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Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
(For help, go to Lesson 2-1.)
Determine whether each relation is a function.
1. {(2, 4), (1, 3), (–3, –1), (4, 6)}
2. {(2, 6), (–3, 1), (–2, 2)}
3. {(x, y)| x = 3}
4. {(x, y)| y = 8}
5. {(x, y)| x = y2}
6. {(x, y)| x2 + y2 = 36}
7. {(a, b)| a = b3}
8. {(w, z)| w = z – 36}
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
Solutions
1. {(2, 4), (1, 3), (–3, –1), (4, 6)}; yes, this is a function because each
element of the domain is paired with exactly one element in the range.
2. {(2, 6), (–3, 1), (–2, 2)}; yes, this is a function because each element
of the domain is paired with exactly one element in the range.
3. {(x, y)| x = 3}; no, this is not a function because it is a vertical line and
fails the vertical line test.
4. {(x, y)| y = 8}; yes, this is a function because it is a horizontal line and
passes the vertical line test.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
Solutions (continued)
5. {(x, y)| x = y2}; no, this is not a function because an element of the
domain is paired with more than one element in the range.
Example: 4 = 22 and 4 = (–2)2
6. {(x, y)| x2 + y2 = 36}; no, this is not a function because it is a circle and
fails the vertical line test.
7. {(a, b)| a = b3}; yes, this is a function because each element of the
domain is paired with exactly one element in the range.
8. {(w, z)| w = z – 36}; yes, this is a function because each element of
the domain is paired with exactly one element in the range.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
Analyze this periodic function. Identify one cycle in two
different ways. Then determine the period of the function.
Begin at any point on the graph.
Trace one complete pattern.
The beginning and ending x-values of
each cycle determine the period of the
function.
Each cycle is 7 units long.
The period of the function is 7.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
Determine whether each function is or is not periodic. If it is,
find the period.
a.
The pattern of y-values in one section repeats exactly in other sections.
The function is periodic.
Find points at the beginning and end of one cycle.
Subtract the x-values of the points: 2 – 0 = 2.
The pattern of the graph repeats every 2 units, so the period is 2.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
(continued)
b.
The pattern of y-values in one section repeats exactly in other sections.
The function is periodic.
Find points at the beginning and end of one cycle.
Subtract the x-values of the points: 3 – 0 = 3.
The pattern of the graph repeats every 3 units, so the period is 3.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
Find the amplitudes of the two functions in
Additional Example 2.
1
a. amplitude = 2 (maximum value – minimum value)
1
= 2 [2 – (–2)]
1
= 2 (4) = 2
Use definition of
amplitude.
Substitute.
Subtract within parentheses and simplify.
The amplitude of the function is 2.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
(continued)
1
b. amplitude = 2 (maximum value – minimum value)
1
= 2 [6 – 0]
1
= 2 (6) = 3
Use definition of
amplitude.
Substitute.
Subtract within parentheses and simplify.
The amplitude of the function is 3.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
The oscilloscope screen below shows the graph of the
alternating current electricity supplied to homes in the United States.
Find the period and amplitude.
1
1 unit on the t-axis = 360 s
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
(continued)
1
One cycle of the electric current occurs from 0 s to 60 s.
The maximum value of the function is 120, and the minimum is –120.
period = 1 – 0
60
= 1
60
Use the definitions.
amplitude = 1 [120 – (–120)]
2
= 1 (240) = 120
Simplify.
2
1
The period of the electric current is 60 s.
13-1
The amplitude is 120 volts.
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
pages 699–702 Exercises
12. 1
1. x = –2 to x = 3, x = 2 to x = 7; 5
13. 2
2. x = 0 to x = 4, x = 5 to x = 9; 4
14.
3. x = 0 to x = 4, x = 2 to x = 6; 4
4. not periodic
5. periodic; 12
15.
6. not periodic
7. not periodic
8. periodic; 8
9. periodic; 7
16. a. y
10. 4
b. x
11. 3
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
22. Check students’ work.
17. Answers may vary. Sample:
Yes; average monthly
temperatures for three years
should be cyclical due to the
variation of the seasons.
23. 3, –3, 4;
18. Answers may vary. Sample:
No; population usually increases
or decreases but is not cyclical.
24. 5, 0, 8;
19. Answers may vary. Sample:
Yes; traffic that passes through
an intersection should be at the
same levels for the same times
of day for two consecutive work days.
20. 60 beats per min
21. a. 1 s
b. 1.5 mV
13-1
25. 4, –4, 8;
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
26. 1 yr
33. a.
27. 2 weeks
28. 3 months
29. 1 hour
30. 1 day
31. 2, 2, 2
32. a. 67
b. 70
c. 70
d. 67
b. 5 s, 1 ft
1
c. Answers may vary. Sample: about 1 3 s
34. a.
b.
c.
d.
24.22 days
0.78 day
0.22 day
Answers may vary. Sample: The calendar year
is meant to predict events in the solar year.
Keeping the difference between the two minimal
is necessary for the calendar year to be useful.
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
35. C
36.
37.
38.
39.
40. [4] 64 seconds. The first two functions
are at the beginning of their cycles
G
together every 6 • 7 = 42 seconds:
B
42, 84, 126, . … The third function
is at the beginning of its cycle every
D
8 seconds, starting at 20 seconds:
[2] A period is the length of 1 cycle,
20, 28, 36, 44, 52, 60, 68, 76, 84, . …
The three functions are all at the
so n seconds = x seconds
m cycles
1 cycle
beginning of their cycles at 84
n
Then xm = n, or x = .
seconds, which is 64 seconds
m
after the third function begins.
n
The period is
seconds.
[3] minor error in calculation
m
[1] answer only with no explanation
[2] incomplete explanation
[1] answer only, with no explanation
13-1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
45. (x + 3)2 = 20(y – 2)
41.
42. 11, 13; an = 2n – 1; explicit or a1 = 1,
an = an – 1 + 2; recursive
43. 14, 16; an = 2n + 2; explicit or a1 = 4,
an = an – 1 + 2; recursive
44. 38, 51; a1 = 3, an = an – 1 + 2n – 1;
recursive or an = n2 + 2; explicit
13-1
(x – 5)2
(y + 3)2
46.
–
5
4
=1
(y – 1)2
(y + 2)2
47.
–
4
16
=1
Exploring Periodic Data
ALGEBRA 2 LESSON 13-1
1. Determine whether the
function shown is periodic.
Find the period and the amplitude of
the periodic functions shown.
3.
yes
6; 1.5
2. Determine whether the
function shown is periodic.
4.
no
3; 0.5
13-1
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
(For help, go to page 54.)
For each measure, draw an angle with its vertex at the origin of the
coordinate plane. Use the positive x-axis as one ray of the angle.
1. 90°
2. 45°
3. 30°
4. 150°
5. 135°
6. 120°
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
Solutions
1.
2.
3.
4.
5.
6.
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
Find the measure of the angle.
The angle measures 60° more than a right angle of 90°.
Since 90 + 60 = 150, the measure of the angle is 150°.
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
Sketch each angle in standard position.
a. 48°
b. 310°
13-2
c. –170°
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
The Aztec calendar stone has 20 divisions for the 20 days in
each month of the Aztec year. An angle on the Aztec calendar shows
the passage of 16 days. Find the measures of the two coterminal
angles that coincide with the angle.
16
The terminal side of the angle is 20 of a full rotation from the initial side.
16
• 360° = 288°
20
To find a coterminal angle, subtract one full rotation.
288° – 360° = –72°
Two coterminal angle measures for an angle on the Aztec calendar
that show the passage of 16 days are 288° and –72°.
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
Find the cosine and sine of 135°.
From the figure, the x-coordinate of point A
is –
2 , so cos 135° = – 2 , or about –0.71.
2
2
Use a 45°-45°-90° triangle to find sin 135°.
opposite leg = adjacent leg
2
= 2
0.71
Substitute.
Simplify.
The coordinates of the point at which the terminal side of a 135° angle
intersects are about (–0.71, 0.71), so cos 135° –0.71 and sin 135° 0.71.
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
Find the exact values of cos (–150°) and sin (–150°).
Step 1: Sketch an angle of –150°
in standard position.
Sketch a unit circle.
Step 2: Sketch a right triangle. Place
the hypotenuse on the
terminal side of the angle.
Place one leg on the x-axis.
(The other leg will be parallel
to the y-axis.)
x-coordinate = cos (–150°)
y-coordinate = sin (–150°)
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
(continued)
The triangle contains angles of 30°, 60°, and 90°.
Step 3: Find the length of each side of the triangle.
hypotenuse = 1
The hypotenuse is a radius of the unit circle.
1
shorter leg = 2
1
longer leg = 2
The shorter leg is half the hypotenuse.
3
3= 2
The longer leg is
3 times the shorter leg.
Since the point lies in Quadrant III, both coordinates are negative. The
longer leg lies along the x-axis, so
cos (–150°) = –
3
1
, and sin (–150°) = – .
2
2
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
pages 708–710 Exercises
11.
8.
1. –315°
2. –135°
3. 240°
12. 25°
9.
13. 215°
4. 115°
14. 315°
5. –110°
15. 4°
6. –340°
7.
16. 140°
10.
17. 150°
18. 55°
19. 180°
20. 220°, –140°
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
3 ; 0.50, –0.87
2
2
– 2 , – 2 ; –0.71, –0.71
2
2
3 , – 1 ; 0.87, –0.50
2
2
– 1 , 3 ; –0.50, 0.87
2
2
3 , 1 ; 0.87, 0.50
2 2
2 , – 2 ; 0.71, –0.71
2
2
3 , – 1 ; 0.87, –0.50
2
2
– 2 , 2 ; –0.71, 0.71
2
2
21. 1 , –
31. 0.71, –0.71
41. 275°, –445°
22.
32. –0.87, 0.50
42. 295°, –65°
33. –0.09, –1.00
43. 573°, –147°
34. 0.98, –0.17
44. 303°, –417°
35. –0.90, 0.44
45. II
36. 0.00, 1.00
46. III
23.
24.
25.
26.
27.
28.
29. 1.00, 0.00
30. 0.85, 0.53
37–44. Answers may vary. 47. negative x-axis
Samples:
48. IV
37. 405°, –315°
49. positive x-axis
38. 235°, –485°
39. 45°, –315°
40. 40°, –320°
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
50. a.
b. II
c. If the terminal side of an
angle is in Quadrants I or II,
then the sine of the angle is
positive; otherwise it is not.
If the terminal side of an
angle is in Quadrants I or
IV, then the cosine of the
angle is positive; otherwise
it is not.
51. a. 0.77, 0.77, 0.77
b. The cosines of the three angles
are equal because the angles
are coterminal.
52. The x-coordinate of the point on the ray
defined by angle is equal to cos ;
similarly for the y-coordinate and sin .
The angles 0°, 180°, and 360° lie on
the x-axis, and thus their sines are all
0 and their cosines are ±1. The angles
90° and 270° lie on the y-axis, so their
cosines are 0 and their sines are ±1.
53.
13-2
1 ,
2
3
2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
3
2
54.
– 1 ,
55.
– 2 ,–
2
56.
1 ,–
2
2
2
2
3
2
2 ,– 2
2
2
57.
1,–
2
58.
59. –
3
2
2 , 2
2
2
60. Answers may vary. Sample:
30°, 150°, –210°, 390°
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
61. No; yes; if the sin and cos
are both negative, is in
Quadrant III. –120° is in
Quadrant III.
67. [2]
The terminal side forms an angle of
30° with the negative x-axis. Using
the unit circle, x = – 3 and y = 1 .
62. a. Check students’ work.
b. –20°
63. A
So cos (–210°) = –
2
3.
2
2
[1] answer only, with no work shown
64. H
65. D
66. H
13-2
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
69. periodic; 3
68. [4]
70. not periodic
71. periodic; 6
72.
The terminal side forms an angle
of 45° with the negative x-axis,
so sin(–135°) = –
2 and cos(–135°) = – 2 .
2
2
Then [sin (–135°)]2 + [cos (–135°)]2 =
–
2
2
2
+
–
2
2
2
= 2 + 2 = 4 = 1.
4
4
4
(OR a convincing argument using x2 + y2 = 1)
[3] one computational error
[2] incomplete explanation with correct answer
[1] answer only, with no work shown
13-2
73.
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
74. (0, 2
5 ), (0, –2
5)
75. (0, 5
5 ), (0, –5
5)
76. (
85, 0), (–
77. (
145, 0), (–
13-2
85, 0)
145, 0)
Angles and the Unit Circle
ALGEBRA 2 LESSON 13-2
Sketch each angle in standard position. Use a right triangle to find the
exact values of the cosine and sine of the angle.
2. –120°
1. 45°
2
2
;
2
2
3
– 1; –
2
2
3. What angle is less than 360° and coterminal with 45°?
–315°
13-2
Radian Measure
ALGEBRA 2 LESSON 13-3
(For help, go to page 870.)
Find the circumference of a circle with the given radius or diameter.
Round your answer to the nearest tenth.
1. radius 4 in.
2. diameter 70 m
3. radius 8 mi
4. diameter 3.4 ft
5. radius 5 mm
6. diameter 6.3 cm
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
Solutions
1. C = 2
2. C =
3. C = 2
4. C =
5. C = 2
6. C =
r=2
d=
(70 m)
r=2
d=
r=2
d=
(4 in.)
25.1 in.
219.9 m
(8 mi)
(3.4 ft)
50.3 mi
10.7 ft
(5 mm)
(6.3 cm)
31.4 mm
19.8 cm
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
a. Find the radian measure of angle of 45°.
45°
180°
45 •
r radians
radians
=
= 180 • r
r=
45 •
180
= 4
0.785
Write a proportion.
Write the cross-products.
Divide each side by 45.
Simplify.
An angle of 45° measures about 0.785 radians.
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
(continued)
b. Find the degree measure of
13
6
radians
13
6
=
d°
180
• 180 =
Write a proportion.
•d
• 180 30
d=
16•
13
= 390°
An angle of
13
6
13
.
6
Write the cross-product.
Divide each side by
Simplify.
radians measures 390°.
13-3
.
Radian Measure
ALGEBRA 2 LESSON 13-3
a. Find the degree measure of an angle of – 3
2
–3
2
180°
= – 3 radians •
2
radians
1
radians •
= –270°
An angle of – 3
2
radians measures –270°.
13-3
90
180°
radians
radians.
Multiply by
180°
.
radians
Radian Measure
ALGEBRA 2 LESSON 13-3
(continued)
b. Find the radian measure of an angle of 54°.
3
54° • 180° radians = 54° • 180° radians
10
3
= 10 radians
3
An angle of 54° measures 10 radians.
13-3
Multiply by 180° radians.
Simplify.
Radian Measure
ALGEBRA 2 LESSON 13-3
Find the exact values of cos
radians •
3
180°
= 60°
radians
and sin
.
3 radians
3 radians
Convert to degrees.
Draw the angle.
Complete a 30°-60°-90° triangle.
The hypotenuse has length 1.
1
The shorter leg is 2 the length of the hypotenuse, and the longer leg
is 3 times the length of the shorter leg.
1
Thus, cos 3 radians = 2
3
and sin 3 radians = 2 .
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
Use this circle to find length s to the nearest tenth.
s = r
Use the formula.
7
7
=6• 6
Substitute 6 for r and 6 for .
=7
Simplify.
22.0
Use a calculator.
The arc has length 22.0 in.
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
Another satellite completes one orbit around Earth every 4 h.
The satellite orbits 2900 km above Earth’s surface. How far does the
satellite travel in 1 h?
1
Since one complete rotation (orbit) takes 4 h, the satellite completes 4
of a rotation in 1 h.
Step 1: Find the radius of the satellite’s orbit.
r = 6400 + 2900 Add the radius of Earth and the distance
from Earth’s surface to the satellite.
= 9300
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
(continued)
Step 2: Find the measure of the central angle the satellite travels
through in 1 h.
=
1
•2
4
=
1
•
2
Step 3: Find s for
s=r
Multiply the fraction of the rotation by the
number of radians in one complete rotation.
Simplify.
= 2.
Use the formula.
= 9300 • 2
14608
Substitute 9300 for r and 2 for .
Simplify.
The satellite travels about 14,608 km in 1 h.
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
pages 715–719 Exercises
11. 90°
1. – 5 , –5.24
12. 270°
2. 5 , 2.62
13.
3
6
3. –
4. –
5. 8
9
6.
9
2
3
, –1.57
, –1.05
, 2.79
, 0.35
7. 540°
8. 198°
9. –120°
10. –172°
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
3 ,1
2 2
3
15. 1 ,
2 2
14.
16. 0, 1
17. – 1 , 3
2 2
18. – 3 , 1
2
2
19. 0, –1
20. 3.1 cm
21. 10.5 m
22. 51.8 ft
25. 43.2 cm
31. III
26.
107 in.
32. II
27.
32 ft
33. positive y-axis
28. a. 11,048 km
b. 33,144 km
c. 27,620 km
d. 276,198 km
e. 18.1 h
29.
34. II
35. negative x-axis
36. III
37.
42.2 in.
30. a. 15°,
b.
c.
12
radians
1036.7 mi
413.6 mi
23. 25.1 in.
0.71, –0.71
24. 4.7 m
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
38.
40.
–0.50, –0.87
1.00, 0.00
39.
41.
–0.87, –0.50
0.00, 1.00
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
42.
0.81, –0.59
43. a–b.
c.
All five triangles are
congruent by SSS. All
have a hypotenuse of 1
unit, a long leg of about
0.81 unit, and a short
leg of 0.59 unit.
cos
5
3
sin
10
cos 4
5
cos 6
5
cos 9
5
44. Check students’ work.
45.
11 radians
46. The student forgot to
include parentheses.
47.
798 ft; 55°, 665°
5
48.
= 0.81, cos 3 = 0.59; 49.
10
= –0.81, sin 4 = 0.59;
5
= –0.81, sin 6 = –0.59;
5
= 0.81, sin 9 = –0.59
5
23.6 in.; – 7 , 17
= 0.81, sin
13-3
= 0.59;
6
6
If two angles measured
in radians are coterminal,
the difference of their
measures will be evenly
divisible by 2 .
Radian Measure
ALGEBRA 2 LESSON 13-3
50.
57. a. 0.5017962; 0.4999646; the first four terms
6.3 cm
51.
2
4
6
8
b. 1 – x + x – x + x – . . .
4008.7 mi
52. – 3
2!
c.
radians
2
53. – 11 radians
3
54. 4 radians
3
55. 35 radians
6
56.
2
2
=
•2 r =
8!
0.951; 18°
59. G
60. D
61. G
2
r
s
r =s
6!
58. C
s
2
4!
r
•2 r
62. [2] For a central angle of 1 radian, the length
of the intercepted arc is the length
of the radius.
[1] incomplete explanation
s =r
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
63.
73. (x – 3)2 + (y – 7)2 = 42.25
66.
74. (x + 8)2 + (y – 4)2 = 9
64.
67.
68. 9.1, 5.41
65.
69. 12.9, 3.53
70. 30, 8.09
71. x2 + y2 = 64
72. x2 + (y + 5)2 = 16
13-3
Radian Measure
ALGEBRA 2 LESSON 13-3
1. Rewrite each angle measure using the other unit, either degrees
or radians.
a. 225°
b.
2
5
4
radians
radians 90°
2. A wrench turns through an angle of 1.5 radians. If the wrench is
14 in. long, what is the distance that the end of the wrench
moves?
21 in.
3. A jogger runs 100 m around a circular track with a radius of 40 m.
Through what angle does the jogger move? Express your answer
in both radians and degrees.
2.5 radians; about 143.2°
13-3
The Sine Function
ALGEBRA 2 LESSON 13-4
(For help, go to Lesson 13-1.)
Use the graph. Find the value(s) of each of the following.
1. the period
2. the domain
3. the amplitude
4. the range
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
Solutions
1. the period: 2 units
2. the domain: all real numbers
3. the amplitude: 2 = 1 unit
2
4. the range: –1 <
–y<
– 1, where y is a real number
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
Use the graph of the sine function.
a. What is the value of y = sin
The value of the function at
for
= 180°?
= 180° is 0.
b. For what other value(s) of from 0° to 360° does the graph of sin
have the same value as for = 180°?
When y = 0,
= 0° and 360°.
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
Estimate each value from the graph. Check your estimate
with a calculator.
a. sin 3
The sine function reaches its median value of 0 at
3.14. The
value of the function at 3 is slightly more than 0, or about 0.1.
sin 3 = 0.1411200081
Use a calculator to check your estimate.
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
(continued)
b. sin
2
The sine function reaches its maximum value of 1 at 2 , so sin 2 = 1.
sin
2
=1
Use a calculator to check your estimate.
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
Use the graph of y = sin 6 .
a. How many cycles occur in this graph?
How is the number of cycles related to
the coefficient of in the equation?
The graph shows 6 cycles.
The number of cycles is equal to the coefficient of .
b. Find the period of y = sin 6 .
2
÷6= 3
Divide the domain of the graph by the number of cycles.
The period of y = sin 6
is 3 .
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
This graph shows the graph of y = a • sin
a = 3 and a = 3.
for values of
4
a. Find the amplitude of each sine curve. How
does the value of a affect the amplitude?
The amplitude of y = sin
3
of y = 4 • sin
is 1, and the amplitude
3
is 4 .
The amplitude of y = 3 • sin
is 3.
In each case, the amplitude of the curve is | a |.
b. How would a negative value of a affect each graph?
When a is negative, the graph is a reflection in the x-axis.
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
a. Sketch one cycle of a sine curve with amplitude 3
and period 4.
Step 1: Choose scales for the y-axis and the xaxis that are about equal ( = 1 unit).
On the x-axis, mark one period (4 units).
Step 2: Mark equal spaces through one cycle by
dividing the period into fourths.
Step 3: Since the amplitude is 3, the maximum
3 and the minimum is –3. Plot the five
points and sketch the curve.
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
(continued)
b. Use the form y = a sin b . Write an equation with
a > 0 for the sine curve in part a.
The amplitude is 3, and a > 0, so a = 3.
2
The period is 4, and 4 = b , so b = 2 .
An equation for the function is y = 3 sin 2 x.
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
Sketch one cycle of y = 5 sin 3 .
3
5
5
| a | = 3 , so the amplitude is 3 .
b = 3, so there are 3 cycles from 0 to 2 .
2
2
2
=
,
so
the
period
is
.
b
3
3
Divide the period into fourths. Using the values of the amplitude and
period, plot the zero-max-zero-min-zero pattern.
Sketch the curve.
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
Find the period of the following sine curve. Then write an
equation for the curve.
According to the graph, one cycle takes 3 units to
complete, so the period is 3.
To write the equation, first find b.
2
period = b
Use the relationship between
the period and b.
2
3= b
Substitute.
2
b= 3
Multiply each side by
2.094
Simplify.
Use the form y = a sin b . An equation for the graph is y = 5 sin
13-4
b
.
3
2
3
.
The Sine Function
ALGEBRA 2 LESSON 13-4
pages 724–727 Exercises
12.
–0.7
13. 3; 2, 2
1. 1
2.
0.7
3
1
14. ; 1, 4
2
3.
0.9
15. 2; 3,
2
4. 0
y = 4 sin 1
2
19.
16.
5.
–0.9
6.
–0.9
7. 1
y = 2 sin 3
17.
8.
0.1
9.
–0.8
10.
18.
–1
11. –1
y = 3 sin
20.
y = sin
y = 1 sin 2
3
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
21.
28. 2 ; y = 2 sin
25.
29. 2 ; y = –3 sin
2
y = 1.5 sin
3
26.
30.
; y = 5 sin 2
32.
; y = –sin 2
2
31.
; y = 1 sin 6
3
2
22.
33. 4; y = 3 sin
23.
34. 1; 1, 2
27.
35. 5; 1, 2
5
36.
24.
; 1, 2
37. 1; 3, 2
38. 1; 5, 2
39. 2 ; 5, 1
13-4
2
The Sine Function
ALGEBRA 2 LESSON 13-4
40. a.
b.
b. As a increases, the
amplitude of the graph
increases.
41. a.
They are reflections of each
other in the x-axis.
They are reflections of each
other in the x-axis.
c. When either a or b is replaced
by its opposite, the graph is a
reflection of the original graph
in the x-axis.
42. a.
b. 4
43. a.
1
440
b. 0.001
c. 880
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
44. • |a| is the amplitude of the function.
, 5
46.
2
• b is the number of cycles in the
interval 0° to 360°.
• 360° is the period of the function.
b
The properties relating to number
47. 1, 2
of cycles and period are affected.
45. 2 , 3.5
5
48. 2 , 0.4
3
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
49. 6, 0.5
53. y = sin 60
58.
,1
54. y = sin 30
55. y = sin 240,000
56. 2 , 1
59. a. days from spring
equinox, hours of
sunlight
50. 12 , 1.2
5
b. 23 h, about 365 days
12
57. 2 , 1
c. y = 23 sin 2
12
x
365
d. 1.1 h
51. Check students’ work.
e. Check students’ work.
52. a. 4, 2
b. y = 4 sin
c. coil B
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
65. [4] The amplitude is 1, so a = 1.
60. C
radians =
radians,
6
180°
so b = 2 ÷ = 2 • 6 = 12.
6
30° •
61. G
62. B
63. G
64. [2] Since sine is always
positive in the first and
second quadrants, a
value of where its
sine is equal to the sin
60° would be 180° – 60°, or 120°.
[1] answer only, with no work shown
13-4
The function is y = sin 12 .
[3] one calculation error
[2] incomplete explanation
[1] answer only, with no work shown
66. – 4
9
67. 5
radians, –1.40 radians
radians, 2.62 radians
6
68. – 4 radians, –4.19 radians
3
The Sine Function
ALGEBRA 2 LESSON 13-4
69. 16
radians, 5.59 radians
9
70. – 5 radians, –7.85 radians
2
71.
49%
72. an = 7 – n; –8
73. an = 12 + 3n; 57
74. an = 0.8n – 0.2; 11.8
13-4
The Sine Function
ALGEBRA 2 LESSON 13-4
1. Sketch the graph of y = 3 sin 2 in the interval from 0 to 2 .
2. Write an equation of the sine function for this graph.
1
x
y = 3 sin 3
3. What is the amplitude of the graph in Question 2?
1
3
13-4
The Cosine Function
ALGEBRA 2 LESSON 13-5
(For help, go to Lesson 13-1 and 13-4.)
Find the x-coordinate of each point on the unit circle.
1. A
2. B
3. C
4. D
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
Solutions
1. x-coordinate of point A: 1
2. x-coordinate of point B: 0
3. x-coordinate of point C: –1
4. x-coordinate of point D: 0
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
Use the graph shown below.
a. Find the domain, period range, and
amplitude of this function.
The domain of the function is all real numbers.
The function goes from its maximum value of 2 and back again in an
interval from 0 to 2 . The period is 2 .
The function has a maximum value of 2 and a minimum value of –2.
The range is –2 <
–y<
– 2.
1
1
1
The amplitude is 2 (maximum – minimum) = 2 (2 – (–2)) = 2 (4) = 2.
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
(continued)
b. Examine the cycle of the cosine function in the interval from 0 to 2 .
Where in the cycle does the maximum value occur? Where does the
minimum occur? Where do the zeros occur?
The maximum value occurs at 0 and 2 .
The minimum value occurs at
.
3
The zeros occur at 2 and at
.
2
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
Sketch the graph of y = –2 cos
in the interval from 0 to 4.
| a | = 2, so the amplitude is 2.
b=
2
, so the graph has 2 full cycles from 0 to 4.
= 2, so the period is 2.
Divide the period into fourths. Plot five points for
the first cycle. Use 2 for the maximum and –2 for
the minimum.
Repeat the pattern for the second cycle.
Sketch the curve.
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
Suppose 8-in. waves occur every 6 s. Write an equation that
models the height of a water molecule as it moves from crest to crest.
The equation will have the form y = a cos b . Find the values for a and b.
8
a= 2
amplitude =
=4
maximum – minimum
2
Simplify.
2
period = b
2
6= b
2
b= b
= 3
Use the formula for the period.
The period is 6. Substitute.
b
Multiply each side by 6 .
Simplify.
An equation that models the height of the water molecule is y = 4 cos 3 .
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
2x
In the function y = –2 cos 3 , for which values of x is the
function equal to 1?
Solve the equation, 1 = –2 cos
2x
, for the interval of 0 to 10.
3
Step 1: Use two equations. Graph the equations
y = 1 and y = –2 cos 2x on the same screen.
3
Step 2: Use the Intersect feature to find the
points at which the two graphs intersect.
The graph show two solutions in the interval. They are x = 3.14 and 6.28.
The solution to the equation for the interval 0 <
– 10 is 3.14 and 6.28, or
–x<
and 2 .
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
6.
pages 732–734 Exercises
1. 2 , 3; max: 0, 2
; min:
; zeros:
2
,3
2
2. 2 , 1; max: 0, 2 , 4 , 2 ;
3
3
5
min: , ,
; zeros: , , 5 , 7 , 3 , 11
6 2 6
3
6
6 2
3
3.
, 1; max: 0, , 2 ; min: , 3 ;
2 2
3
zeros: ,
,5 ,7
4 4 4
4
4. 2 , 2; max: ; min: 0, 2 ; zeros: , 3
2 2
3
7.
8.
5.
9.
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
cos 2
22. 2 , –3 <
–y <
– 3, 3
23. , –1 <
–y <
– 1, 1
cos
24. 4 , –2 <
–y<
– 2, 2
10. y = 2 cos 2
11. y =
12. y =
2
3
1, 1
25. 4 , – 1 <
y<
–
–
3
3 3
26. 6 , –3 <
–y <
– 3, 3
13. y = –3 cos 2
14. y = 2 cos
4
15. y = 4 cos 2
3
27. 2
1,1
,–1 <
y<
–
–
3
2
2 2
28. 4 , –16 <
–y <
– 16, 16
16. 0.52, 2.62, 3.67, 5.76
3
18. 0.55, 1.45, 2.55, 3.45, 4.55, 5.45
29. 2, –0.7 <
–y <
– 0.7, 0.7
30. 0.64, 2.50
19. 2.52
31. 1.83, 2.88, 4.97, 6.02
20. 0.00
32. 0.50, 2.50, 4.50
17. 1.98, 4.30
21. 0.86, 5.14
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
33. a. 3.79, 5.64
b. 10.07, 11.92; these values
are the sums of the values
from part (a) and 2 .
35. a. 5.5 ft; 1.5 ft
b. about 12 h 22 min
c. y = 1.5 cos 2
t
742
d. anytime except between
7:49 A.M. and 12:39 P.M.
34. a.
36. a. 4 s; 6 ft
b. y = –6 cos 2 t
c.
b. Answers may vary. Sample:
0 s, 4 s, 8 s, 12 s
c. 2 s; 2 s
5
d. –6 cos
2
t=3
e. No; at 13.5 s you are right of the
puddle and moving to the right.
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
38. y = cos
37. a.
b.
x or y = –cos
12
x
39. On the unit circle, the x-values
of – are equal to the x-values
of , so cos(– ) = cos . –cos
is the opposite of cos , so these
graphs are reflections of each
other over the x-axis.
shift of 2 units to the right
40. A
41. A
They are the same.
c. To write a sine function as a
cosine function, replace sin
with cos and replace with –
12
42. C
43. D
2
.
44. C
45. C
13-5
The Cosine Function
ALGEBRA 2 LESSON 13-5
46.
49. about 1111
50. about 204
51. about 83
y = sin 6
47.
52. an = 10 • 3n – 1; 10, 30, 90, 270, 810
53. an = 12(–0.3)n – 1; 12, –3.6, 1.08, –0.324, 0.0972
54. an = 900 – 1
3
y = 2.5 sin 2
48.
y = 4 sin 2
13-5
n – 1;
900, –300, 100, – 100 , 100
3
9
The Cosine Function
ALGEBRA 2 LESSON 13-5
1. Sketch the graph of y = 2 cos 2 in the interval from 0 to 6.
2. What is the amplitude of the graph in Question 1?
2
3. Write an equation of the cosine function for this graph.
2x
y = 3 cos 3
13-5
The Tangent Function
ALGEBRA 2 LESSON 13-6
(For help, go to Lessons 13-4 and 13-5.)
Use a calculator to find the sine and cosine of each value of . Then
calculate the ratio sin .
cos
1.
3
radians
3. 90 degrees
5.
5
radians
2
2. 30 degrees
4.
5
radians
6
6. 0 degrees
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
Solutions
1. sin
3
sin 3
0.866; cos = 0.5;
cos
3
3
2. sin 30° = 0.5; cos 30°
0.866
0.5
0.866; sin 30°
cos 30°
1.73
0.5
0.866
3. sin 90° = 1; cos 90° = 0; sin 90° = 1 , undefined
cos 90°
13-6
0
0.58
The Tangent Function
ALGEBRA 2 LESSON 13-6
Solutions (continued)
4. sin 5 = 0.5; cos 5
6
6
5
sin 6
–0.866;
5
cos 6
5
sin 2
5
5
5. sin
= 1; cos
= 0;
2
2
5
cos 2
= 1 , undefined
0
6. sin 0° = 0; cos 0° = 1; sin 0° = 0 = 0
cos 0°
0.5
–0.866
1
13-6
–0.58
The Tangent Function
ALGEBRA 2 LESSON 13-6
Use the graph of y = tan
to find each value.
a. tan –45°
tan –45° = –1
b. tan 0°
tan 0° = 0
c. tan 45°
tan 45° = 1
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
Sketch two cycles of the graph y = tan
period = b
Use the formula for the period.
= 1 =2
Substitute 2 for b and simplify.
2
2
.
1
One cycle occurs in the interval –
Asymptotes occur every 2
to
units, at
.
=– ,
, and 3 .
Sketch the asymptotes.
Plot three points in each cycle.
Sketch the curve.
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
What is the height of the triangle, in the design from
Example 3, when = 18°? What is the height when = 20°?
Step 1: Sketch the graph.
Step 2: Use the TABLE feature.
When
= 18°, the height of the triangle is about 32.5 ft.
When
= 20°, the height of the triangle is about 36.4 ft.
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
pages 737–740 Exercises
12. 2 ,
=–
13.
=–
3
1. 0
2. 0
14.
3. –1
4. undefined
4
3
,
2
2
,
17.
,
3 3
,
8 8
3
=–
2
4
,
2
3
4
15.
18.
16.
19.
5. 1
6. 0
7. 1
8. undefined
9.
10.
11.
2
5
,
=–
50, undefined, –50
,
10 10
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
20.
23. 6
–100, undefined, 100
21.
24. 2
5
51.8, 125,
301.8
22. a.
25. 2
b.
c.
2
3
14.3 ft
20.2 ft
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
26. 1.11, 4.25
30. a. 140.4 ft2
b.
27. 2.03, 5.18
28. 0.08, 1.65, 3.22, 4.79
29. a.
c.
d.
b. Check students’ work; doubling
the coefficient of the tangent
function also doubles the output.
c. Answers may vary. Sample:
the values of y = 600 tan x will
be three times greater than the
values of y = 200 tan x.
13-6
1.7 in., 5.2 in.
5.2 in.2, 15.6 in.2
3888 tiles, 1296 tiles
31. Check students’ work.
32. The asymptotes occur at x = –
and x =
2b
2b
; adding or subtracting
multiples of their difference, b, will
give other asymptote values.
The Tangent Function
ALGEBRA 2 LESSON 13-6
33. 200
44. a. Check students’ work.
b. The new pattern is asymptote
—(–a)—zero—(a)—
asymptote.
42. a.
34. 0
35. 135
36. –162
b.
c.
37. 70
38. y = tan 1 x
6.9 ft
27.7 ft2
166.3 ft2
43. a.
2
they are similar by AA. sin
39. y = –tan 1 x
2
AP = BQ = tan
OA OB
1
40. y = –tan x or
y = tan (–x)
41. y = tan (2x)
45. Answers may vary. Sample:
Triangles OAP and OBQ both
share the angle and each
triangle has a right angle, so
b.
c.
130 ft
61,500 ft2
13-6
= tan .
cos
. Thus sin
cos
=
The Tangent Function
ALGEBRA 2 LESSON 13-6
46. C
47. H
48. D
49. G
50. D
51. [2] There is no discussion
of the amplitude of the
tangent function because
the tangent function has
no max. or min. value.
[1] incomplete explanation
52. [4] (Student graphs y = tan and y = sin
correctly, showing 2 periods of y = tan
and 1 period of y = sin .) The period of
y = tan is half the period of y = sin .
[3] statement correct and graph accurate,
but doesn’t illustrate 2 periods of
y = tan
[2] accurate graph with incorrect answer
OR inaccurate graph with correct answer
[1] answer only, with no graph
53. 1.32, 4.97
54. 1.77, 4.51
55. 1.93, 4.35
56. 6.15
57. 2.95, 5.43
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
58. 0.44, 1.56, 2.44, 3.56, 4.44, 5.56
59.
5.9, 6, 4 and 6
60.
42.6, 42, 42
61.
8.0, 8, 7.9 and 8.5
62. 83
63. –227
64. 66
65. –8.3
66. 145
67. –332
13-6
The Tangent Function
ALGEBRA 2 LESSON 13-6
1. Sketch the graph of y =
1
tan 2 in the interval from 0 to 2 .
2
2. What is the period of the graph in Question 1?
2
3. If you know that the function y = tan x is undefined for x = a, what
does that tell you about the graph at the value a?
The graph tends toward the vertical asymptote x = a.
13-6
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
(For help, go to Lesson 2-6.)
Graph each pair of equations on the same coordinate plane. Identify
each translation as horizontal, vertical, or diagonal.
1. y = 2x, y = 2x + 5
2. g(x) = | x |, ƒ(x) = | x + 3 |
3. y = –x, y = –x – 1
4. g(x) = | x |, h(x) = | x | – 4
5. y = –| x |, y = –| x – 2 | + 1
6. y = x2, y = (x + 3)2 – 2
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
Solutions
1. y = 2x
y = 2x + 5
vertical or
horizontal
translation
2. g(x) = | x |
f(x) = | x + 3|
horizontal
translation
3. y = –x
y = –x – 1
horizontal,
vertical, or
diagonal
translation
4. g(x) = | x |
h(x) = | x | – 4
vertical
translation
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
Solutions (continued)
5. y = –| x |
y = –| x – 2| + 1
diagonal
translation
6. y = x2
y = (x + 3)2 – 2
diagonal
translation
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
What is the value of h in each translation? Describe each
phase shift.
a. g(x) = ƒ(x + 3)
b. y = | x – 2 |
h = –3;
h = 2;
the phase shift is 3
units to the left.
the phase shift is 2
units to the right.
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
Use the graph of y = sin x. Sketch each translation of the
graph in the interval 0 <
–x<
–2 .
a. y = sin x + 2
Translate the graph y = sin x 2 units up.
b. y = sin (x –
)
Translate the graph y = sin x
13-7
units to the right.
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
Use the graph of y = sin x. Sketch the translation
y = sin x + 2 – 1 in the interval 0 <
–x <
–2 .
Translate the parent function
the left and 1 unit down.
13-7
2
units to
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
Sketch the graph of y = sin 3 x – 2 – 1 in the interval 0 to 2 .
2
Since a = 1 and b = 3, the graph is a translation of y = sin 3x.
Step 1: Sketch one cycle of y = sin 3x.
Use five points in the pattern
zero-max-zero-min-zero.
Step 2: Since h =
2
and k = – 1 , translate the graph
1
unit down.
2
2
2
units to the right and
Extend the period pattern from 0 to 2 .
Sketch the graph.
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
Write an equation for each translation.
a. y = cos x,
b. y = –sin x, 3 units to the right
units up
A shift up or down means k =
.
Up means use a plus sign.
An equation is y = cos x +
A shift left or right means h = 3.
Right means use a minus sign.
An equation is y = –sin (x – 3).
.
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
Use the following graph, which shows the model for the data
given in Example 6, to draw some conclusions about the weather in
New Orleans.
One can draw the conclusion that the temperature
for New Orleans is not steady, but fluctuates
throughout the year.
More exactly that the temperature varies approximately through 48 degrees.
One can also determine that the hottest temperature for the city never
gets above 100 degrees.
Nor does the coldest temperature ever get lower than 50 degrees.
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
pages 746–748 Exercises
8.
11.
9.
12.
10.
13.
1. –1; 1 unit to the left
2. –2; 2 units to the left
3. 1.6; 1.6 units to the right
4. 3; 3 units to the right
5. –
;
6. 5
; 5
7
units to the left
7
units to the right
7.
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
14.
17. 3, 2
18. 4,
19. 1, 2
; 1 unit up
; 1 unit left and 2 units down
;
2
units left and 2 units up
20. 1, 2; 3 units right and 2 units up
15.
21.
22.
16.
23.
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
24.
27.
30.
25.
28.
31. y = sin (x +
32. y = cos x –
)
2
33. y = sin x + 3
34. y = cos (x – 1.5)
26.
29.
35. y = cos
3
2
36. y = sin x – 3
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
42. y = –10 cos
37. a.
43. a.
2
b. –
10
; y = 10 sin
; sin x = cos x –
2
b.
(x – 228) + 77.5
y = 8.5 cos 2
365
38. y = sin (x – 2) – 4
39. y = cos (x + 3) +
40. y = sin x –
+ 3.5
2
41. y = 2 cos x –
y = 2 sin x +
3
6
x–
2
2
; cos x = sin x +
44. a. 14.5 sin 2
10
2
(x – 105.75) + 76.5
365
b. The difference between the two
models is the horizontal shift.
c. about 66°F
d. March 20 (day 79)
45. a. Check students’ work.
b. g(x) = ƒ(x + 4) – 3
– 1;
46. a. y = 3 sin 2(x – 2) + 1
b. 3, ; 2 units right and 1 unit up
–1
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
47.
50.
53.
48.
51.
54.
49.
52.
55. C
56. G
57. B
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
58. [2] If the function y = cos
is
shifted to the right by
60. [4]
sin
2
–
2
= 3 sin
– sin
0 = sin
the same as y = sin .
2
– sin
0 = 2 sin
2
2
, which is
So a = 1, and b = –
= 3 sin
0 = 2 sin
radians, the result is
y = cos
sin
sin = 0 at – , 0, , and 2
[3] one calculation error
[2] incomplete answer
[1] answer only, with no work shown
.
[1] answer only, with no
work shown
59. [2] y = 4 sin x has an amplitude
61. ; = – ,
of 4, but a min. value of –1.
6
12 12
Shift the graph up 5 units
62. 4 ; = –2 , 2
so the min. value is 4. The
function is y = 4 sin + 5.
[1] answer only, with no work shown
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
63. 2
3
64. 6 ;
=–
;
,
3 3
= –3 , 3
65. 0.0064
66. 0.3456
67.
0.136
68.
0.198
69. 62
70. 16,383
71. –335,923
72. 96.09375
13-7
Translating Sine and Cosine Functions
ALGEBRA 2 LESSON 13-7
1. Sketch the graph of y = 1 sin x – 1 in the interval from 0 to 2 .
2
2. What is the value of h in the translation g(x) = ƒ(x + 4)?
h = –4
3. Describe the phase shift in the translation g(x) = ƒ(x + 4).
four units to the left
4. Write an equation for the translation of y = sin x,
y = sin x – 2
13-7
2
units down.
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
(For help, go to Lesson 7-8.)
Find the reciprocal of each fraction.
1.
9
13
2. – 5
8
3.
1
2
4. 4m
5. – 14
15
t
Graph each pair of functions on the same coordinate plane.
6. y = x, y = –x
7. y = x2, y = ±
8. y = | 2x |, y = –| 2x |
9. y = –6x2, y = ±
13-8
x
6x
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
(For help, go to Lesson 7-8.)
Solutions
1. The reciprocal of 9 is 13 .
2. The reciprocal of – 5 is – 8 .
3. The reciprocal of 1 is 2 or 2 .
4. The reciprocal of 4m is 15 .
13
9
8
1
2
15
5. The reciprocal of – 14 is – t .
t
14
7. y = x2
y=± x
6. y = x
y = –x
13-8
5
4m
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
(For help, go to Lesson 7-8.)
Solutions (continued)
9. y = –6x2
y = ± 6x
8. y = | 2x|
y = –| 2x|
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
a. Find csc 45°.
Use a calculator in degree mode.
Use the definition.
1
csc 45° = sin 45°
b. Suppose cos
sec
1
= cos
=
=
1
4
5
5
4
4
= 5 . Find sec .
Use the definition.
Substitute.
Simplify.
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
Find the exact value of csc 45°.
Use the unit circle to find the exact value of sin 45°. Then write the reciprocal.
The y-coordinate of point P is
1
csc 45° = sin 45°
2
.
2
Use the definition.
1
=
=
=
2
2
2
2
2
Substitute.
Simplify.
Rationalize the denominator.
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
Evaluate each expression. Use your calculator’s radian
mode. Round to the nearest thousandth.
1
a. cot = 5 =
tan
cot
5
1.376
b. sec (–2) =
1
cos (–2)
5
sec (–2)
13-8
–2.403
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
Graph y = cos x and y = sec x in the interval from 0 to 2 .
Step 1: Make a table of values.
x
0
cos x
1
sec x
1
2
2
3
0.9 .5
0
–.5 –0.9 –1 –0.9 –.5
0
.5 0.9 1
1.2
–
–2 –1.2 –1 –1.2 –2
–
2
6
3
2
5
6
7
6
4
3
3
2
5
3
Step 2: Plot the points and sketch the graphs.
y = sec x will have a vertical asymptote
whenever its denominator (cos x) is 0.
13-8
11
6
2
1.2 1
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
Use the graph of y = sec x to find the value of sec 13°.
1
Step 1: Use degree mode. Graph y = cos x .
Step 2: Use the TABLE feature.
sec 13°
1.0263
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
Use the function y = 6 sec
to form an angle of 55°.
y = 6 sec
1
= 6 cos
6
= cos
to find the length of string needed
Use the definition of secant.
Simplify.
Graph the function. Use the Value feature.
To form an angle of 55°, the string must be about 10.5 ft long.
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
11. 0
23. 1
1. 1.02
12. –1
24. –1
2. 1.02
13. undefined
25. –1.25
3. –0.70
14. 2
26. 17.13
4. –1.06
15. undefined
27. 1.73
5. 3
16.
28. 1.02
pages 752–755 Exercises
3
3
3
4
6. 18
13
17. undefined
7. – 5
3
8. – 3
4
19. 2
29.
18. 2
20.
2
9.
2
21. –7.02
10.
3
3
22. –1
13-8
30.
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
31.
39. 1.7321
44. –1
40. 0.5774
45. –1
41. a.
46.
32.
33. 1.1547
34. 5.7588
35. –2.9238
47.
b.
c.
d.
28.3 ft
23.1 ft
20.7 ft
36. 2
42. –1
37. 1.0642
43. 2
38. 1.3054
3 ; 1.15
3
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
48.
49.
54. a. domain: all real numbers
except multiples of ; range:
all real numbers >
– 1 or <
– –1;
period 2
b. 1
c. –1
55. a. Reciprocals have the same sign.
b. The reciprocal of –1 is –1.
56. csc 180° is undefined because
sin 180° is 0 and csc
50. B
51. C
52. A
53. D
=
1
sin
.
57. sec 90° is undefined because
cos 90° is 0 and sec
=
1 .
cos
58. cot 0° is undefined because
tan 0° is 0 and cot
13-8
=
1 .
tan
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
59. a.
b.
The domain of y = tan x is
all real numbers except odd
multiples of
b. 63.9 ft
c. 69.3 ft
d. 41.4°; 60.9 ft
60. a.
c.
2
asymptotes. The domain of
y = cot x is all real numbers
except multiples of , which
are its asymptotes. The range
of both functions is all real numbers.
The graphs have the same period
and range. Their asymptotes
are shifted by
d.
2
.
Answers may vary. Sample:
x=
13-8
, which are its
4
,x= 3
4
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
61.
63.
65.
1 unit down
3 units up
4 units right
64.
62.
66.
2
2
units right
units left
2 units left, 1 unit down
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
b. y = –cos x—domain: all real numbers;
range: all real numbers between –1 and 1,
inclusive; period: 2 ; y = –sec x—domain:
67.
units left, 3 units up
68.
all real numbers except odd multiples of ;
2
range: all real numbers except those
between –1 and 1; period: 2
c. Multiples of
±1 = cos (n
6
69. a.
units right, 2 units down
; sec (n
)=
1
cos (n
)
= 1 =
±1
).
d. Answers may vary. Sample: The graphs
have the same period and their signs are
always the same. However, they have no
range values in common except 1 and –1.
e. The signs of –sec x and –cos x are the same
because reciprocals have the same sign.
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
70. a.
b.
73. a.
II
I
71. y = sec x and y = csc x are
not parabolas because
parabolas are not restricted
by asymptotes, whereas the
branches of y = sec x and
y = csc x are between
asymptotes.
72. y = cos 3x has 3 cycles for
each cycle of y = cos x. Thus,
for each cycle of y = sec x,
y = sec 3x has 3 cycles. Each
b. Answers may vary. Sample:
Given y = cot bx, as |b| decreases,
the period increases; as |b| increases,
the period decreases. If b < 0, cot x
begins each branch with negative
y-values and ends with positive
y-values; the opposite is true for b > 0.
cycle of y = sec 3x is 1 as wide
3
as one cycle of y = sec x.
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
74. a.
b. Answers may vary. Sample:
Given y = b sec x, as |b|
increases, the branches move
further from the x-axis, but the
asymptotes do not change.
If b < 0, it is a reflection in the
x-axis of y = |b| sec x.
75. 40
9
76. 41
9
77. 9
40
78. 41
40
79. 1.35
80. 0.6
81. 1.25
82. 3.4
83. 2, 2
; 5 units down
84. 1, 2
; 4 units left, 7 units down
85. 3, 2
; units left, 4 units up
86. 5, 2; 1.5 units right, 8 units down
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
91. a. 1(3) + 1(3) + 1(0); 6 units2
b. 1(4) + 1(4) + 1(3); 11 units2
87.
92. a. 1(7) + 1(7) + 1(4); 18 units2
b. 1(8) + 1(8) + 1(7); 23 units2
88.
89. a. 1(0) + 1(0) + 1(3); 3 units2
b. 1(3) + 1(3) + 1(12); 18 units2
90. a. 1(4) + 1(4) + 1(5); 13 units2
b. 1(5) + 1(5) + 1(8); 18 units2
13-8
Reciprocal Trigonometric Functions
ALGEBRA 2 LESSON 13-8
1. What are the equations of the two lines that define the maximum
and minimum values for the graphs of y = sin x and y = cos x?
y = 1 and y = –1
2. Which basic trigonometric functions have vertical asymptotes?
y = tan x, y = cot x,
y = sec x, y = csc x
3. What is the range of y-values that y =/ sec x and y =/ csc x cannot
have?
–1 < y < 1
13-8
Periodic Functions and Trigonometry
ALGEBRA 2 CHAPTER 13
Page 760
1. periodic; 4, 2
12. 1 cycle; 2, 2
2. not periodic
14. Check students’ work.
3. 328°
by 180° . Example:
5. 15°
7. 2
3
8. 10
3
13. 1 cycle; 3, 4
2
4
2
15. Answers may vary. Sample:
Multiply the radian measure
4. 131°
6. – 5
18. 2,
, –3.93
, 2.09
2
3
radians • 180° = 120°
19.
3
,5
,2
20.
3
; 1.05, 5.24
; 1.05, 2.09
3
21.
, 7 , 13 , 19 ;
12
12
12 12
3
16. 42 in.
17. 4,
0.26, 1.83, 3.40, 4.97
, 10.47
22. 0
9. 150°
10. –450°
11. 46°
13-A
Periodic Functions and Trigonometry
ALGEBRA 2 CHAPTER 13
23.
29. y = sin x – 1
26.
30. y = cos (x – 7.5)
31. y = sin (x + 3) – 1.5
32. y = cos x –
24.
27.
33. 1
25.
2
34. 1
2
35. 1
2
28.
36. 2
37. undefined
38.
13-A
3
2
+8
Periodic Functions and Trigonometry
ALGEBRA 2 CHAPTER 13
39.
2
2
40. –
43.
3
3
41.
44.
42.
13-A