cos 29, and tan 20.
S in ce sin 9 = ^
o n the in terv al
y ^ j, on e p o in t o n the
term in al sid e o f 9 h as y -co o rd in ate —7 and a d istan ce o f 25 u n its
from the o rigin , as show n. T h e x -co o rd in ate o f this p o in t is
th erefore —V 2 5 2 — 7 2 or —24. U sin g this p o in t, w e find that
co s 0 = — or —~
25
r
and
tan 9 = — or
x
25
U se th ese v alu es and the d o u b le-an g le id e n tities for sin e and co sin e to fin d sin 20 and cos 29.
T h en find tan 29 u sin g eith er the tan g en t d o u b le-an g le id en tity o r the d efin itio n o f tangent.
sin 29 = 2 sin 9 cos 9
Method 1
tan 29 =
cos 2 9 = 2 c o s 2 0 — 1
2 ta n ^
1 - tan2 0
Method 2 tan 2 0 = - ^ ^
cos 20
2(1
\2 /
1
► GuidedPractice
1.
If cos 0 =
o r 336
I 7 \2
527
U !4)
on the in terv al |o, y j , find sin 20, cos 29, and tan 29.
V..................................................................................................... .................
346
| Lesson 5-5
_
336
625 o r ^ 6
527
527
625
StudyTip
More than One Identity Notice
that there are three identities
associated with cos 29. While
there are other identities that
could also be associated with
sin 28 and tan 29, those
associated with cos 29 are worth
memorizing because they are
more commonly used.
H I B 3 ! E 3 Solve an Equation Using a Double-Angle Identity
Solve sin 2 0 — sin 0 — 0 on the interval [0, 2tt].
U se the sin e d o u b le-an g le id en tity to rew rite the eq u atio n as a fu n ctio n o f a sin gle angle.
Original equation
sin 29 — sin d = 0
Sine Double-Angle identity
2 sin 6 cos 0 — sin 9 = 0
Factor.
sin 9 (2 cos 9 — 1) = 0
sin 0 = 0
or
Zero Product Property
2 cos 0 — 1 = 0
cos 9 =
I = 0 o r tt
1
T h erefore, 0 = -?• o r -
T h e solu tio n s o n th e in terv al [0, 2tt] are 0 = 0, y ,
^ Gllided Practice
tt ,
3 '
o r -y-.
Solve each equation on the interval [0, 2ir].
2A. cos 2a = —s in 2 a
2B. ta n 2/3 = 2 ta n /3
T h e d o u ble an g le id en tities can b e u sed to d eriv e the p o w er-red u cin g id en tities below . T h ese
id en tities m ak e calcu lu s-related m an ip u latio n s o f fu n ctio n s lik e y = c o s 2 x m u ch easier.
K eyC on cept Power-Reducing Identities
ta n 2 0 =
COS 2 0 = 1 + C ° S 2 *
s in 2 e = 1 - C20S 2 e
j ~ cos ^
1 + COS 20
Proof Power-Reducing Identity for Sine
1 - cos 2 0
1 - (1 - 2 s in 2 6)
2
2
2 s in 2
2
= S in 2
Cosine Double-Angle Identity
0
Subtract.
Simplify.
9
„
_ _ _ _ _ _
-/
_
You will prove the power-reducing identities for cosine and tangent in Exercises 82 and 83.
Use an Identity to Reduce a Power
Rewrite sin4 x in terms with no power greater than 1.
Frangois Viete
(1540- 1603)
Born in a village in western
France, Viete was called to Paris
to decipher messages for King
Henri III. Extremely skilled at
manipulating equations, he used
double-angle identities for sine
and cosine to derive triple-,
quadruple-, and quintuple-angle
identities.
s in 4 x = (sin 2 x )2
_
—
(sin2
— cos2xj
Sine
1 — 2 cos 2x + cos2 2x
4
1 — 2 cos 2x +
x )2 = sin4 x
Power-Reducing Identity
... ,
Multiply.
1 + cos 4x
2
4
2 — 4 cos 2x + 1 + cos 4x
= -i (3 — 4 cos 2x + cos 4x)
Cosine Power-Reducing Identity
Common denominator
Factor.
Guided Practice
Rewrite each expression in terms with no power greater than 1.
3A. c o s 4 x
3B. s in 3 0
^j^connectE^mcgra^^^^^m1
347
B Q S 3 3 3 3 & Solve an Equation Using a Power-Reducing Identity
Solve cos2 x — cos 2 x = 4 .
2
Solve Algebraically
cos2 X ■
1 + cos 2 x
2
cos 2x =
Original equation
2
_ 1
Cosine Power-Reducing identity
— cos
— 2—x = 2
Multiply each side by 2.
1 + cos 2x — 2 cos 2 x = 1
Subtract 1 from each side.
cos 2x — 2 cos 2x = 0
Subtract like terms.
—cos 2x = 0
Multiply each side by —1.
cos 2x = 0
TV
2 x = fo r f
4
4
Solutions for double angle in [0,2it]
Divide each solution by 2.
T h e grap h o f y = cos 2x h as a p erio d o f t t , s o the so lu tio n s are x = ^
37V
Yl'K O f — — b Yl'K, n 6 .
Support Graphically
T h e g rap h o f y = c o s 2 x — cos 2x — j h as zero s at ^ and
o n the in terv al [0, 7r].
Ztro
X=.7B£3SBifi Y = - i . i E - i 3
[0,
p-
GuidedPractice
tv]
s c l: ~ b y [ - 1 . 5 , 1 . 5 ] scl: 1
Solve each equation.
4A. cos4 a — sin 4 a = -j
4B. s in 2 3/3 = s in 2 /3
By replacing 6 w ith — in e ach o f the p o w er-red u cin g id en tities, y o u can d eriv e each o f the
fo llo w in g h alf-an g le identities. T h e sig n o f e ach id e n tity th at in v o lv es the ± sy m b ol is d eterm in ed
by ch eck in g the q u ad ran t in w h ich the term in al sid e o f — lies.
WatchOut!
Determining Signs To determine
which sign is appropriate when
using a half-angle identity, check
KeyConcept Half-Angle Identities
>
- cos 6
+ cos 0
tan
sinf = ± ^ f * *
-fL
the quadrant in which - | lies, not
the quadrant in which 0 lies.
ta n ! =
1 — cos 9
sine
tan 1 =
2
sine
1 + cose
Proof Half-Angle Identity for Cosine
/1 + cos 0
2 "
I
11 + cos|
_,_\l
H
± V
2
+ cos 2x
)
Rewrite 0 as 2 •
Substitute x -
6
= ±Vcos^ x
Cosine Power-Reducing Identity
= cos X
Simplify.
= cos|
Substitute.
You will prove the half-angle identities for sine and tangent in Exercises 66-68.
348
| Lesson 5-5 | M ultiple-Angle and Product-to-Sum Identities
Evaluate an Expression Involving a Half Angle
Find the exact value of cos 112.5°.
N o tice th at 112.5° is h a lf o f 225°. T h erefore, ap p ly the h alf-an g le id e n tity fo r cosine, n o tin g
th at sin ce 112.5° lies in Q u ad ran t II, its cosin e is n eg ativ e.
225°
112.5° =
cos 112.5° = c o s ;
+ cos 225°
-
{
225°
Cosine Half-Angle Identity (Quadrant II angle)
k
V2
cos 225° =
2-V2
-{
Subtract and then divide.
V 2 -V 2
7=—
V4
V 2 -V 2
o r
Quotient Property of Square Roots
5----------
2
CHECK U se a calcu lato r to su p p o rt y o u r assertio n th at co s 112.5° = — ^
cos 112.5° « -0 .3 8 2 6 8 3 4 3 2 4
and
V 2 - V2
-0 .3 8 2 6 8 3 4 3 2 4 ✓
p GuidedPractice
Find the exact value of each expression.
StudyTip
Tangent Half-Angle Identities
When evaluating the tangent
function for half-angle values, it is
usually easiest to use the form of
the tangent half-angle identity
tan » =
sinceits
2
sin 9
denominator has only one term.
5B. ta n ^ |
5A. sin 75°
R ecall th at yo u can use su m and d ifference id en tities to so lv e equ ation s. H alf-an g le id en tities can
also b e u sed to solv e equ ations.
m
m
m
Solve an Equation Using a Half-Angle Identity
Solve sin2 x = 2 cos2 j on the interval [0, 2 ir].
Original equation
s in 2 x = 2 cos
+ COS x
s in 2 *
• 2
"|/1 + COS
sin x = 2 ^
j
Cosine Half-Angle Identity
Simplify.
Multiply.
sin2 x = 1 + cos x
Pythagorean Identity
1 — cos2 x — 1 + cos x
Subtract 1 from each side.
—cos2 x — cos x = 0
Factor.
cos z (—cos x — 1) = 0
cos x = 0
or
Zero Product Property
—cos x — 1 = 0
* = T2 o r ^2
cos x = —1; therefore, x = tt.
T h e solu tio n s o n the in terv al [0, 2tv] are x = y , -y -, o r
Solutions in [0 ,2 ir]
tt.
p GuidedPractice
Solve each equation on the interval [0, 2 t v ] .
6A. 2 s in 2 ^ + cos x = 1 + sin x
6B. 8 tan -J + 8 cos x tan £
connectED.m cgraw-hill.com |
349
2
Use Product-to-Sum Identities
Product-to-Sum Identities
K eyC on cep t
sin
asm [3 = l[c o s (a -
cos
a cos (3 =
f3) - cos (a + /3)]
(a -
-|-[cos
sin a cos f3 = -|{sin ( a + /3) + sin ( a - /?)]
p) + cos (a + f3)]
cos a sin /3 = l[s in (a + /3) - sin ( a - /3)]
P ro o f Product-to-Sum Identity for sin
StudyTip
Proofs Remember to work the
more complicated side first when
proving these identities.
To w o rk w ith fu n ctio n s su ch as y = cos 5x sin 3x in
calcu lu s, y o u w ill n eed to ap p ly one o f th e fo llo w in g p ro d u ct-to -su m identities.
a
cos
f3
|[s in ( a + P ) + sin ( a - f3)]
= l(s in
a cos (3 +
cos
More complicated side of identity
a sin (3 +
sin
a
cos /3 - cos a sin /3)
Sum and Difference Identities
= -1(2 sin a cos (3)
= sin
a
Combine like terms.
cos /3
Multiply.
J
You will prove the remaining three product-to-sum identities in Exercises 8 4 -8 6 .
■ - i S
B
Use an Identity to Write a Product as a Sum or Difference
Rewrite cos 5 x sin 3 x as a sum or difference.
cos 5x sin 3x = -j[s in (5x + 3x) — sin (5x — 3x)]
y
= —(sin 8 x — sin 2 x )
Simplify.
1
1
= 2 sin 8 x - - sin 2x
Distributive Property
GuidedPractice
Rewrite each product as a sum or difference.
7B. sin 7 x sin 6 x
7A. sin 4 0 cos 6
V
T h ese p ro d u ct-to -su m id en tities h av e corresp on d in g su m -to -p rod u ct id entities.
K ey C o n cep t
Sum-to-Product Identities
sin a + sin j3 = 2 sin
cos | —2
sin a — sin f3 = 2 cos
sin (
'
cos a + cos /3 = 2 cos | t t ^ ) cos | a g
~)
cos a — cos 0 = - 2 sin j —
P ro o f
Sum-to-Product Identity for sin
2 sm(
2 1 cos(
2
I
+
sin
(3
......
a
+ [3
,
Substitute x = — —
2 and jy
= 2 s in x c o s y
= 2 | | [ s i n (x + y) + sin (x - y )]j
. ( a -f /3
= sm ( 2 +
a
a —(3\
=
a —f3
— —
2
Product-to-Sum Identity
. / a + /3
2 ) + s in (
si n | a ~ ^ |
2
a —f3\
2 J
= sin( ¥ ) + sin( f )
= sin a + sin (3
Substitute and simplify.
Combine fractions.
Simplify.
J
You will prove the remaining three sum -to-product identities in Exercises 8 7 -8 9 .
350
| Lesson 5-5
Multiple-Angle and Product-to-Sum Identities
B E 2 E 0 3 0 Use a Product-to-Sum or Sum-to-Product Identity
Find the exact value of sin
(
12
5tt
12 2
+ sin
12
I 5tt
IT
12 I COS
7T '
12
\
12
2
Sum-to-Product Identity
,
= 2 sin -?■ cos
Simplify.
-(#)(#)
sinf = T - andcosf = T -
V6
2
Simplify.
►GuidedPractice
Find the exact value of each expression.
8A. 3 cos 37.5° cos 187.5°
8B. cos
- cos Y J
You can also use su m -to -p ro d u ct id e n tities to solv e so m e trig o n o m etric equ atio ns.
PJ2EIEE13E3 Solve an Equation Using a Sum-to-Product Identity
Solve cos 4x + cos 2x = 0.
Solve Algebraically
Original equation
cos 4x 4- cos 2x = 0
2 cos ( 4 ^ ± 2 £ ) c o s ( i l ^ ) = 0
Cosine Sum -to-Product Identity
Simplify.
(2 cos 3x)(co s x) = 0
S et e ach facto r equ al to zero and fin d solu tio n s on the in te rv al [ 0 , 2ir].
First factor set equal to 0
2 cos 3x = 0
Divide each side by 2.
cos 3x = 0
3x = f o r f
WatchOut
Periods for M ultiple Angle
Trigonometric Functions Recall
from Lesson 4 -4 that the periods
of y = sin to and y =
cos for are ~
k
* =? “ ?
>
Second factor set equal to 0
cos x = 0
2
2
Solutions in [ 0 ,2 it]
Multiple angle solutions in [0 ,2 it]
Divide each solution by 3.
T h e p erio d o f y = cos 3x is — , so the so lu tio n s are
— + — n,
2
not 27^.
3
■+ 2 t m , or
'
37T
+ 2 tm, n €
Support Graphically
/TT- 'TT C/TT*
T h e grap h o f y = cos 4x + cos 2x h as zeros at —,
6
2
and
6
on the in terv al [0 , 27t]. ✓
Z tro
X =.E 23E 9B 7B
p GuidedPractice
V=0
0, 2tt] s c l : b y [-3 , 3] scl: 1
Solve each equation.
9A. sin x + sin 5x = 0
9B. cos 3 x — cos 5x = 0
connected
351
Exercises
= Step-by-Step Solutions begin on page R29.
Find the values of sin 20, cos 26, and tan 26 for the given
value and interval. (Example 1)
1. cos 6 = f , (270°, 360°)
2. tan 6 =
26. s in 2 9 — 1 = c o s 2 6
15
(90°, 180°)
4. sin 6 =
24. 1 — s in 2 6 — cos 29 = j
25. c o s2 6 — j cos 26 = 0
(180°, 270°)
3. cos 6 =
Solve each equation. (Example 4)
27. c o s2 6 — sin 9 = 1
y - , 2tt)
28.
5. t a n 0 = - | , ( ^ 2 i r )
MACH NUMBER T h e an g le 9 at the v ertex o f the
con e-sh ap ed sh o ck w av e p ro d u ced b y a p lan e break in g
the sou n d b arrie r is related to the m ach n u m b er M
d escrib in g th e p la n e 's sp eed b y the h alf-an g le eq u atio n
6. ta n 0 = V 3 , ( o , f )
f)
1
sin Tr = ~h- (Example 5)
2
M
7. s i n 0 . § , ( f , * )
8. cos 0 = - A ( * , f )
.a a il
La
Solve each equation on the interval [0, 2 tt]. (Example 2)
9. sin 29 = cos 6
a. E xp ress the m ach n u m b er o f the p lan e in term s
10. cos 26 = cos 6
o f cosine.
b. U se the exp ressio n fo u n d in p a rt a to find the m ach
11. cos 29 — sin 9 = 0
n u m b er o f a p lan e if cos 6 =
12.' t a n 29 — t a n 2 6 ta n 2 6 = 2
13. sin 29 csc 6 = 1
Find the exact value of each expression.
14. ^fcos 2 0 + 4 cos 6 = —3
( 15) GOLF A g olf b all is h it w ith an in itial v elo city o f 88 feet
p er second . T h e d istan ce the b all trav els is foun d by
v02 sin 20
d=
—
> w here vQis the in itial velocity, 6 is the angle
th at the p ath o f the b a ll m akes w ith the gro u n d , and 32 is
in feet p er second squared . (Example 2)
29.
sin 67.5°
30.
31.
tan 157.5°
32. sin
cos^j
H tt
12
Solve each equation on the interval [0, 2ir]. (Example 6)
33. sin — + cos 0 = 1
34. tan — = sin —
35. 2 sin
oc
■ 2 0— — cos 0— =
36. 11 — sin
= sin 0
Rewrite each product as a sum or difference. (Example 7)
a. If the b all travels 242 feet, w h at is 6 to the nearest
degree?
b. U se a d o u b le-an gle id en tity to rew rite the eq u atio n
for d.
Rewrite each expression in terms with no power greater
than 1. (Example 3)
16. c o s3 6
17. ta n 3 9
18. s ec4 9
19. c o t3 6
20. cos
•sin
22. s in 2 6 — c o s2 9
352 | Lesson 5-5
21. s in 2 6 c o s 3
23.
sin4 0
cos2
0
37.
cos 3 0 cos 0
38.cos 12.r sin 5x
39.
sin 3x cos 2x
40.sin 8 0 sin 0
Find the exact value of each expression. (Example 8)
41. 2 sin 135° sin 75°
42. cos
43. | sin 172.5° sin 127.5°
44. sin 142.5° cos 352.5°
45. sin 75° + sin 195°
46. 2 cos 105° + 2 cos 195°
at
o
■
17tC
r\
•
TT
47. 3 sm - j y - 3 sin —
48. cos ^
+ cos
+ cos
Solve each equation. (Example 9)
49. cos 0 — cos 3 0 = 0
50. 2 cos 4 0 + 2 cos 2 0 = 0
5 1 . sin 3 0 + sin 5 0 = 0
52. sin 2 0 — sin 0 = 0
53. 3 cos 60 — 3 cos 4 0 = 0
54. 4 sin 0 + 4 sin 30 = 0
M u ltip le -A n gle and Product-to-Sum Identities
Rewrite each expression in terms of cosines of multiple
angles with no power greater than 1.
Simplify each expression.
55 . J 1 + c° s6*
56.
71.
s in 6 9
72. s in 8 9
73.
cos7 9
74. sin 4 9 c o s4 9
Write each expression as a sum or difference.
57. cos (a + b) cos (a —b)
59. sin (b -I- 9) cos ( b + tt)
58. sin (0 — 7r) sin (9 + 7t)
75.
MULTIPLE REPRESENTATIONS In th is p ro blem , you w ill
in v estig ate h o w g rap h s o f fu n ctio n s can b e u sed to
60. cos (a — b) sin (b — a)
find identities.
a. GRAPHICAL U se a g rap h in g calcu lato r to grap h
61. MAPS A M ercato r p ro jectio n is a flat p ro jectio n o f the
f( x ) = 4|sin 9 cos ~ — co s 9 sin
globe in w h ich the d istan ce b etw een the lin es o f latitu d e
on the interval
[—27V, 2tt].
increases w ith th eir d istan ce from the equator.
b. ANALYTICAL W rite a sin e fu n ctio n h(x) th at m od els the
g rap h o t f( x ) . T h e n v e rify th at/ (x) = h(x) algebraically.
C. GRAPHICAL U se a g rap h in g calcu lato r to graph
g(x) = c o s 2 I# — y j — s in 2 ( o — y j on the interval
[—2tv, 2tt],
d. ANALYTICAL W rite a co sin e fu n ctio n k(x) th at m od els
the g rap h o f g(x). T h e n v e rify th at g(x) = k(x)
algebraically.
H.O.T. Problem s
The calcu latio n o f a p o in t o n a M ercato r p ro jectio n
Use Higher-Order Thinking Skills
76. CHALLENGE V erify the fo llo w in g identity,
con tain s the exp ressio n tan ^45° + j^J, w here £ is the
latitu d e o f the point.
sin 29 co s 9 — co s 29 sin 9 = sin 9
a. W rite the exp ressio n in term s o f sin i and cos 1.
REASONING Consider an angle in the unit circle. Determine
w hat quadrant a double angle and half angle would lie in
if the terminal side of the angle is in each quadrant.
b. Find the v alu e o f this exp ressio n if i = 60°.
62.
BEAN BAG TOSS Iv an con stru cted a b e a n b a g to ssin g gam e
as sh ow n in the figu re below .
77. I
78. II
79. Ill
CHALLENGE Verify each identity.
80. sin 4 0 = 4 sin 9 co s 0 — 8 s in 3 0 cos 9
(81^) cos 4 0 = 1 — 8 s in 2 0 c o s 2 0
PROOF Prove each identity.
a. E xactly h o w far w ill th e b a ck ed g e o f th e b o ard b e
b.
from the ground ?
83. tan2 0 = j ~ cos^
1 -I- cos 29
E xactly h o w lo n g is the entire setu p?
PROOF Prove each identity.
63. cos 29 = c o s2 9 — s in 2 9
64. cos 2 9 = 2 c o s 2 9 — 1
65. tan 2 9 =
66 . sin -
2 tan 8.
1 - tan2 9
67. ta n | = ± \ / | ^ i |
2
V 1 + cos 9
68. ta n | =
2
82. cos2 0 = 1 + c2os2e
cos 9
sin 9 a
1 + cos e
84.
cos a cos (3 = l[c o s (a —/3)+ cos (a + /3)]
85.
sina
86.
cos a sin /3 = -^-[sin (a + [3) — sin (a — (3)]
87.
cos a
cos /3 = -^-[sin ( a + f3)+ sin (a — f3)]
+ cos (3 = 2 cos cos
88. sin a — sin /3 = 2 cos
j sin
Verify each identity by using the pow er-reducing identities
and then again by using the product-to-sum identities.
89. cos a — cos (3 = —2 sin |—
69. 2 c o s2 59 — 1 = cos 100
90. WRITING IN MATH D escrib e th e step s th at y o u w ou ld u se to
70. c o s2 29 — s in 2 29 = cos 49
sin
j
find th e e x a ct v alu e o f cos 8 0 if cos 0 =
1
353
Spiral Review
Find the exact value of each trigonometric expression. (Lesson 5-4)
91 . cos y j
92-cos ^
93. sin ^
94. sin ^
95.cos
96. sin
97. GARDENING E liza is w aitin g for the first d ay o f sp rin g in w h ich there w ill b e 1 4 h o u rs of
d aylig h t to start a flow er garden. The n u m b er o f h o u rs o f d ay lig h t H in h e r to w n can be
m o d eled b y H = 1 1 . 4 5 + 6 . 5 sin ( 0 . 0 1 6 8 d — 1 . 3 3 3 ) , w here d is the d ay o f the year, d = 1
rep resen ts Jan u ary 1 , d = 2 represents Ja n u a ry 2 , and so on. O n w h at d ay w ill E liza b egin
gard ening ? (Lesson 5-3)
Find the exact value of each expression. If undefined, write undefined. (Lesson 4-3)
98. csc
99. tan
( - j )
210°
100. sin ^
101. cos
(-3 7 8 0 ° )
Graph and analyze each function. Describe the domain, range, intercepts, end behavior, continuity,
and where the function is increasing or decreasing. (Lesson 2-1)
1 I
i
102./(x) = - j r x 3
103. fi x ) = 4 x 4
1 0 4 ./(x) = —3 x 6
105./ (x) = 4 x 5
Skills Review for Standardized Tests
106. REVIEW Id en tify the eq u ation for the graph.
107. REVIEW F ro m a lo o k o u t p o in t o n a cliff abo v e a lake,
the an g le o f d ep ressio n to a b o a t on the w ater is 12°.
T h e b o a t is 3 k ilo m eters from the shore ju s t b elo w the
cliff. W h a t is th e h e ig h t o f the cliff from the su rface o f
the w ater to the lo o k o u t point?
A y = 3 cos 29
1
B y = cos 29
3
C
¥
¥
J
C y = 3 cos
1
D y=
3
COS
3
ta n 12°
3
tan
12°
FREE RESPONSE U se the grap h to an sw er each o f the follow ing.
108.
a. W rite a fu n ctio n o f the fo rm f{ x ) = a cos {bx + c) + d that
corresp on d s to the graph.
b. R ew rite/ (x) as a sin e fun ction.
C.
R ew rite/ (x) as a cosin e fu n ctio n o f a sin gle angle.
d. F in d all solu tio n s o f f( x ) =
0.
e. H o w do the solu tio n s th at y ou fo u n d in p art d relate to the g rap h o f/ (x)?
V.
354
| Lesson 5-5 | M u ltip le -A n gle and Product-to-Sum Identities
Study Guide and Review
Chapter Summary
KeyConcepts
Trigonometric Identities
•
•
KeyVocabulary
cofunction
(Lesson 5- 1 )
(p. 337)
Trigonom etric identities are identities th a t involve trigonom etric
difference identity
functions and can be used to find trigonom etric values.
double-angle identity
Trigonom etric expressions can be sim plified by w riting the expression
half-angle identity
in te rm s of one trigonom etric function or in te rm s of sine and cosine
•
p. 314)
(p. 346)
(p. 348)
(p. 312)
only.
identity
The m ost com m on trigonom etric
odd-even identity
identity is the Pythagorean
power-reducing identity
p. 347)
product-to-sum identity
(p. 350)
identity sin2 9 + cos2 6 = 1.
(p. 314)
Pythagorean identity
quotient identity
Verifying Trigonometric Identities
(p. 312)
reduction identity
(p. 340)
(p. 337)
trigonometric identity
•
Start with the more complicated side of the identity and work
to transform it into the simpler side.
•
Use reciprocal, quotient, Pythagorean, and other basic
trigonometric identities.
•
Use algebraic operations such as combining fractions, rewriting
fractions as sums or differences, multiplying expressions, or
factoring expressions.
•
Convert a denominator of the form 1 ± u or u ± 1 to a single term
using its conjugate and a Pythagorean Identity.
•
Work each side separately to reach a common expression.
Solving Trigonometric Equations
(p. 312)
reciprocal identity
sum identity
(Lesson 5- 2)
(p. 313)
verify an identity
ip. 312)
(P -320)
VocabularyCheck
Complete each identity by filling in the blank. Then name the identity.
1. s e c f l = ______________
(Lesson 5-3)
•
Algebraic techniques that can be used to solve trigonometric equations
include isolating the trigonometric expression, taking the square root
of each side, and factoring.
•
Trigonometric identities can be used to solve trigonometric equations
by rewriting the equation using a single trigonometric function or
by squaring each side to obtain an identity.
2.
sin 9
cos 9
3.
1 = sec2 9
4. cos (9 0 ° - 9) ■
5. tan ( — 0 ) = ___
Sum and Difference Identities
(Lesson 5-4)
6 . s in ( o + /3 ) = s i n a .
•
Sum and difference identities can be used to find exact values
of trigonometric functions of uncommon angles.
7.
•
Sum and difference identities can also be used to rewrite a
trigonometric expression as an algebraic expression.
8.
Multiple-Angle and Product-Sum Identities
•
(Lesson 5-5)
Trigonometric identities can be used to find the values of expressions
that otherwise could not be evaluated.
9.
10.
cos a .
. = cos2 a - sin 2 a
+ cos 9
1 - cos 29
. = ^-[cos ( a - (3) + cos (a + /3 )j
.. .
y
/H connectED .m cgraw -hill.com |
355
'V
Study Guide and Review
Continued
Lesson-by-Lesson Review
Trigonometric Identities (pp. 312-319)
Find the value of each expression using the given information.
Example 1
11. sec 0 and cos 0; tan 0 = 3, cos 9 > 0
If sec 0 = —3 and sin 0 > 0, find sin 0.
1
12. cot 9 and sin 0; cos i
tan 9 < 0
13. csc 0 and tan 0; cos 9 = -f, sin 9 < 0
0
14. cot 9 and cos 9; tan 9 =
cos 0 =
1
3
sec 9 < 0
Now you can use the Pythagorean identity that includes sin 9 and
cos 9 to find sin 9.
cos‘
17. sin2 ( - x ) + cos2 ( - x )
21.
Pythagorean Identity
= 1
sin20 + ( _ l ) 2 =
= 1
Simplify each expression.
18. sin2 x + cos2 x + cot2 *
sec2 x - tan2 x
COS
sec 0 = - 3
3
sin‘
19.
Reciprocal Identity
sec 9
csc 9 > 0
15. sec 9 and sin 0; cot 0 = - 2 , csc 0 < 0
16. cos 9 and sin 0; cot 9 =
Since sin 0 > 0 and sec 9 < 0 , 9 must be in Quadrant II. To find sin
O, first find cos 0 using the Reciprocal Identity for sec 9 and cos 9.
20 .
(-X )
1
22 .
1 - sin x
1
= 1
Multiply.
sin2 9 = ^
Subtract.
sin2 9
tan2x + 1
COS 0 =
. 0„ = _V8
sm
cosx
1 + sec x
2V2
o r _
1
Simplify.
Verifying Trigonometric Identities (pp. 3 2 0 - 3 2 6 )
Example 2
Verify each identity.
sin 9
+ i ±cose = 2csc0.
1 + cos 0
sin 0
23.
■+ . sin - „ = 2 csc i
1 - cos 9 1 + cos (
Verify that
24.
cos 0 , sin 9■= 1
sec 9
csc 9
The left-hand side of this identity is more complicated, so start
with that expression.
25.
cot 9
, 1 + csc 9
1 + csc 9
cot 9
26.
cos 9
1 - sin 9
1 + sin 9
cos
27.
cot2 9
1 + csc 9
csc 0 — 1
28.
S +Hf=SK(,+csc'
29.
sec 9 + csc 1
1 + tan 9
-
2 sec 9
CSC I
30. cot 9 csc 9 + sec 9 = csc2 9 sec 9
31.
sin 9
sin 9 + cos 9
32.
COS4
tan 9
1 + tan 9
sin4 9 =
1 - tan2 9
sec‘
1 + cos 9 _ sin2 0 + (1 + cos t
sin 9
•+
sin 0
sin 0(1 + cos 9)
1 + cos
_ sin2 0 + 1 + 2 cos 9 + cos2 9
sin 0(1 + cos 0)
_ sin2 0 + cos2 0 + 1 + 2 cos 9
sin 0(1 + cos 0)
_ 1 + 1 + 2 cos 9
~ sin 0(1 + cos 0)
_ 2 + 2 COS 0
~ sin 0(1 + cos 0)
2(1 + cos 0)
~ sin 0(1 + cos 0)
2
sin 0
= 2 csc 0
356
C h a p te r 5
I Study G uide and Review
Lesson-by-Lesson Review
Solving Trigonometric Equations
.............................................
(pp. 3 2 7 - 3 3 3 )
Find all solutions of each equation on the interval
[ 0 ,2 - * ] .
Example 3
Solve the equation sin 9 = 1 — cos 6 for all values of 6.
33. 2 s i n x = V 2
34. 4 c o s 2 x = 3
35. ta n 2 x - 3 = 0
36. 9 + c o t2 x = 1 2
37. 2 s in 2 x = sin x
38. 3 cos a-+ 3 = s in 2 *
sin
6 = 1 - cos 6
Original equation.
s in 2
9 = (1 - cos 9 )2
Square each side.
s in 2
9 = 1 - 2 cos 9 + c o s 2 9
Expand.
1 - co s 2 0 = 1 -
Solve each equation for all values of x.
2 cos
0 = 2 co s 2
39. sin 2 x - sin x = 0
9 + cos2 9
Pythagorean Identity
9 - 2 cos 9
Subtract.
0 = 2 cos 0 (c o s 0 - 1 )
40. ta n 2 x = tan x
Factor.
Solve fo r x on [ 0 , 27r],
41. 3 cos x = cos x - 1
cos 0 = 0
or
cos 0 = 1
42. sin 2 x = sin x + 2
0 = cos-1 0
9 = cos'“ 1 1
43. sin2 x = 1 - cos x
0 = | : Or ^
0 = o
Q'TTA c h eck show s th a t ~ is an extraneous solution. So the solutions
44. sin x = cos x + 1
are 0 = y + 2 itk or 0 = 0 + 2/77T.
J
v_____
-------------------------------
---------------
bum ana uitterence identities (pp. 3 3 6 - 3 4 3 )
Find the exact value of each trigonometric expression.
Exam ple 4
45. cos 1 5 °
46. sin 3 4 5 °
47. t a n ^ y
Find the exact value of tan
48. s i n ^ |
49. c o s - ^
50. t a n - ^ |
.
tan 1 7 " = ' = " ( t
tan ^
Simplify each expression.
+
4
tan
Sum Identity
3
1 - V3
52. cos 2 4 ° cos 3 6 ° - sin 2 4 ° sin 3 6 °
1 - (-V 3 )
53. sin 9 5 ° cos 5 0 ° - cos 9 5 ° sin 5 0 °
1 - V3
5 4 . cos I j f cos J L + sin f
sin %
Verify each identity.
55. cos (0 + 3 0 ° ) - sin (0 + 6 0 °) = - s i n 0
^
(cos 0 - sin 0 )
Evaluate for tangent.
Simplify.
1 + \/3
1 - V3
1
56. cos ( 0 +
2 3 lt _ 5 tt , 2 it
12
4
3
)
t
+ tan ^
1 - tan
ta n f + ta n - f
51 ■ -------- --------------- 5 1 -te n -ta n L .
OQ-rr
+
V3
1 - V3
Rationalize the denominator.
1 - V3
4-2V 3
1 -3
Multiply.
z W
3
Simplify.
57. cos ( 0 - y j + cos ( 0 + y j = cos 0
58. tan ( 0 + 3? ) = Jan e ~ ]
\
4 1
tan 6 + 1
v
j
connectED.m cgraw-hill.com |
357
H Study Guide and Review
l l
■
a
Continued
■
Lesson-by-Lesson Review
Multiple-Angle and Product-Sum Identities (pp. 346-354)
Example 5
Find the values of sin 29, cos 26, and tan 26 for the given value
and interval.
Find the values of sin 29, cos 26, and tan 26 if 6 is in the
59. cos 6 = 1 (0°, 90°)
60.tan 6 = 2, (180°, 270°)
fourth quadrant and tan 6 = — - y .
61. sin 6 = j , ( y , if )
62.sec 0 =
9 is in the fourth quadrant, so cos 9
2nj
sin 2 9 = 2 sin 9 cos 9
Find the exact value of each expression.
64. cos
65. tan 67.5°
66. c o s -^
67. s i n ^
68. tan
cos 20 = 2 cos2 9 - 1
11 TT
63. sin 75°
^ a n d s in 0 = _ | .
= 2(i)
12
tan 29 =
13tt
■ (-*)
2 tan e
1 - tan2
_ / _ 2 4 \2
I 7I
12
48
' 7 or 336
527
_ 527
49
Applications and Problem Solving
69. CONSTRUCTION Find the tangent of the angle that the ramp makes
with the building if sin 0 =
(Lesson 5-1)
V t4 5
and cos i
145
12VT45
145 '
72. PROJECTILE MOTION A ball thrown with an initial speed v0 at an
angle 9 that travels a horizontal distance d w ill remain in the air t
d
Suppose a ball is thrown with an
seconds, where t v0 cos 9'
initial speed of 50 feet per second, travels 100 feet, and is in the
air for 4 seconds. Find the angle at which the ball was thrown.
(Lesson 5-3)
73. BROADCASTING Interference occurs when two waves pass
70. LIGHT The intensity of light that emerges from a system of two
/n
polarizing lenses can be calculated by / = / „ - •
0
CSC2
-, where l0 is
9’
the intensity of light entering the system and 9 is the angle of the
axis of the second lens with the first lens. Write the equation for the
light intensity using only tan 9. (Lesson 5-1
71. MAP PROJECTIONS Stenographic projection is used to project the
contours of a three-dimensional sphere onto a two-dimensional
map. Points on the sphere are related to points on the map using
sin a . Verify that r = 1 + c o s a . (Lesson 5-2)
r=
sin a
1 - cos a
through the same space at the same time. It is destructive
if the amplitude of the sum of the waves is less than the
amplitudes of the individual waves. Determine whether the
interference is destructive when signals modeled by y = 20 sin
(31 + 45°) and y = 20 sin (3f + 225°) are combined. (Lesson 5-4)
74. TRIANGULATION Triangulation is the process of measuring
a distance d using the angles a and (3 and the distance I
using £ = d , d (Lesson 5-5)
tan a + tan /3
d
Aa
I
a. Solve the formula for d.
t sin a sin (3
sin a cos (3 + cos a sin f3'
b. Verify that d = —
c. Verify that d =
e sin a sin (3
sin (a + f3) '
d. Show that if a = (3, then d = 0.5£ tan a .
358
C h a p te r s
Study G uide and Review
ill!
Practice Test
Find the value of each expression using the given information.
Find the exact value of each trigonometric expression.
1. sin 9 and cos 9, csc 6 = - 4 , cos 9 < 0
19.
tan 165°
2. csc 9 and sec 9 , tan 9 =
20 .
cos
5
csc 9 < 0
12
21. sin 75°
Simplify each expression.
3.
22. cos 465° - cos 15°
sin (90° - x)
tan (90° - x)
23. 6 sin 675° - 6 sin 45°
4. sec2 x - 1
tan2 x + 1
5.
24. MULTIPLE CHOICE Which identity is true?
sin 9 (1 + cot2 9)
F cos (9 +
1
tt
sec‘ 9
H sin ^0 - - y - j = cos 9
= 1
J sin
cos 9
, 1 - sin9
2 cos9
1 + sin 9
cos <
1 + sin 9
1
1 + cos 9
= - s in
G cos (7x - 9) = cos 9
Verify each identity.
csc‘
csc‘ 9
tt)
1
1 -
cos(
: 2 CSC
9. - s e c 2 0 sin2 9 = cos2 6 ~ 1
cos2 9
A tan (-9 ) = - ta n 9
B tan (—0) =
1
cot (- 9 )
C tan ( - 9 ) ■
sin (-9)
cos (-9)
+ 9) = sin 9
Simplify each expression.
25. cos ^ cos
26.
10. sin4 x - cos4 x = 2 sin2 x - 1
11. MULTIPLE CHOICE Which expression is not true?
(tt
- sin ~ sin
tan 135° - tan 15°
1 + tan 135° tan 15°
27. PHYSICS A soccer ball is kicked from ground level with an initial
speed of v at an angle of elevation 9.
A.0
D tan ( -9 ) + 1 = sec (-9 )
-A Find all solutions of each equation on the interval [ 0 , 2n],
a. The horizontal distance dthe ball will travel can be determined
i/2 cin O/i
using d =
^
, where g is the acceleration due to gravity.
12. V 2 sin 0 + 1 = 0
13. sec2 0 = |
Verify that this expression is the same as | v2 (tan 9 - tan 9
sin2 9).
Solve each equation for all values of 6.
b. The maximum height h the object will reach can be determined
using h = v ™
14. tan2 9 - tan 9 = 0
15.
16.
1 - sin i
COS I
1
sec 9 - 1
attained to the horizontal distance traveled.
: COS I
1
= 2
sec 9 + 1
17. s e c 0 - 2 t a n 0 = O
Find the values of sin 20, cos 26, and tan 26 for the given value and
interval.
28.
18. CURRENT The current produced by an alternator is given
by / = 40 sin 135-jrf, where I is the current in amperes and t is
the time in seconds. At what time t does the current first reach
20 amperes? Round to the nearest ten-thousandths.
Find the ratio of the maximum height
tan 9 = - 3 ,
2 tt|
29. cos 9 = i , (0°, 90°)
5
30. cos 9 =
Connect to AP Calculus
Rates of Change for Sine and Cosine'
L,.*...................
Objective
Approximate rates of
change for sine and
cosine functions using the
difference quotient.
**
In Chapter 4, you learned that many
real-world situations exhibit periodic
behavior over time and thus, can be
modeled by sinusoidal functions.
Using transformations of the parent
functions sin xand cos x, trigonometric
models can be used to represent data,
analyze trends, and predict future values.
Hours of Daylight
1
16
i* *
__
>
12
V)
k.
3
X 8
i
__
*
(
__
__
<
* i»
n
4
0
J
M
M
J
Month
S
N
While you are able to model real-world situations using graphs of
sine and cosine, differential calculus can be used to determine the
rate that the model is changing at any point in time. Your knowledge
of the difference quotient, the sum identities for sine and cosine, and
the evaluation of limits now makes it possible to discover the rates
of change for these functions at any point in time.
Activity 1 Approximate Rate of Change
Approxim ate the rate of change of fix ) = sin x at several points.
Substitute/(x) = sin x into the difference quotient.
m ■
FfflTO
f{x + h) - f( x )
h
sin (x + h) —sin x
h
Approximate the rate of change of f i x ) at x =
Repeat Steps 1 and 2 for x = 0 and for x =
f
TT
Let h = 0.1,0.01, 0.001, and 0.0001.
tt .
A n a ly ze the Results
1. Use tangent lines and the graph of f i x ) = sin x to interpret the values found in Steps 2 and 3.
2. What will happen to the rate of change
of
f i x ) as x increases?
V
....... ... ...............
Unlike the natural base exponential function g(x) = ex and the natural logarithmic function h{x) = In x, an expression
to represent the rate of change of fix) = sin x a t any point is not as apparent. However, we can substitute f(x) into the
difference quotient and then simplify the expression.
fix + h) - fix)
m
Difference quotient
h
sin {x + h) - sin x
h
(sin xcos h + cos / sin h) - sin x
sin xcos h - sin x
= sinx ( “
360
C h a p te r 5
i ^ l ) + c o s x (^ )
fix) = sin x
Sine Sum Identity
Group terms with sin xand cos x.
Factor sin x and cos x.
We now have two expressions that involve h, sin x
—
-j
and cos x
obtainan
accurateapproximation
of the rate of change of f(x) at a point, we want h to be as close to 0 as possible. Recall that in Chapter 1, we were
able to substitute h = 0 into an expression to find the exact slope of a function at a point. However, both of the
fractional expressions are undefined at h = 0 .
sin x
^
1j
expressions cosx
Original
undefined
undefined
We can approximate values for the two expressions by finding the limit of each as h approaches 0 using techniques
discussed in Lesson 1 -3.
Activity 2 Calculate Rate of Change
Find an expression for the rate of change of f(x ) = sin x.
cos h —1
Use a graphing calculator to estimate lim ■
°
h-> 0
h
'K
Verify the value found in Step 1 by using the
T A B L E function of your calculator.
E5HEI
sin h
Repeat Steps 1 and 2 to estimate lim —
h->0 h
FIHfH
Substitute the values found in Step 2 and Step 3
into the slope equation
K=.(i66H4
[ - I T , tt]
Vi
-.0 0 3 0
-.0 0 2 0
B rans
0.0000
.00100
.00200
HHfH
V = - .0 3 3 H 1
scl:-?- by [-1.5,1.5] scl: 1
.00300
Simplify the expression in Step 4.
.00150
i.O E-3
E.0E-H
ERROR
"EE "H
- I E "3
-.001E
X= -.0 0 1
p A n a ly z e the Results
3. Find the rate of change of/(x) = sin x at x =
2, and
4. Make a conjecture as to why the rates of change for all trigonometric functions must be
modeled by other trigonometric functions.
M odel and Apply
5.
In this problem, you will find an expression for the rate
of change o f f (x) = cos x at any point x.
a. Substitute/(x) = cos x into the difference quotient.
b. Simplify the expression from part a.
C.
Use a graphing calculator to find the limit of the two
fractional expressions as h approaches 0.
d. Substitute the values found in part c into the slope equation found in part b.
e. Simplify the slope equation in part d.
f.
Find the rate of change of f(x ) = cos x at x = 0,
tt ,
3 tt
and
2 '
361