Copyright © Big Ideas Learning, LLC Algebra 2 447All rights reserved. Worked-Out Solutions
Chapter 9
Chapter 9 Maintaining Mathematical Profi ciency (p. 459)
1. ∣ 4 ∣ = 4 ∣ 6 + 4 ∣ = ∣ 10 ∣ = 10
∣ 2 − 9 ∣ = ∣ −7 ∣ = 7 − ∣ 7 ∣ = −7
So, the order is − ∣ 7 ∣ , ∣ 4 ∣ , ∣ 2 − 9 ∣ , and ∣ 6 + 4 ∣ .
2. ∣ 9 − 3 ∣ = ∣ 6 ∣ = 6 ∣ −4 ∣ = 4
∣ 0 ∣ = 0 ∣ −5 ∣
— ∣ 2 ∣
= 5 — 2 = 2.5
So, the order is ∣ 0 ∣ , ∣ −5 ∣
— ∣ 2 ∣
, ∣ −4 ∣ , and ∣ 9 − 3 ∣ .
3.
∣ −83 ∣ = ∣ −512 ∣ = 512 ∣ 9 − 1 ∣ = ∣ 8 ∣ = 8
∣ −2 ⋅ 8 ∣ = ∣ −16 ∣ = 16 ∣ 9 ∣ + ∣ −2 ∣ − ∣ −1 ∣ = 9 + 2 − 1 = 10
So, the order is ∣ 9 − 1 ∣ , ∣ 9 ∣ + ∣ −2 ∣ − ∣ −1 ∣ , ∣ −2 ⋅ 8 ∣ , and ∣ −83 ∣ .
4.
∣ −4 + 20 ∣ = ∣ 16 ∣ = 16 ∣ 5 ∣ − ∣ 3 ⋅ 2 ∣ = 5 − ∣ 6 ∣ = 5 − 6 = −1
− ∣ 42 ∣ = − ∣ 16 ∣ = −16 ∣ −15 ∣ = 15
So, the order is − ∣ 42 ∣ , ∣ 5 ∣ − ∣ 3 ⋅ 2 ∣ , ∣ −15 ∣ , and ∣ −4 + 20 ∣ .
5. a2 + b2 = c2
122 + 52 = c2
144 + 25 = c2
169 = c2
13 = c
So, the length is 13 meters.
6. a2 + b2 = c2
72 + b2 = 252
49 + b2 = 625
b2 = 576
b = 24
So, the length is 24 feet.
7. a2 + b2 = c2
9.62 + 7.22 = c2
92.16 + 51.84 = c2
144 = c2
12 = c
So, the length is 12 millimeters.
8. a2 + b2 = c2
a2 + 212 = 352
a2 + 441 = 1225
a2 = 784
a = 28
So, the length is 28 kilometers.
9. a2 + b2 = c2
a2 + 42 = ( 12 1 —
3 ) 2
a2 + 16 = 152 1 —
9
a2 = 136 1 —
9
a = 11 2 —
3
So, the length is 11 2 —
3 inches.
10. a2 + b2 = c2
( 3 — 10
) 2 + b2 = ( 1 — 2 ) 2
9 —
100 + b2 =
1 —
4
b2 = 4 —
25
b = 2 —
5
So, the length is 2 —
5 yard, or 0.4 yard.
11. Yes, the triangle is a right triangle. The line passing through
the points ( x1, y1 ) and ( x2, y1 ) is horizontal. The line passing
through the points ( x2, y1 ) and ( x2, y2 ) is vertical. Horizontal
and vertical lines are perpendicular, so the triangle formed
by the line segments connecting ( x1, y1 ) , ( x2, y1 ) , and ( x2, y2 )
contains a right angle.
Chapter 9 Mathematical Practices (p. 460)
1. Because θ = 135°, (x, y) lies on the line y = −x.
x2 + y2 = 1
x2 + (−x)2 = 1
2x2 = 1
x2 = 1 — 2
x = − 1 —
√—
2 = −
√—
2 —
2
The coordinates of (x, y) are ( − √
— 2 —
2 ,
√—
2 —
2 ) .
2. Because θ = 315°, (x, y) lies on the line y = −x.
x2 + y2 = 1
x2 + (−x)2 = 1
2x2 = 1
x2 = 1 — 2
x = 1 —
√—
2 =
√—
2 —
2
The coordinates of (x, y) are ( √—
2 —
2 , −
√—
2 —
2 ) .
448 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
3. Because θ = 225°, (x, y) lies on the line y = x.
x2 + y2 = 1
x2 + x2 = 1
2x2 = 1
x2 = 1 — 2
x = − 1 —
√—
2 = −
√—
2 —
2
The coordinates of (x, y) are ( − √
— 2 —
2 , −
√—
2 —
2 ) .
9.1 Explorations (p. 461)
1. sin 30° = 1 — 2 cos 30° =
√—
3 —
2 tan 30° =
√—
3 —
3
sin 45° = √
— 2 —
2 cos 45° =
√—
2 —
2 tan 45° = 1
sin 60° = √
— 3 —
2 cos 60° = 1 —
2 tan 60° = √
— 3
2. a. This is true because (90° − θ) represents the other acute
angle in the triangle, cos(90° − θ) = opp —
hyp , which is the
same ratio as sin θ.
b. This is true because (90° − θ) represents the other acute
angle in the triangle, sin(90° − θ) = adj —
hyp , which is the
same ratio as cos θ.
c. This is true by defi nition because cosecant is the
reciprocal of sine.
d. This is true by defi nition because cotangent is the
reciprocal of tangent.
e. (sin θ)2 + (cos θ)2 = ( opp —
hyp ) 2 + ( adj
— hyp
) 2
= opp2
— hyp2 + adj2
— hyp2
= 1 —
hyp2 + (opp2 + adj2)
= 1 —
hyp2 ⋅ hyp2
= 1
So, (sin θ)2 + (cos θ)2 = 1.
f. (sec θ)2 − (tan θ)2 = ( hyp —
adj ) 2 − ( opp
— adj
) 2
= hyp2
— adj2
− opp2
— adj2
= hyp2 − opp2
—— adj2
= adj2 —
adj2
= 1
So, (sec θ)2 − (tan θ)2 = 1.
3. When the side lengths of a right triangle are known,
the ratios of the side lengths can be used to fi nd the
trigonometric function of an acute angle.
4. sin 25° = y —
1 cos 25° =
x —
1
sin 25° = y cos 25° = x
0.42 ≈ y 0.91 = x
So, x ≈ 0.91 and y ≈ 0.42.
9.1 Monitoring Progress (pp. 463–465)
1. From the Pythagorean Theorem, the length of the hypotenuse is
hyp. = √—
42 + 32
= √—
25
= 5.
Using adj. = 4, opp. = 3, and hyp. = 5, the values of the six
trigonometric functions of θ are:
sin θ = 3 — 5 cos θ = 4 — 5 tan θ = 3 — 4
csc θ = 5 — 3 sec θ = 5 — 4 cot θ = 4 — 3
2. From the Pythagorean Theorem, the length of the adjacent
side is
adj. = √—
172 − 152
= √—
64
= 8.
Using adj. = 8, opp. = 15, and hyp. = 17, the values of the
six trigonometric functions of θ are:
sin θ = 15 —
17 cos θ = 8 — 17 tan θ = 15
— 8
csc θ = 17 —
15 sec θ = 17
— 8 cot θ = 8 — 15
3. From the Pythagorean Theorem, the length of the opposite
side is
opp. = √——
( 5 √—
2 ) 2 − 52
= √—
25
= 5.
Using adj. = 5, opp. = 5, and hyp. = 5 √—
2 , the values of the
six trigonometric functions of θ are:
sin θ = √
— 2 —
2 cos θ =
√—
2 —
2 tan θ = 1
csc θ = √—
2 sec θ = √—
2 cot θ = 1
Copyright © Big Ideas Learning, LLC Algebra 2 449All rights reserved. Worked-Out Solutions
Chapter 9
4. Step 1 Draw a right triangle with acute angle θ such that
the leg adjacent θ has length 7 and the hypotenuse
has length 10.
opp. = 51
7
10
θ
Step 2 Find the length of the opposite side. By the
Pythagorean Theorem, the length of the other leg is
opp. = √—
102 − 72 = √—
51 .
Step 3 Find the values of the remaining fi ve trigonometric
functions. Because cos θ = 7 — 10
, sec θ = hyp —
adj = 10
— 7 .
The others are:
sin θ = √
— 51 —
10 tan θ =
√—
51 —
7
csc θ = 10 √
— 51 —
51 cot θ =
7 √—
51 —
51
5. Write an equation using a trigonometric function that
involves the ratio of x and 6. Solve the equation for x.
cos 45° = adj —
hyp
√
— 2 —
2 = x —
6
3 √—
2 = x
The length of the side is x = 3 √—
2 ≈ 4.24.
6. Because the triangle is a right triangle, A and B are
complementary angles. So, A = 90° − 45° = 45°. Next,
write two equations using trigonometric functions, one that
involves the ratio of a and 5, and one that involves b and 5.
Solve the fi rst equation for a and the second for b.
sin 45° = opp —
hyp cos 45° = adj
— hyp
sin 45° = a — 5 cos 45° = b —
5
5(sin 45°) = a 5(cos 45°) = b
5 √
— 2 —
2 = a 5 √
— 2 —
2 = b
So, A = 45°, a = 5 √—
2 —
2 , and b = 5 √
— 2 —
2 .
7. Because the triangle is a right triangle, A and B are
complementary angles. So, B = 90° − 32° = 58°. Next,
write two equations using trigonometric functions, one that
involves the ratio of a and 10, and one that involves the ratio
of c and 10.
tan 32° = opp —
adj sec 32° = hyp
— adj
tan 32° = a — 10
sec 32° = c — 10
10 tan 32° = a 10 sec 32° = c
6.25 ≈ a 11.79 ≈ c
So, B = 58°, c ≈ 11.79, and a ≈ 6.25.
8. Because the triangle is a right triangle, A and B are
complementary angles. So, B = 90° − 71° = 19°. Next,
write two equations using trigonometric functions, one that
involves the ratio of a and 20, and one that involves the ratio
of b and 20.
cos 19° = adj —
hyp sin 19° = opp
— hyp
cos 19° = a — 20
sin 19° = b — 20
20 cos 19° = a 20 sin 19° = b
18.91 ≈ a 6.51 ≈ b
So, B = 19°, a ≈ 18.91, and b ≈ 6.51.
9. Because the triangle is a right triangle, A and B are
complementary angles. So, A = 90° − 60° = 30°. Next,
write two equations using trigonometric functions, one that
involves the ratio of b and 7, and one that involves the ratio
of c and 7.
cot 30° = adj —
opp sec 30° = hyp
— adj
cot 30° = b — 7 sec 30° = c —
7
7 cot 30° = b 7 sec 30° = c
12.12 ≈ b 14 = c
So, A = 30°, b ≈ 12.12, and c = 14.
10. sec 76° = x — 2
2(sec 76°) = x
8.3 ≈ x
The length is about 8.3 miles.
11.
h72 ft
38°
Write and solve an equation to fi nd the height h.
sin 38° = h — 72
72(sin 38°) = h
44.3 ≈ h
The height of the parasailer above the boat is about 44.3 feet.
450 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
9.1 Exercises (pp. 466–468)
Vocabulary and Core Concept Check
1. In a right triangle, the two trigonometric functions of θ that
are defi ned using the lengths of the hypotenuse and the side
adjacent to θ are cosine and secant.
2. The angle of elevation and angle of depression are always
equal.
3. To solve a right triangle, the missing angles and side lengths
must be found.
4. “What is the ratio of the side opposite θ to the hypotenuse?”
is different. This ratio is 4 —
6 , or
2 —
3 .
csc θ = hyp.
— opp. = 6 —
4 =
3 —
2 ; 1
— sin θ =
1 —
opp.
— hyp.
=
1 —
4 —
6
= 6 —
4 =
3 —
2 ;
= hyp. —
opp. =
6 —
4 =
3 —
2
Monitoring Progress and Modeling with Mathematics
5. From the Pythagorean Theorem, the length of the
hypotenuse is
hyp. = √—
92 + 122
= √—
225
= 15.
Using adj. = 9, opp. = 12, and hyp. = 15, the six
trigonometric functions of θ are:
sin θ = opp.
— hyp.
= 12
— 15
= 4 —
5 cos θ =
adj. —
hyp. =
9 —
15 =
3 —
5
tan θ = opp.
— adj.
= 12 —
9 = 4 —
3 csc θ = hyp.
— opp.
= 15 —
12 = 5 —
4
sec θ = hyp. —
adj. = 15
— 9 = 5 —
3 cot θ = adj.
— opp.
= 9 — 12
= 3 — 4
6. From the Pythagorean Theorem, the length of the
hypotenuse is
hyp. = √—
82 + 62
= √—
100
= 10.
Using adj. = 8, opp. = 6, and hyp. = 10, the values of the
six trigonometric functions of θ are:
sin θ = opp.
— hyp.
= 6 —
10 =
3 —
5 cos θ = adj.
— hyp.
= 8 — 10
= 4 — 5
tan θ = opp. —
adj. = 6 —
8 = 3 —
4 csc θ = hyp.
— opp.
= 10 —
6 = 5 —
3
sec θ = hyp. —
adj. = 10
— 8 = 5 —
4 cot θ = adj.
— opp.
= 8 — 6 = 4 —
3
7. From the Pythagorean Theorem, the length of the adjacent
side to θ is
adj. = √—
72 − 52
= √—
24
= 2 √—
6 .
Using adj. = 2 √—
6 , opp. = 5, and hyp. = 7, the values of the
six trigonometric functions of θ are:
sin θ = opp.
— hyp.
= 5 —
7 cos θ =
adj. —
hyp. =
2 √—
6 —
7
tan θ = opp. —
adj. =
5 —
2 √—
6 =
5 √—
6 —
12 csc θ =
hyp. — opp. =
7 —
5
sec θ = hyp.
— adj.
= 7 —
2 √—
6 = 7 √
— 6 —
12 cot θ =
adj. — opp. =
2 √—
6 —
5
8. From the Pythagorean Theorem, the length of the adjacent
side to θ is
adj. = √—
92 − 32
= √—
72
= 6 √—
2 .
Using adj. = 6 √—
2 , opp. = 3, and hyp. = 9, the values of the
six trigonometric functions of θ are:
sin θ = opp.
— hyp.
= 3 —
9 =
1 —
3
cos θ = adj.
— hyp.
= 6 √
— 2 —
9 =
2 √—
2 —
3
tan θ = opp.
— adj.
= 3 —
6 √—
2 =
3 √—
2 —
12 =
√—
2 —
4
csc θ = hyp.
— opp. = 9 —
3 = 3
sec θ = hyp.
— adj.
= 9 —
6 √—
2 =
9 √—
2 —
12 =
3 √—
2 —
12
cot θ = adj.
— opp. = 6 √
— 2 —
3 = 2 √
— 2
Copyright © Big Ideas Learning, LLC Algebra 2 451All rights reserved. Worked-Out Solutions
Chapter 9
9. From the Pythagorean Theorem, the length of the opposite
side to θ is
opp. = √—
182 − 102
= √—
224
= 4 √—
14 .
Using adj. = 10, opp. = 4 √—
14 , and hyp. = 18, the values of
the six trigonometric functions of θ are
sin θ = opp.
— hyp.
= 4 √
— 14 —
18 =
2 √—
14 —
9
cos θ =
adj. —
hyp. = 10
— 18
= 5 — 9
tan θ = opp.
— adj.
= 4 √
— 14 —
10 =
2 √—
14 —
5
csc θ = hyp.
— opp. = 18 —
4 √—
14 =
18 √—
14 —
56 =
9 √—
14 —
28
sec θ = hyp.
— adj
= 18 —
10 = 9 —
5
cot θ = adj.
— opp. = 1c —
4 √—
14 =
10 √—
14 —
56 =
5 √—
14 —
28
10. From the Pythagorean Theorem, the length of the opposite
side to θ is
opp. = √—
262 − 142
= √—
480
= 4 √—
30 .
Using adj. = 14, opp. = 4 √—
30 , and hyp. = 26, the values of
the six trigonometric functions of θ are
sin θ = opp.
— hyp.
= 4 √
— 30 —
26 =
2 √—
30 —
13
cos θ = adj.
— hyp.
= 14
— 26
= 7 —
13
tan θ = opp.
— adj.
= 4 √
— 30 —
14 =
2 √—
30 —
7
csc θ = hyp.
— opp. = 26 —
4 √—
30 =
26 √—
30 —
120 =
13 √—
30 —
60
sec θ = hyp.
— adj.
= 26
— 14
= 13
— 7
cot θ = adj.
— opp. = 14 —
4 √—
30 =
14 √—
30 —
120 =
7 √—
30 —
60
11.
9
497
θ
sin θ = opp.
— hyp.
= 4 —
√—
97 =
4 √—
97 —
97 cot θ =
adj. — opp. =
9 —
4
cos θ = adj.
— hyp.
= 9 —
√—
97 =
9 √—
97 —
97 csc θ =
hyp. — opp. =
√—
97 —
4
12. sin(90° − θ) = cos θ cos(90° − θ) = sin θ tan(90° − θ) = cot θ cot(90° − θ) = tan θ sec(90° − θ) = csc θ csc(90° − θ) = sec θ The expression 90° − θ relates a trigonometric function with
its cofunction.
13. Step 1 Draw a right triangle with acute angle θ such that
the leg opposite θ has length 7 and the hypotenuse
has length 11.
Step 2 Find the length of the adjacent side. By the
Pythagorean Theorem, the length of the other leg is
adj. = √—
112 − 72 = 6 √—
2 .
adj. = 6 2
711
θ
Step 3 Find the values of the remaining fi ve trigonometric
functions. Because sin θ = 7 —
11 , csc θ =
11 —
7 . The
other values are:
cos θ = adj.
— hyp.
= 6 √
— 2 —
11
tan θ = opp.
— adj.
= 7 —
6 √—
2 =
7 √—
2 —
12
sec θ = hyp. —
adj. =
11 —
6 √—
2 =
11 √—
2 —
12
cot θ = adj.
— opp. = 6 √
— 2 —
7
14. Step 1 Draw a right triangle with acute angle θ such that
the leg adjacent θ has length 5 and the hypotenuse
has length 12.
Step 2 Find the length of the opposite side. By the
Pythagorean Theorem, the length of the other leg is
opp. = √—
122 − 32 = √—
119 .
opp. = 119
5
12
θ
452 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
Step 3 Find the values of the remaining fi ve trigonometric
functions. Because cos θ = 5 —
12 , sec θ =
12 —
5 . The
other values are:
sin θ = opp.
— hyp
= √—
119 —
12
tan θ = opp.
— adj.
= √—
119 —
5
csc θ = hyp.
— opp. = 12 —
√—
119 =
12 √—
119 —
119
cot θ = adj.
— opp. = 5 —
√—
119 =
5 √—
119 —
119
15. Step 1 Draw a right triangle with acute angle θ such that
the leg adjacent θ has length 6 and the leg opposite
θ has length 7.
Step 2 Find the length of the hypothenuse. By the Pythagorean
Theorem, the length of the hypothenuse is
hyp. = √—
62 + 72 = √—
85 .
hyp. = 85
6
7
θ
Step 3 Find the values of the remaining fi ve trigonometric
functions. Because tan θ = 7 —
6 , cot θ = 6 —
7 . The other
values are:
sin θ = opp.
— hyp.
= 7 —
√—
85 =
7 √—
85 —
85
cos θ = adj.
— hyp.
= 6 —
√—
85 =
6 √—
85 —
85
csc θ = hyp.
— opp = √
— 85 —
7
sec θ = hyp.
— adj.
= √—
85 —
6
16. Step 1 Draw a right triangle with acute angle θ such that
the leg opposite θ has length 8 and the hypotenuse
has length 15.
Step 2 Find the length of the adjacent side. By the
Pythagorean Theorem, the length of the adjacent
side is
adj. = √—
152 − 82 = √—
161 .
adj. = 161
815
θ
Step 3 Find the value of the remaining fi ve trigonometric
functions. Because csc θ = 15
— 8 , sin θ =
8 —
15 . The
other values are:
cos θ = adj.
— hyp.
= √—
161 —
15
tan θ = opp.
— adj.
= 8 —
√—
161 =
8 √—
161 —
161
sec θ = hyp.
— adj.
= 15 —
√—
161 =
15 √—
161 —
161
cot θ = adj.
— opp.
= √—
161 —
8
17. Step 1 Draw a right triangle with acute angle θ such that
the leg adjacent θ has length 9 and the hypotenuse
has length 14.
Step 2 Find the length of the opposite side. By the
Pythagorean Theorem, the length of the opposite
side is
opp. = √—
142 − 92 = √—
115 .
opp. = 115
9
14
θ
Step 3 Find the value of the remaining fi ve trigonometric
functions. Because sec θ = 14
— 9 , cos θ =
9 —
14 . The
other values are:
sin θ = opp.
— hyp.
= √—
115 —
14
tan θ = opp.
— adj.
= √—
115 —
9
csc θ = hyp.
— opp. = 14 —
√—
115 =
14 √—
115 —
115
cot θ = adj.
— opp. = 9 —
√—
115 =
9 √—
115 —
115
Copyright © Big Ideas Learning, LLC Algebra 2 453All rights reserved. Worked-Out Solutions
Chapter 9
18. Step 1 Draw a right triangle with acute angle θ such that
the leg adjacent θ has length 16 and the leg opposite
θ has length 11.
Step 2 Find the length of the hypotenuse. By the
Pythagorean Theorem, the length of the
hypotenuse is
hyp. = √—
162 + 112 = √—
377 .
hyp. = 377
16
11
θ
Step 3 Find the value of the remaining fi ve trigonometric
functions. Because cot θ = 16
— 11
, tan θ = 11
— 16
. The
other values are:
sin θ = opp.
— hyp.
= 11 —
√—
377 =
11 √—
377 —
377
cos θ = adj.
— hyp.
= 16 —
√—
377 =
16 √—
377 —
377
csc θ = hyp.
— opp. = √—
377 —
11
sec θ = hyp. —
adj. =
√—
377 —
16
19. The adjacent side was used instead of the opposite.
sin θ = opp.
— hyp.
= 8 —
17
20. The reciprocal of csc θ is 1 —
sin θ .
csc θ = 1 —
sin θ =
1 —
6 √
— 2 —
11
= 11 √
— 2 —
12
21. Write an equation using a trigonometric function that
involves the ratio of x and 9. Solve the equation for x.
cos 60° = adj.
— hyp.
1 —
2 = x —
9
9 —
2 = x
The length of the side is x = 9 — 2 = 4.5.
22. Write an equation using a trigonometric function that
involves the ratio of x and 6. Solve the equation for x.
cos 60° = adj.
— hyp.
1 —
2 = x —
6
3 = x
The length of the side is x = 3.
23. Write an equation using a trigonometric function that
involves the ratio of x and 12. Solve the equation for x.
sin 30° = opp.
— hyp.
1 —
2 = x —
12
6 = x
The length of the side is x = 6.
24. Write an equation using a trigonometric function that
involves the ratio of x and 13. Solve the equation for x.
sin 30° = opp.
— hyp.
1 —
2 = x —
13
13
— 2 = x
The length of the side is x = 13 —
2 = 6.5.
25. Write an equation using a trigonometric function that
involves the ratio of x and 8. Solve the equation for x.
tan 45° = opp.
— adj.
1 = x — 8
8 = x
The length of the side is x = 8.
26. Write an equation using a trigonometric function that
involves the ratio of x and 7. Solve the equation for x.
tan 45° = opp.
— adj.
1 = x — 7
7 = x
The length of the side is 7.
27. cos 14° ≈ 0.9703
28. tan 31° ≈ 0.6009
29. csc 59° = 1 —
sin 59° ≈ 1.1666
30. sin 23° ≈ 0.3907
31. cot 6° = 1 —
tan 6° ≈ 9.5144
454 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
32. sec 11° = 1 —
cos 11° ≈ 1.0187
33. Because the triangle is a right triangle, A and B are
complementary angles. So, A = 90° − 36° = 54°. Next,
write two equations using trigonometric functions, one that
involves the ratio of b and 23, and one that involves c and 23.
Solve the fi rst equation for b and the second for c.
tan 36° = opp.
— adj.
sec 36° = hyp.
— adj.
tan 36° = b — 23
sec 36° = c — 23
23(tan 36°) = b 23 ( 1 —
cos 36° ) = c
16.71 ≈ b 28.43 ≈ c
So, A = 54°, b ≈ 16.71, and c ≈ 28.43.
34. Because the triangle is a right triangle, A and B are
complementary angles. So, B = 90° − 27° = 63°. Next,
write two equations using trigonometric functions, one that
involves the ratio of a and 9, and one that involves c and 9.
Solve the fi rst equation for a and the second equation for c.
tan 27° = opp.
— adj.
sec 27° = hyp.
— adj.
tan 27° = a — 9 sec 27° = c —
9
9(tan 27°) = a 9 ( 1 —
cos 27° ) = c
4.59 ≈ a 10.10 ≈ c
So, B = 63°, a ≈ 4.59, and c ≈ 10.10.
35. Because the triangle is a right triangle, A and B are
complementary angles. So, B = 90° − 55° = 35°. Next,
write two equations using trigonometric functions, one that
involves the ratio of b and 17, and one that involves c and 17.
Solve the fi rst equation for b and the second equation for c.
cos 55° = adj.
— hyp.
csc 55° = hyp.
— opp.
cos 55° = b — 17
csc 55° = c — 17
17(cos 55°) = b 17 ( 1 —
sin 55° ) = c
11.90 ≈ b 20.75 ≈ c
So, B = 35°, b ≈ 11.90, and c ≈ 20.75.
36. Because the triangle is a right triangle, A and B are
complementary angles. So, A = 90° − 16° = 74°. Next,
write two equations using trigonometric functions, one that
involves the ratio of a and 14, and one that involves c and 14.
Solve the fi rst equation for a and the second equation for c.
cot 16° = adj.
— opp. csc 16° = hyp.
— opp.
cot 16° = a — 14
csc 16° = c — 14
14 ( 1 —
tan 16° ) = a 14 ( 1
— sin 16°
) = c
48.82 ≈ a 50.79 ≈ c
So, A = 74°, a ≈ 48.82, and c ≈ 50.79.
37. Because the triangle is a right triangle, A and B are
complementary angles. So, B = 90° − 43° = 47°. Next,
write two equations using trigonometric functions, one that
involves the ratio of a and 31, and one that involves c and 31.
Solve the fi rst equation for a and the second equation for c.
tan 43° = opp.
— adj.
sec 43° = hyp.
— adj.
tan 43° = a — 31
sec 43° = c — 31
31(tan 43°) = a 31 ( 1 —
cos 43° ) = c
28.91 ≈ a 42.39 ≈ c
So, B = 47°, a ≈ 28.91, and c ≈ 42.39.
38. Because the triangle is a right triangle, A and B are
complementary angles. So, A = 90° − 31° = 59°. Next,
write two equations using trigonometric functions, one that
involves the ratio of b and 23, and one that involves the ratio
of c and 23. Solve the fi rst equation for b and the second
equation for c.
tan 31° = opp.
— adj.
sec 31° = hyp.
— adj.
tan 31° = b — 23
sec 31° = c — 23
23(tan 31°) = b 23 ( 1 —
cos 31° ) = c
13.82 ≈ b 26.83 ≈ c
So, A = 59°, b ≈ 13.82, and c ≈ 26.83.
Copyright © Big Ideas Learning, LLC Algebra 2 455All rights reserved. Worked-Out Solutions
Chapter 9
39. Because the triangle is a right triangle, A and B are
complementary angles. So, A = 90° − 72° = 18°. Next,
write two equations using trigonometric functions, one that
involves the ratio of a and 12.8, and one that involves b and
12.8. Solve the fi rst equation for a and the second equation
for b.
cos 72° = adj.
— hyp.
sin 72° = opp.
— hyp.
cos 72° = 9 —
12.8 sin 72° = b
— 12.8
12.8(cos 72°) = a 12.8 (sin 72°) = b
3.96 ≈ a 12.17 ≈ b
So, A = 18°, a ≈ 3.96, and b ≈ 12.17.
40. Because the triangle is a right triangle, A and B are
complementary angles. So, B = 90° − 64° = 26°. Next,
write two equations using trigonometric functions, one that
involves the ratio of b and 7.4, and one that involves c and
7.4. Solve the fi rst equation for b and the second equation
for c.
cot 64° = adj.
— opp. csc 64° = hyp.
— opp.
cot 64° = b —
7.4 csc 64° = c
— 7.4
7.4 ( 1 —
tan 64° ) = b 7.4 ( 1
— sin 64°
) = c
3.61 ≈ b 8.23 ≈ c
So, B = 26°, b ≈ 3.61, and c ≈ 8.23.
41. tan 79° = w —
100
100(tan 79°) = w
514 ≈ w
The width is about 514 meters.
42. tan 52° = h —
458
458(tan 52°) = h
586 ≈ h
The height is about 586 feet.
43.
h
75
80°
Write and solve an equation to fi nd h.
tan 80° = h — 75
75(tan 80°) = h
425 ≈ h
The height of the building is about 427 meters (including the
person’s eye level above the ground).
44.
h800 ft
30°
Write and solve an equation to fi nd the height h.
sin 30° = h —
800
800(sin 30°) = h
400 ≈ h
The height is about 400 feet.
45. a. tan 24° = b —
1000
1000(tan 24°) = b
445 ≈ b
The height is about 451 feet (including your eye level of
5.5 feet).
b. The elevation at the top of Mount Rushmore is about
5280 + 451 = 5731 feet.
46. Answers will vary.
47. a. The radius of the Tropic of Cancer is
3960 cos (23.5°) ≈ 3631. So, the circumference is
about 2π ⋅ 3631 ≈ 22,818 miles.
b. The distance between the two points is the length of
the diameter of the Tropic of Cancer. So, it is about
2 ⋅ 3960 cos (23.5°) ≈ 7263 miles.
48. a. The side adjacent θ is x.
b. The side opposite θ is y.
c. Yes, they are equal.
cos θ = adj.
— hyp.
= x — h and sin (90° − θ) =
opp. —
hyp. = x —
h
49. a. sec(90° − 25°) = d —
25,000
25,000 ( 1 ——
cos(90° − 25°) ) = d
59,155 ≈ d
The distance is about 59,155 feet.
b. 25,0002 + x2 ≈ 59,1552
x ≈ √—— 59,1552 − 25,0002
x ≈ 53,613
The distance is about 53,613 feet.
c. tan(90° − 15°) = x + y —
25,000
25,000 tan (90° − 15°) = x + y
25,000 tan (90° − 15°) − 53,613 ≈ y
39,688 ≈ y The distance between the towns is about 39,688 feet.
Use the tangent function to fi nd the horizontal distance,
x + y, from the airplane to the second town and subtract
53,613 feet to fi nd the distance between the two towns.
456 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
50.
h
x50 m32° 53°
tan 32° = h —
50 + x and tan 53° = h —
x
(50 + x) tan 32° = h x tan 53° = h
(50 + x) tan 32° = x tan 53° 50 tan 32° + x tan 32° = x tan 53° 50 tan 32° = x(tan 53° − tan 32°)
50 tan 32° ——
tan 53° − tan 32° = x
44.495 ≈ x
Thus, h = 44.495 tan 53° ≈ 59. So, the height of the
building is about 59 meters.
51. Your friend is correct. The triangle must be a 45-45-90
triangle because both acute angles would be the same and
have the same cosine value.
52. The triangle on the right is equilateral, so all of the angles
must be 60°. Using geometry, it can be shown that θ = 30° and that the two triangles form a larger 30-60-90 triangle.
Because the large triangle is a right triangle, the six
trigonometric ratios can be found. The legs of the triangle are
1 and √—
3 , and the hypotenuse is 2.
sin θ = 1 — 2 cos θ =
√— 3 —
2 tan θ =
√— 3 —
3
csc θ = 2 sec θ = 2 √—
3 —
3 cot θ = √
— 3
53. a. sin 30° = x — 1
sin 30° = x
1 —
2 = x
The perimeter is 6 units.
b. Sample answer: Each side is part of two right triangles
with opposing angles ( 180° — n ) . So, each side length is
2 sin ( 180° — n ) , and there are n sides.
c. 2n sin ( 180° — n ) ≈ 2π
n sin ( 180° — n ) ≈ π
When n = 50, π = 50 sin ( 180° — 50
) ≈ 3.14.
Maintaining Mathematical Profi ciency
54. ( 5 yr —
1 ) ( 365 days
— 1 yr
) ( 24 hr —
1 day ) ( 60 min
— 1 hr
) ( 60 sec —
1 min ) = 157,680,000 sec
55. ( 12 pt —
1 ) ( 1 gal
— 8 pt
) = 1.5 gal
56. ( 5.6 m —
1 ) ( 1000 mm
— 1 m
) = 5600 mm
57. C = 2π(6) ≈ 37.7 cm,
A = π(6)2 ≈ 113.1 cm2
58. C = 2π(11) ≈ 69.1 in.
A = π(11)2 ≈ 380.1 in.2
59. C = π(14) ≈ 44.0 ft,
A = π ( 14 — 2 ) 2 ≈ 153.9 ft2
9.2 Explorations (p. 469)
1. a. Degreemeasure 0° 45° 90° 135° 180° 225° 270° 315° 360°
Radianmeasure
0 π — 4
π — 2
3π — 4 π
5π — 4
3π — 2
7π — 4 2π
b. Degreemeasure 30° 60° 120° 150° 210° 240° 300° 330°
Radianmeasure
π — 6
π — 3
2π — 3
5π — 6
7π — 6
4π — 3
5π — 3
11π — 6
Sample answer: To convert degrees to radians, multiply
degrees by π radians
— 180°
.
2. Radianmeasure
2π — 9
4π —
9
5π —
9
7π —
9
11π —
9
13π —
9
14π —
9
16π —
9
Degreemeasure 40° 80° 100° 140° 220° 260° 280° 320°
Sample answer: To convert radians to degrees, multiply
radians by 180° —
π radians .
3. Sample answer: To convert degrees into radians, multiply
degrees by π radians
— 180°
.
4. 30 radians ⋅ 180° —
π radians ≈ 1719°.
30 radians
— 30 degrees
≈ 1719° — 30°
= 57.3
Sample answer: 1 radian is about 57.3°, so 30 radians is
about 57.3 times greater than 30°, or about 1719°.
Copyright © Big Ideas Learning, LLC Algebra 2 457All rights reserved. Worked-Out Solutions
Chapter 9
9.2 Monitoring Progress (pp. 470–473)
1.
x
y
65°
2. Because 300° is 30° more than 270°, the terminal side is 30° counterclockwise past the negative y-axis.
x
y
300°
3. Because −120° is negative, the terminal side is 30° clockwise from the negative y-axis.
x
y
−120°
4. Because −450° is negative, the terminal side is the negative
y-axis.
x
y
−450°
5. 80° + 360° = 440° 80° − 360° = −280°
6. 230° + 360° = 590° 230° − 360° = −130°
7. 740° − 2 ⋅ 360° = 20° 740° − 3 ⋅ 360° = −340°
8. −135° + 360° = 225° −135° − 360° = −495°
9. 135° = 135 degrees ( π radians —
180 degrees )
= 3π — 4
10. −40° = (−40 degrees) ( π radians —
180 degrees )
= − 2π — 9
11. 5π — 4 = 5π —
4 radians ( 180°
— π radians
) = 225°
12. −6.28 = (−6.28 radians) ( 180° —
π radians )
≈ −359.8°
13. S = rθ = 220 ( π —
2 )
= 110π
≈ 346
The length of the outfi eld fence is about 346 feet. The area of
the fi eld is about 38,013 square feet.
9.2 Exercises (pp. 474–476)
Vocabulary and Core Concept Check
1. An angle is in standard position when its vertex is at the
origin and its initial side lies on the positive x-axis.
2. When the angle is positive, its rotation is counterclockwise
and when the angle is negative, its rotation is clockwise.
3. Sample answer: A radian is a measure of an angle that is
approximately equal to 57.3° and there are 2π radians in a
circle.
4. The expression −90° does not belong because it has a
different terminal side than the other three angles.
A = 1 — 2 r 2θ
= 1 — 2 (220)2 ( π —
2 )
= 12,100π≈ 38,013
458 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
Monitoring Progress and Modeling with Mathematics
5. Because 110° is 30° more than 90°, the terminal side is 30° counterclockwise past the positive y-axis.
x
y
110°
6. Because 450° is 90° more than 360°, the terminal side is the
positive y-axis.
x
y
450°
7. Because −900° is negative and 180° less than −720°, the
terminal side is the negative x-axis.
x
y
−900°
8. Because −10° is negative, the terminal side is 10° clockwise
from the positive x-axis.
x
y
−10°
9. 70° + 360° = 430° 70° − 360° = −290°
10. 255° + 360° = 615° 255° − 360° = −105°
11. −125° + 360° = 235° −125° − 360° = −485°
12. −800° + 3 ⋅ 360° = 280° −800° + 2 ⋅ 360° = −80°
13. 40° = 40 degrees ( π radians —
180 degrees )
= 2π — 9
14. 315° = 315 degrees ( π radians —
180 degrees )
= 7π — 4
15. −260° = (−260 degrees) ( π radians —
180 degrees )
= − 13π — 9
16. −500° = (−500 degrees) ( π radians —
180 degrees )
= − 25π — 9
17. π — 9 = ( π —
9 radians ) ( 180° —
π radians )
= 20°
18. 3π — 4 = 3π —
4 radians ( 180° —
π radians )
= 135°
19. −5 = (−5 radians) ( 180° — π radians
) ≈ −286.5°
20. 12 = 12 radians ( 180° — π radians
) ≈ 687.5°
21. A full revolution is 360° or 2π radians. The terminal side
rotates one-sixth of a revolution from the positive x-axis, so
multiply by 1 —
6 to get
1 —
6 ⋅ 360° = 60° and
1 —
6 ⋅ 2π = π —
3 .
22. The angles are 15π —
4 and −
π — 4 .
Sample answer: The angle 315° is equivalent to 7π — 4 radians,
and 7π — 4 + 2π = 15π —
4 and 7π —
4 − 2π = − π —
4 .
23. B; Because 600° is 240° more than 360°, the terminal side
makes one complete rotation 360° counterclockwise plus
240° more.
Copyright © Big Ideas Learning, LLC Algebra 2 459All rights reserved. Worked-Out Solutions
Chapter 9
24. D; − 9π — 4 = − 9π —
4 radians ( 180° —
π radians ) = −405°;
Because −405° is negative and 45° more than −360°, the
terminal side makes one complete rotation 360° clockwise
plus 45° more.
25. A; 5π — 6 = 5π —
6 radians ( 180° —
π radians ) = 150°; Because 150° is 60°
more than 90°, the terminal side is 60° counterclockwise past
the positive y-axis.
26. C; Because −240° is negative and 60° more than −180°, the
terminal side is 60° clockwise past the negative x-axis.
27. s = rθ
= 10 ( π — 2 )
= 5π ≈ 15.7
The length of the safety rail is about 15.7 yards and the area
of the deck is about 78.5 square yards.
28. a. s = rθ = 21.89 ( 34.92° ⋅
π radians —
180° )
≈ 13.3
The arc length is about 13.3 meters.
b. A = 1 — 2 r 2θ
= 1 — 2 (21.89)2 ( 34.92° ⋅
π radians —
180° )
≈ 146
The area of the sector is about 146 square meters.
29. The wrong conversion was used.
24° = 24 degrees ( π radians —
180 degrees )
= 24π radians —
180
= 0.42 radian
30. The angle was not converted to radians.
40° = 40 degrees ( π radians —
180 degrees ) =
2π — 9 radians
A = 1 —
2 (6)2 ( 2π —
9 ) ≈ 12.57 cm2
31. 200 ⋅ 360° = 72,000° 200 ⋅ 2π = 400π The angle is 72,000° or 400π radians.
32.
x
y
5:00
9:00
The hour hand has moved 8 —
12 of a revolution.
8 —
12 ⋅ 360° = 240°
8 —
12 ⋅ 2π = 4π —
3
The angle measure is 240° or 4π — 3 radians.
Sample answer: The minute hand would generate an angle of
2880° or 16π.
33. cos 4π — 3 = − 0.5
34. sin 7π — 8 ≈ 0.383
35. csc 10π — 11
= 1 —
sin 10π — 11
≈ 3.549
36. cot ( − 6π — 5 ) =
1 —
tan ( − 6π — 5 )
≈ −1.376
37. cot (−14) = 1 —
tan (−14) ≈ −0.138
38. cos 6 ≈ 0.960
39. 120° = 120 degrees ( π radians —
180 degrees ) = 2 —
3 π radians
A = 1 — 2 r1
2θ − 1 — 2 r2
2θ
= 1 — 2 (25)2 ( 2 —
3 π ) − 1 —
2 (11)2 ( 2 —
3 π )
= 625 —
2 π − 121
— 3 π
= 168π ≈ 528
The area is about 528 square inches.
40. a. 15 revolutions per minute is equivalent to 15
— 60
revolutions
per second.
2π ⋅ 15 —
60 = 1 —
2 π
There are π — 2 radians of rotation per second.
b. s = rθ
= ( 58 —
2 ) ( π —
2 )
≈ 45.6
The arc length is about 45.6 feet.
A = 1 —
2 r2θ
= 1 — 2 (10)2 ( π —
2 )
= 25π≈ 78.5
460 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
41. Four hours is 1 —
6 of a day. So, the rotation is
1 —
6 of Earth, or 60°
or π — 3 radians.
42. Using s = r (π − θ), the arc length of the small sector can be
found to be 1. Therefore, π − θ = 1 and θ = π − 1. So, the
angle is π − 1 radians.
43. Full sector: 1 —
2 ( 6
5 —
8 ) 2 ( 2π —
20 ) ≈ 6.89
Double region: 1 —
2 ( 6
5 —
8 ) 2 ( 2π —
20 ) −
1 —
2 ( 6
5 —
8 −
3 —
8 ) ( 2π —
20 ) ≈ 0.76
Triple region: 1 —
2 ( 3
3 —
4 +
3 —
8 ) 2 ( 2π —
20 ) −
1 —
2 ( 3
3 —
4 ) 2 ( 2π —
20 ) ≈ 0.46
The area of the entire sector is about 6.89 square inches, the
area of the double region is about 0.76 square inches, and the
area of the triple region is about 0.46 square inches.
44. Sample answer: This continued fraction (which is irrational)
gives rise to a sequence of rational approximations for π.
When the next fraction is added, the value gets closer to the
value of π = 3.1415926535 . . ., as shown.
3 = 3
3 + 1 —
7 =
22 —
7 = 3.142857143 . . .
3 + 1 —
7 + 1 —
15
= 333
— 106
= 3.141509434 . . .
3 + 1 ——
7 + 1 —
15 + 1 —
1
= 355
— 113
= 3.141592920 . . .
45. Your friend is correct. When the arc length is equal to the
radius, the equation s = rθ shows that θ = 1 and A = 1 —
2 r 2θ
is equivalent to A = s2
— 2 for r = s and θ = 1.
46. a. s = rθ = 42 ⋅ π — 8 ≈ 16.49
The length of the arc is about 16.49 inches.
b. The angle you would rotate through is 15π —
8 radians.
c. A = 1 —
2 r 2θ =
1 —
2 (42)2 ( π —
8 ) = 441π —
4 ≈ 346.4
Each step has 441π — 4 square inches of area, so
15 ⋅ 441π — 4 ≈ 5195.4 square inches of carpeting is needed.
47. a. 0.55 ⋅ 60 = 33
So, the measure is 70°33′. b. The measure is 110.76° because
110 + 45
— 60
+ 36 —
3600 = 110.76°.
Maintaining Mathematical Profi ciency
48. d = √——
(x2 − x1)2 + (y2 − y1)
2
= √——
(3 − 1)2 + (6 − 4)2
= √—
22 + 22
= √—
8
≈ 2.83
49. d = √——
(x2 − x1)2 + (y2 − y1)
2
= √———
[10 − (−7)]2 + [8 − (−13)]
2
= √—
172 + 212
= √—
730
≈ 27.02
50. d = √——
(x2 − x1)2 + (y2 − y1)
2
= √———
[ −3 − (−3) ] 2 + (16 − 9)2
= √—
02 + 72
= √—
49
= 7
51. d = √——
(x2 − x1)2 + (y2 − y1)
2
= √——
(8 − 2)2 + (−5 − 12)2
= √——
62 + (−17)2
= √—
325
≈ 18.03
52. d = √——
(x2 − x1)2 + (y2 − y1)
2
= √————
[ −20 − (−14) ] 2 + [ −32 − (−22) ] 2
= √——
(−6)2 + (−10)2
= √—
136
≈ 11.66
53. d = √——
(x2 − x1)2 + (y2 − y1)
2
= √——
(−1 − 4)2 + (34 − 16)2
= √——
(−5)2 + 182
= √—
349
≈ 18.68
9.3 Explorations (p. 477)
1. a. sin θ = √—
3 —
2 cos θ = − 1 —
2 tan θ = − √
— 3
b. sin θ = √
— 2 —
2 cos θ = −
√—
2 —
2 tan θ = −1
c. sin θ = −1 cos θ = 0 tan θ is undefi ned.
d. sin θ = − √—
3 —
2 cos θ =
1 —
2 tan θ = − √
— 3
Copyright © Big Ideas Learning, LLC Algebra 2 461All rights reserved. Worked-Out Solutions
Chapter 9
e. sin θ = − √
— 2 —
2 cos θ =
√—
2 —
2 tan θ = −1
f. sin θ = 0 cos θ = −1 tan θ = 0
2. Sample answer: The coordinates of the point on the unit
circle can be used to fi nd the ratios of the six trigonometric
functions where sin θ = y — r , cos θ =
x — r , tan θ =
y — x , csc θ =
r — y ,
sec θ = r — x , and cot θ =
x — y .
3. a. tan θ = y —
x is undefi ned when x = 0, which occurs when
θ = π — 2 + nπ, where n is an integer.
b. cot θ = x —
y is undefi ned when y = 0, which occurs when
θ = nπ, where n is an integer.
c. sec θ = r —
x is undefi ned when x = 0, which occurs when
θ = π — 2 + nπ, where n is an integer.
d. csc θ = r —
y is undefi ned when y = 0, which occurs when
θ = nπ, where n is an integer.
9.3 Monitoring Progress (pp. 479–481)
1. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √—
32 + (−3)2
= √—
18
= 3 √—
2
Using x = 3, y = −3, and r = 3 √—
2 , the values of the six
trigonometric functions of θ are:
sin θ = y —
r = −
√—
2 —
2 csc θ =
r —
y = − √
— 2
cos θ = x —
r =
√—
2 —
2 sec θ =
r —
x = √
— 2
tan θ = y —
x = −1 cot θ =
x —
y = −1
2. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √——
(−8)2 + 152
= √—
289
= 17
Using x = −8, y = 15, and r = 17, the values of the six
trigonometric functions of θ are:
sin θ = y —
r =
15 —
17 csc θ =
r —
y =
17 —
15
cos θ = x —
r = − 8 —
17 sec θ =
r —
x = − 17
— 8
tan θ = y —
x = − 15
— 8 cot θ =
x —
y = − 8 —
15
3. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √——
(−5)2 + (−12)2
= √—
169
= 13
Using x = −5, y = −12, and r = 13, the values of the six
trigonometric functions of θ are:
sin θ = y —
r = − 12
— 13
csc θ = r —
y = − 13
— 12
cos θ = x —
r = − 5 —
13 sec θ =
r —
x = − 13
— 5
tan θ = y —
x =
12 —
5 cot θ =
x —
y =
5 —
12
4. Draw a unit circle with the angle θ = 180° in standard
position.
Identify the point where the terminal side of θ intersects the
unit circle. The terminal side of θ intersects the unit circle at
(−1, 0).
x
y
(−1, 0)
θ
Find the values of the six trigonometric functions. Let
x = −1 and y = 0 to evaluate the trigonometric functions.
sin θ = y —
r = 0 csc θ =
r —
y =
1 —
0 undefi ned
cos θ = x —
r = −1 sec θ =
r —
x = −1
tan θ = y —
x = 0 cot θ =
x —
y =
−1 —
0 undefi ned
5.
x
y
210°
30°
The terminal side lies in Quadrant III. So, the reference angle
is 210° − 180° = 30°.
462 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
6.
x
y
−260°
80°
The terminal side lies in Quadrant II. So, the reference angle
is 180° − 100° = 80°.
7.
x
y
−79
π29π
The terminal side lies in Quadrant III. So, the reference angle
is 11π —
9 − π = 2π —
9 .
8.
x
y
154π
4π
The terminal side lies in Quadrant IV. So, the reference angle
is 2π − 7π — 4 = π —
4 .
9. The angle −210° is coterminal with 150°. The reference
angle is 180° − 150° = 30°. The cosine function is negative
in Quadrant II, so cos (−210°) = −cos (30°) = − √—
3 —
2 .
10. The angle 11π —
4 is coterminal with
3π — 4 . The reference angle is
π − 3π — 4 =
π — 4 . The secant function is negative in Quadrant II, so
sec 11π —
4 = −sec
π — 4 = − √
— 2 .
11. d = v 2
— 32
sin 2θ
= 272
— 32
sin(2 ⋅ 20°)
≈ 14.6
The horizontal distance traveled is about 14.6 feet.
9.3 Exercises (pp. 482–484)
Vocabulary and Core Concept Check
1. A quadrantal angle is an angle in standard position whose
terminal side lies on an axis.
2. After fi nding the reference angle θ′, evaluate − cosθ′ (cosine
is negative in Quadrant III).
Monitoring Progress and Modeling with Mathematics
3. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √—
42 + (−3)2
= √—
25
= 5
Using x = 4, y = −3, and r = 5, the values of the six
trigonometric functions of θ are:
sin θ = y —
r = − 3 —
5 csc θ =
r —
y = − 5 —
3
cos θ = x —
r =
4 —
5 sec θ =
r —
x = 5 —
4
tan θ = y —
x = − 3 —
4 cot θ =
x —
y = − 4 —
3
4. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √——
52 + (−12)2
= √—
169
=13
Using x = 5, y = −12, and r = 13, the values of the six
trigonometric functions of θ are:
sin θ = y —
r = − 12
— 13
csc θ = r —
y = − 13
— 12
cos θ = x —
r = 5 —
13 sec θ =
r —
x = 13
— 5
tan θ = y —
x = − 12
— 5 cot θ =
x —
y = − 5 —
12
5. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √——
(−6)2 + (−8)2
= √—
100
= 10
Using x = −6, y = −8, and r = 10, the values of the six
trigonometric functions of θ are:
sin θ = y —
r = − 4 —
5 csc θ =
r —
y = − 5 —
4
cos θ = x —
r = − 3 —
5 sec θ =
r —
x = − 5 —
3
tan θ = y —
x =
4 —
3 cot θ =
x —
y =
3 —
4
Copyright © Big Ideas Learning, LLC Algebra 2 463All rights reserved. Worked-Out Solutions
Chapter 9
6. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √—
32 + 12
= √—
10
Using x = 3, y = 1, and r = √—
10 , the values of the six
trigonometric functions of θ are:
sin θ = y —
r =
√—
10 —
10 csc θ =
r —
y =
√—
10 —
1 = √
— 10
cos θ = x —
r =
3 —
√—
10 =
3 √—
10 —
10 sec θ =
r —
x =
√—
10 —
3
tan θ = y —
x =
1 —
3 cot θ =
x —
y =
3 —
1 = 3
7. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x 2 + y 2
= √——
(−12)2 + (−9)2
= √—
225
= 15
Using x = −12, y = −9, and r = 15, the values of the six
trigonometric functions of θ are:
sin θ = y —
r = − 3 —
5 csc θ =
r —
y = − 5 —
3
cos θ = x —
r = − 4 —
5 sec θ =
r —
x = − 5 —
4
tan θ = y —
x =
3 —
4 cot θ =
x —
y =
4 —
3
8. Use the Pythagorean Theorem to fi nd the length of r.
r = √— x 2 + y 2
= √— 12 + (−2)2
= √—
5
Using x = 1, y = −2, and r = √—
5 , the values of the six
trigonometric functions of θ are:
sin θ = y —
r =
−2 —
√—
5 = − 2 √
— 5 —
5 csc θ =
r —
y = −
√—
5 —
2
cos θ = x —
r =
1 —
√—
5 =
√—
5 —
5 sec θ =
r —
x =
√—
5 —
1 = √
— 5
tan θ = y —
x =
−2 —
1 = −2 cot θ =
x —
y = −
1 —
2
9. Draw a unit circle with the angle θ = 0° in standard position.
Identify the point where the terminal side of θ intersects the
unit circle. The terminal side of θ intersects the unit circle
at (1, 0).
x
y
(1, 0)
Find the values of the six trigonometric functions. Let x = 1
and y = 0 to evaluate the trigonometric functions.
sin θ = y —
r =
0 —
1 = 0 csc θ =
r —
y =
1 —
0 undefi ned
cos θ = x —
r =
1 —
1 = 1 sec θ =
r —
x =
1 —
1 = 1
tan θ = y —
x =
0 —
1 = 0 cot θ =
x —
y =
1 —
0 undefi ned
10. Draw a unit circle with the angle θ = 540° in standard
position.
Identify the point where the terminal side of θ intersects the
unit circle. The terminal side of θ intersects the unit circle
at (−1, 0).
x
y
(−1, 0)θ
Find the values of the six trigonometric functions. Let x = −1
and y = 0 to evaluate the trigonometric functions.
sin θ = y —
r =
0 —
1 = 0 csc θ =
r —
y =
1 —
0 undefi ned
cos θ = x —
r =
−1 —
1 = −1 sec θ =
r —
x =
1 —
−1 = −1
tan θ = y —
x =
0 —
−1 = 0 cot θ =
x —
y =
−1 —
0 undefi ned
464 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
11. Draw a unit circle with the angle θ = π — 2 in standard position.
Identify the point where the terminal side of θ intersects
the unit circle. The terminal side of θ intersects the unit
circle at (0, 1).
x
y(0, 1)
θ
Find the values of the six trigonometric functions. Let x = 0
and y = 1 to evaluate the trigonometric functions.
sin θ = y —
r =
1 —
1 = 1 csc θ =
r —
y =
1 —
1 = 1
cos θ = x —
r =
0 —
1 = 0 sec θ =
r —
x =
1 —
0 undefi ned
tan θ = y —
x = 1 —
0 undefi ned cot θ =
x —
y =
0 —
1 = 0
12. Draw a unit circle with the angle θ = 7π — 2 in standard
position.
Identify the point where the terminal side of θ intersects
the unit circle. The terminal side of θ intersects the unit
circle at (0, −1).
x
y
(0, −1)
θ
Find the values of the six trigonometric functions. Let x = 0
and y = −1 to evaluate the trigonometric functions.
sin θ = y —
r =
−1 —
1 = −1 csc θ =
r —
y =
1 —
−1 = −1
cos θ = x —
r =
0 —
1 = 0 sec θ =
r —
x =
1 —
0 undefi ned
tan θ = y —
x = −1
— 0 undefi ned cot θ =
x —
y =
0 —
−1 = 0
13. Draw a unit circle with the angle θ = −270° in standard
position.
Identify the point where the terminal side of θ intersects
the unit circle. The terminal side of θ intersects the unit
circle at (0, 1).
x
y(0, 1)
θ
Find the values of the six trigonometric functions. Let x = 0
and y = 1 to evaluate the trigonometric functions.
sin θ = y —
r =
1 —
1 = 1 csc θ =
r —
y =
1 —
1 = 1
cos θ = x —
r =
0 —
1 = 0 sec θ =
r —
x =
1 —
0 undefi ned
tan θ = y —
x =
1 —
0 undefi ned cot θ =
x —
y =
0 —
1 = 0
14. Draw a unit circle with the angle θ = −2π in standard
position.
Identify the point where the terminal side of θ intersects
the unit circle. The terminal side of θ intersects the unit
circle at (1, 0).
x
y
(1, 0)θ
Find the values of the six trigonometric functions. Let x = 1
and y = 0 to evaluate the trigonometric functions.
sin θ = y —
r =
0 —
1 = 0 csc θ =
r —
y =
1 —
0 undefi ned
cos θ = x —
r =
1 —
1 = 1 sec θ =
r —
x =
1 —
1 = 1
tan θ = y —
x =
0 —
1 = 0 cot θ =
x —
y =
1 —
0 undefi ned
15.
x
y
−100°80°
The terminal side lies in Quadrant III. So, the reference angle
is 260° − 180° = 80°.
Copyright © Big Ideas Learning, LLC Algebra 2 465All rights reserved. Worked-Out Solutions
Chapter 9
16.
x
y
150°30°
The terminal side lies in Quadrant II. So, the reference angle
is 180° − 150° = 30°.
17.
x
y
320°
40°
The terminal side lies in Quadrant IV. So, the reference angle
is 360° − 320° = 40°.
18.
x
y
−370°
10°
The terminal side lies in Quadrant IV. So, the reference angle
is 360° − 350° = 10°.
19.
x
y
154π
4π
The terminal side lies in Quadrant IV. So, the reference angle
is 2π − 7π — 4 =
π — 4 .
20.
x
y
83π
3π
The terminal side lies in Quadrant II. So, the reference angle
is π − 2π — 3 =
π — 3 .
21.
x
y
−56
π6π
The terminal side lies in Quadrant III. So, the reference angle
is 7π — 6 − π =
π — 6 .
22.
x
y
−136
π
6π
The terminal side lies in Quadrant IV. So, the reference angle
is 2π − 11π — 6 =
π — 6 .
23. The equation for tangent is tan θ = y —
x .
tan θ = y —
x = − 2 —
3
24. The angle found is the angle between the terminal side and
the y-axis, instead of the x-axis.
θ is coterminal with 290°, whose terminal side lies in
Quadrant IV. So, θ′ = 360° − 290° = 70°.
25. The angle 135° has the reference angle 180° − 135° = 45°.
The secant function is negative in Quadrant II, so
sec 135° = −sec 45° = − √—
2 .
466 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
26. The angle 240° has reference angle 240° − 180° = 60°.
The tangent function is positive in Quadrant III, so
tan 240° = tan 60° = √—
3 .
27. The angle −150° is coterminal with 210°. The reference
angle is 210° − 180° = 30°. The sine function is negative in
Quadrant III, so sin(−150°) = −sin 30° = − 1 — 2 .
28. The angle −420° is coterminal with 300°. The reference angle
is 360° − 300° = 60°. The cosecant function is negative in
Quadrant IV, so csc (−420°) = −csc 60° = − 2 √
— 3 —
3 .
29. The angle −3π —
4 is coterminal with
5π — 4 . The reference angle
is 5π — 4 − π =
π — 4 . The tangent function is positive in
Quadrant III, so tan ( −3π — 4 ) = tan
π — 4 = 1.
30. The angle −8π —
3 is coterminal with
4π — 3 . The reference angle
is 4π — 3 − π =
π — 3 . The cotangent function is positive in
Quadrant III, so cot ( −8π — 3 ) = cot
π — 3 =
√— 3 —
3 .
31. The angle 7π — 4 has reference angle 2π −
7π — 4 =
π — 4 . The
cosine function is positive in Quadrant IV, so
cos 7π — 4 = cos
π — 4 =
√—
2 —
2 .
32. The angle 11π —
6 has reference angle 2π −
11π — 6 =
π — 6 .
The secant function is positive in Quadrant IV, so
sec 11π —
6 = sec π —
6 =
2 √—
3 —
3 .
33. d = v2
— 32
sin 2θ
= 492
— 32
sin(2.60°)
≈ 65
The horizontal distance traveled is about 65 feet.
34. d = v2
— 32
sin 2θ
= 142
— 32
sin(2.45°)
≈ 6.1
The distance of a jump is about 6.1 feet.
35. d = v2
— 32
sin 2θ
5 = v2
— 32
sin(2 ⋅ 18°)
160 —
sin(36°) = v2
v ≈ 16.5
The initial speed needs to be about 16.5 feet per second.
36. You win the competition because your javelin travels
712
— 32
sin(2 ⋅ 40°) ≈ 155.1 feet.
37. First, determine the distance the top of the treadmill is above
the midpoint. The reference angle is 180° − 110° = 70°,
and by using the sine function, sin 70° = opp.
— hyp.
= opp.
— 5 , or
opp. = 5 sin 70° ≈ 4.7. So, the top of the treadmill is about
4.7 feet above the midpoint. This puts the top of the treadmill
about 4.7 + 6 = 10.7 feet above the ground.
38. Use the following diagram.
x
y
10 ft
75 ft
75 ft
255°15°
First, determine the height above the center of the Ferris wheel.
This gives sin 15° = opp.
— 75
or opp. = 75 sin 15° ≈ 19 feet. So,
the total height is about 19 + 75 + 10 = 104 feet. If the radius
is doubled the height is not doubled.
Sample answer: The initial height that the Ferris wheel is
above the ground is not doubled so the entire height is not
doubled.
39. a. Angle of sprinkler, θ
Horizontal distance water travels, d
30° 16.9
35° 18.4
40° 19.2
45° 19.5
50° 19.2
55° 18.4
60° 16.9
b. The maximum distance is 45°. Because v2
— 32
is constant in this
situation, the maximum distance traveled will occur when
sin 2θ is as large as possible. The maximum value of
sin 2θ occurs when 2θ = 90°, that is, when θ = 45°.
c. The distances are the same.
40. Determine the x-value as described in the diagram,
x = 50 cos 300° = 25. So, the stopping position is
50 − 25 = 25 feet farther from the goal line than the
starting position. So, the stopping position is 125 feet from
the goal line.
Copyright © Big Ideas Learning, LLC Algebra 2 467All rights reserved. Worked-Out Solutions
Chapter 9
41.
x
y√22
(1, 0)
(0, 1)
(−0, 1)
(−1, 0)
,( )12
√32
,( )√22
( )12
, √32√2
2
, )12
√32
,( )√22
)12
, √32−
√22
,( )√22
−√22
,( )√22
−−
(−
, )12
√32( −, )1
2√32( −−
(−
)12
, √32( −)1
2, √3
2( −−
0°
360°
45°135°
225° 315°
180°
90°
270°
30°
60°
150°
120°
240° 300°
330°210°
42. a. This is not possible. If sin θ > 0 and cos θ < 0, then
tan θ < 0.
b. If sin θ > 0, cos θ < 0, and tan θ < 0, then the terminal
side of the angle in standard position lies in Quadrant II.
So, 90° < θ < 180°.
43. tan θ = y —
x =
y —
r —
x —
r =
sin θ — cos θ
sin 90° = 1 and cos 90° = 0, so tan 90° is undefi ned because
you cannot divide by 0, but cot 90° = 0 —
1 = 0.
44. Sine and cosecant are negative because the y-coordinate
is negative in Quadrant IV. Cosine and secant are positive
because the x-coordinate is positive in Quadrant IV. Tangent
and cotangent are negative because the y-coordinate is
negative and the x-coordinate is positive.
45. A point on the line will have the form (x, y). Using this and
the origin, the slope of the line is m = y —
x . Because tan θ =
y —
x ,
tan θ = m.
46. Your friend is incorrect. θ = 240° is also a solution and any
angle coterminal to 60° and 240° are also solutions.
47. a. The reference angle for 117° is 180° − 117° = 63°. So,
sin 63° = opp.
— hyp.
= opp.
— 128
and cos 63° = adj.
— hyp.
= adj.
— 128
, or
opp. = 128 sin 63° ≈ 114 and adj. = 128 cos 63° ≈ 58.1.
Because the point is in Quadrant II, the point is
(−58.1, 114).
b. d = √—— (x2 − x1)
2 + (y2 − y1)2
≈ √——— [128 − (−58.1)]2 + (0 − 114)2
= √—— 186.12 + (−114)2
≈ 218
The distance is about 218 picometers.
48. When C is the circumference of Earth, then the equation is
1 — 2 C = C sin P. So,
1 —
2 = sin P. Because the angle is between
0° and 90°, this happens when P = 60°.
Maintaining Mathematical Profi ciency
49. Find the rational zeros of f. Because f is a polynomial
function, the possible rational zeros are ±1, ±2, ±3, ±4,
±6, and ±12. Use synthetic division.
−3 1 2 1 8 −12
−3 3 −12 12
1 −1 4 −4 0
1 1 2 1 8 −12
1 3 4 12
1 3 4 12 0
So, −3 and 1 are zeros.
So, f (x) = (x − 1)(x + 3)(x2 + 4) and x = −3 and x = 1 are
the only real zeros.
50. Find the rational zeros of f. Because f is a polynomial
function, the possible rational zeros are ±1, ±2, ±3, ±6,
and ±18. Use synthetic division.
−6 1 4 −14 −14 −15 −18
−6 12 12 12 18
1 −2 −2 −2 −3 0
−1 1 4 −14 −14 −15 −18
−1 −3 17 −3 18
1 3 −17 3 −18 0
3 1 4 −14 −14 −15 −18
3 21 21 21 18
1 7 7 7 6 0
So, −6, −1 and 3 are zeros.
So, f (x) = (x − 3)(x + 1)(x + 6)(x2 + 1) and x = 6,
x = −1, and x = 3 are the only real zeros.
468 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
51. Step 1 Plot the x-intercepts. Because −3 and 1 are zeros
of f, plot (−3, 0) and (−1, 0).
Step 2 Plot points between and beyond the x-intercepts.
x −4 −2 0 2
y −10 −6 −18 50
Step 3 Determine the end behavior. Because f has three
factors of the form x − k and a constant factor
of 2, it is a cubic function with positive leading
coeffi cient. So, f(x) → −∞ as x → −∞ and
f(x) → +∞ as x → +∞.
Step 4 Draw the graph so that it passes through the plotted
points and has the appropriate end behavior.
x
y
4
2
6
−8
−6
−10
−20
−4
2 3 4 5 6−4 −1−5−6−7
52. Step 1 Plot the x-intercepts. Because 4, 5, and −9 are zeros
of f, plot (4, 0), (5, 0), and (−9, 0).
Step 2 Plot points between and beyond the x-intercepts.
x −10 0 3 9 — 2
y − 70 — 3 −60 −32 171
— 8
Step 3 Determine the end behavior. Because f has three
factors of the form x − k and a constant factor
of 1 —
3 , it is a cubic function with positive leading
coeffi cient. So, f(x) → −∞ as x → −∞ and
f(x) → +∞ as x → +∞.
Step 4 Draw the graph so that it passes through the plotted
points and has the appropriate end behavior.
x
y
10
20
−20
−60
−70
82 6−6 −2−10
53. Step 1 Plot the x-intercepts. Because 0, −1, and 2 are zeros
of f, plot (0, 0), (−1, 0), and (2, 0).
Step 2 Plot points between and beyond the x-intercepts.
x −2 − 1 — 2 1 3
y 16 − 5 — 16 −2 36
Step 3 Determine the end behavior. Because f has three
factors of the form x − k, it is a quartic function
with positive leading coeffi cient. So, f(x) → +∞
as x → −∞ and f (x) → +∞ as x → +∞.
Step 4 Draw the graph so that it passes through the plotted
points and has the appropriate end behavior.
x
y
8
4
2
6
10
12
−8
−6
−10
−12
−14
−16
−18
−20
−4
1 3 4−1−2−3−4
9.4 Explorations (p. 485)
1. a. x −2π −
7π — 4 −
3π — 2 −
5π — 4 −π
y = sin x 0 √
— 2 —
2 1
√—
2 —
2 0
x π — 4
π — 2
3π — 4 π
5π — 4
y = sin x √
— 2 —
2 1
√—
2 —
2 0 −
√—
2 —
2
x −
3π — 4 −
π — 2 −
π — 4 0
y = sin x − √
— 2 —
2 −1 −
√—
2 —
2 0
x 3π — 2
7π — 4 2π
9π — 4
y = sin x −1 − √
— 2 —
2 0
√—
2 —
2
Copyright © Big Ideas Learning, LLC Algebra 2 469All rights reserved. Worked-Out Solutions
Chapter 9
b.
x
y1
π−2
y = sin x
c. The x-intercepts are x = −2π, π, 0, π, 2π. The local
maximums are at x = − 3π — 2 ,
π — 2 . The local minimums
are at x = − π — 2 ,
3π — 2 . The function is increasing when
−2π < x < − 3π — 2 , −
π — 2 < x <
π — 2 , and
3π — 2 < x < 2π and
decreasing when − 3π — 2 < x < −
π — 2 and
π — 2 < x <
3π — 2 . The
sine function is an odd function.
2. a. x −2π −
7π — 4 −
3π — 2 −
5π — 4 −π
y = cos x 1 √
— 2 —
2 0
− √—
2 —
2 −1
x π — 4
π — 2
3π — 4 π
5π — 4
y = cos x √
— 2 —
2 0 −
√—
2 —
2 −1 −
√—
2 —
2
x −
3π — 4 −
π — 2 −
π — 4 0
y = cos x − √
— 2 —
2 0
√—
2 —
2 1
x 3π — 2
7π — 4 2π
9π — 4
y = cos x 0 √
— 2 —
2 1
√—
2 —
2
b.
x
y1
π π2
−1
π−
y = cos x
c. The x-intercepts are x = − 3π — 2 , −
π — 2 , π —
2 , 3π —
2 . The local
maximums are at x = −2π, 0, 2π. The local minimums
are at x = −π, π. The function is increasing when
−π < x < 0 and π < x < 2π and decreasing when
−2π < x < −π and −π < x < π. The cosine function is
an even function.
3. Sample answer: The minimum value of each function is −1
and the maximum value is 1. Both functions have a repeating
pattern. The x-intercepts for y = sin x occur when x = 0,
±π, ±2π, ±3π . . ., and the x-intercepts for y = cos x occur
when x = ± π — 2 , ±
3π — 2 , ±
5π — 2 . . .
4. As x increases (or decreases) without bound, the graph
of y = sin x does not approach a single value. The graph
oscillates between y = 1 and y = −1.
9.4 Monitoring Progress (pp. 488−490)
1. The function is of the form g(x) = a sin bx, where a = 1 —
4 and
b = 1. So, the amplitude is a = 1 —
4 and the period is
2π — b =
2π — 1 = 2π.
Intercepts: (0, 0); ( 1 — 2 ∙ 2π, 0 ) = (π, 0); (2π, 0)
Maximum: ( 1 — 4 ∙ 2π,
1 —
4 ) = ( π —
2 ,
1 —
4 )
Minimum: ( 3 — 4 ∙ 2π, −
1 —
4 ) = ( 3π —
2 , −
1 —
4 )
x
y
π− ππ−2
1
−1
g(x) = sin x14
The graph of g is a vertical shrink by a factor of 1 —
4 of the
graph of f (x) = sin x.
2. The function is of the form g(x) = a cos bx, where a = 1 and
b = 2. So, the amplitude is a = 1 and the period is
2π — b =
2π — 2 = π.
Intercepts: ( π — 4 , 0 ) ; ( 3π —
4 , 0 ) ; ( 5π —
4 , 0 )
Maximum: (0, 1); (π, 1)
Minimum: ( π — 2 , −1 ) ; ( 3π —
2 , −1 )
x
y1
π
−1
g(x) = cos (2x)
The graph of g is a horizontal shrink by a factor of 1 —
2 of the
graph of f (x) = cos x.
470 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
3. The function is of the form g(x) = a sin bx, where a = 2
and b = π. So, the amplitude is a = 2 and the period is
2π — b =
2π — π
= 2.
Intercepts: (0, 0); (1, 0); (2, 0)
Maximum: ( 1 — 2 , 2 )
Minimum: ( 3 — 2 , −2 )
x
y2
1
0.5 1 1.5
−2
−1
g(x) = 2sin xπ
The graph of g is a horizontal shrink by a factor of 1 — π
and a vertical stretch by a factor of 2 of the graph of
f (x) = sin x.
4. The function is of the form g(x) = a cos bx, where a = 1 —
3
and b = 1 —
2 . So, the amplitude is a =
1 —
3 and the period is
2π — b =
2π — 1 —
2
= 4π.
Intercepts: (π, 0); (3π, 0)
Maximum: ( 0, 1 —
3 ) ; ( 4π,
1 —
3 )
Minimum: ( 2π, − 1 —
3 )
x
y
−
π2
13
13
g(x) = cos x13
12
The graph of g is a horizontal stretch by a factor of 2
followed by a vertical shrink by a factor of 1 —
3 of the graph
of f (x) = cos x.
5. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 4
Step 2 Draw the midline of the graph, y = 4.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , 4 ) ; ( 3π —
2 , 4 )
Maximum: (0, 5); (2π, 5)
Minimum: (π, 3)
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
π
6. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 —
2 Horizontal shift: h =
π — 2
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 0
Step 2 Draw the midline of the graph, y = 0.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , 0 ) ; ( 3π —
2 , 0 )
Maximum: ( π, 1 —
2 )
Minimum: ( 0, − 1 —
2 ) ; ( 2π, −
1 —
2 )
Step 4 Draw the graph through the key points.
x
y1
π
−1
Copyright © Big Ideas Learning, LLC Algebra 2 471All rights reserved. Worked-Out Solutions
Chapter 9
7. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = −π
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = −1
Step 2 Draw the midline of the graph, y = −1.
Step 3 Find the fi ve key points.
On y = k: (0, −1); (π, −1)
Maximum: ( − π — 2 , 0 ) ; ( 3π —
2 , 0 )
Minimum: ( π — 2 , −2 )
Step 4 Draw the graph through the key points.
x
y
π
−1
−2
8. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = − π — 2
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 0
Step 2 Draw the midline of the graph. Because k = 0, the
midline is the x-axis.
Step 3 Find the fi ve key points of f (x) = ∣ −1 ∣ cos ( x − π — 2 ) .
On y = k: (−π, 0); (0, 0); (π, 0)
Maximum: ( − π — 2 , 1 )
Minimum: ( π — 2 , −1 )
Step 4 Refl ect the graph. Because a < 0, the graph is
refl ected in the midline y = 0. So, ( − π — 2 , 1 ) becomes
( − π — 2 , −1 ) and ( π —
2 , −1 ) becomes ( π —
2 , 1 ) .
Step 5 Draw the graph through the key points.
x
y1
−1
9. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −3 ∣ = 3 Horizontal shift: h = 0
Period: 2π — b
= 2π — 1 —
2
= 4π Vertical shift: k = 2
Step 2 Draw the midline of the graph. Because k = 2, the
midline is y = 2.
Step 3 Find the fi ve key points of g(x) = −3 sin 1 —
2 x + 2.
On y = k: (0, 2); (2π, 2)
Maximum: (−π, 5); (3π, 5)
Minimum: (π, −1)
Step 4 Draw the graph through the key points.
x
y
1
2
4
5
π π2 π3
10. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −2 ∣ = 2 Horizontal shift: h = 0
Period: 2π — b =
2π — 4 =
π — 2 Vertical shift: k = −1
Step 2 Draw the midline of the graph. Because k = −1, the
midline is y = −1.
Step 3 Find the fi ve key points of g(x) = −2 cos 4x − 1.
On y = −1: ( π — 8 , −1 ) ; ( 3π —
8 , −1 )
Maximum: ( π — 4 , 1 )
Minimum: (0, −3); ( π — 2 , −3 )
Step 4 Draw the graph through the key points.
x
y1
−3
9.4 Exercises (pp. 491−494)
Vocabulary and Core Concept Check
1. The shortest repeating portion of the graph of a periodic
function is called a cycle.
2. The amplitude of the fi rst function is 1 —
2 and the amplitude of
the second function is 3. The period of the fi rst function is 2π
and the period of the second function is π.
472 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
3. A phase shift is a horizontal translation of a periodic
function.
Sample answer: y = sin ( x − π — 2 ) is a phase shift of y = sin x.
4. The midline of y = 2 sin 3(x + 1) − 2 is y = −2.
5. The function is a periodic function with a period of 2.
6. The function is a periodic function with a period of 2π.
7. The function is not a periodic function.
8. The function is not a periodic function.
9. The amplitude is 1 and the period is 6π.
10. The amplitude is 1 —
2 and the period is 2.
11. The amplitude is 4 and the period is π.
12. The amplitude is 3 and the period is 8π.
13. The function is of the form g(x) = a sin bx, where a = 2
and b = 1. So, the amplitude is a = 3 and the period is
2π — b =
2π — 1 = 2π.
Intercepts: (0, 0); (π, 0); (2π, 0)
Maximum: ( π — 2 , 3 )
Minimum: ( 3π — 2 , −3 )
x
y
3
4
2
1
π
−3
−4
The graph of g is a vertical stretch by a factor of 3 of the
graph of f (x) = sin x.
14. The function is of the form g(x) = a sin bx, where a = 2
and b = 1. So, the amplitude is a = 2 and the period is
2π — b =
2π — 1 = 2π.
Intercepts: (0, 0); (π, 0); (2π, 0)
Maximum: ( π — 2 , 2 )
Minimum: ( 3π — 2 , −2 )
x
y2
1
−1
−2
The graph of g is a vertical stretch by a factor of 2 of the
graph of f (x) = sin x.
15. The function is of the form g(x) = a cos bx, where a = 1
and b = 3. So, the amplitude is a = 1 and the period is
2π — b =
2π — 3 .
Intercepts: ( π — 6 , 0 ) ; ( π —
2 , 0 )
Maximum: (0, 1); ( 2π — 3 , 1 )
Minimum: ( π — 3 , −1 )
x
y1
π
−1
π3
2π3
The graph of g is a horizontal shrink by a factor of 1 —
3 of the
graph of f (x) = cos x.
16. The function is of the form g(x) = a cos bx, where a = 1
and b = 4. So, the amplitude is a = 1 and the period is
2π — b =
2π — 4 =
π — 2 .
Intercepts: ( π — 8 , 0 ) ; ( 3π —
8 , 0 )
Maximum: (0, 1); ( π — 2 , 1 )
Minimum: ( π — 4 , −1 )
x
y1
π
−1
The graph of g is a horizontal shrink by a factor of 1 —
4 of the
graph of f (x) = cos x.
Copyright © Big Ideas Learning, LLC Algebra 2 473All rights reserved. Worked-Out Solutions
Chapter 9
17. The function is of the form g(x) = a sin bx, where a = 1
and b = 2π. So, the amplitude is a = 1 and the period is
2π — b =
2π — 2π
= 1.
Intercepts: (0, 0); (0.5, 0); (1, 0)
Maximum: (0.25, 1)
Minimum: (0.75, −1)
x
y1
0.5
−1
0.8 1.20.4 1.6
The graph of g is a horizontal shrink by a factor of 1 —
2π of the
graph of f (x) = sin x.
18. The function is of the form g(x) = a sin bx, where a = 3
and b = 2. So, the amplitude is a = 3 and the period is
2π — b =
2π — 2 = π.
Intercepts: (0, 0); ( π — 2 , 0 ) ; (π, 0)
Maximum: ( π — 4 , 3 )
Minimum: ( π — 4 , −3 )
x
y
1
2
3
4
π
−3
−4
−2
−1
The graph of g is a horizontal shrink by a factor of 1 —
2 and a
vertical stretch by a factor of 3 of the graph of f (x) = sin x.
19. The function is of the form g(x) = a cos bx, where a = 1 —
3
and b = 4. So, the amplitude is a = 1 —
3 and the period is
2π — b =
2π — 4 =
π — 2 .
Intercepts: ( π — 4 , 0 ) ; ( 3π —
4 , 0 )
Maximums: ( 0, 1 —
3 ) ; ( π,
1 —
3 )
Minimum: ( π — 2 , −
1 —
3 )
x
y
π
−
13
13
The graph of g is a horizontal shrink by a factor of 1 —
4 and a
vertical shrink by a factor of 1 —
3 of the graph of f (x) = cos x.
20. The function is of the form g(x) = a cos bx, where a = 1 —
2
and b = 4π. So, the amplitude is a = 1 —
2 and the period is
2π — b =
2π — 4π
= 1 —
2 .
Intercepts: ( 1 — 8 , 0 ) ; ( 3 —
8 , 0 )
Maximums: ( 0, 1 —
2 ) ; ( 1 —
2 ,
1 —
2 )
Minimum: ( 1 — 4 , −
1 —
2 )
x
y
−
12
12
0.25 0.5 0.75 1
The graph of g is a horizontal shrink by a factor of 1 —
4π and a
vertical shrink by a factor of 1 —
2 of the graph of f (x) = cos x.
21. B; D; The function y = −4 sin πx is of the form
g(x) = a sin bx, where a = −4 and b = π. So, the amplitude
is ∣ a ∣ = ∣ −4 ∣ = 4 and the period is 2π — b =
2π — π
= 2. The
function y = 4 cos πx is of the form g(x) = a cos bx, where
a = 4 and b = π. So, the amplitude is a = 4 and the period
is 2π — b =
2π — π
= 2.
474 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
22. a. The period is 5, so 2π — b = 5 or b =
2π — 5 . The equation is
y = sin 2π — 5 x.
b. The period is 4, so 2π — b = 4 or b =
π — 2 . The equation is
y = 10 sin π — 2 x.
c. The period is 2π, so 2π — b = 2π or b = 1. The equation is
y = 2 sin x.
d. The period is 3π, so 2π — b = 3π or b =
2 —
3 . The equation is
y = 1 —
2 sin
2 —
3 x.
23. The period is 1 —
4 and represents the amount of time, in
seconds, that it takes for the pendulum to go back and forth
and return to the same position. The amplitude is 4 and
represents the maximum distance, in inches, the pendulum
will be from its resting position.
t
d
0.25 0.5 0.75 1
2
3
4
1
−1
−2
−3
−4
24. The period is 6 and represents
t
y
1 2 3 4 5 6
1.75
−1.75
the amount of time, in seconds,
that it takes for the buoy to bob
up and down and return to the same
position. The amplitude is 1.75 and
represents the maximum distance,
in feet, the buoy will be from
its midline.
25. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 2
Step 2 Draw the midline of the graph, y = 2.
Step 3 Find the fi ve key points.
On y = k: (0, 2); (π, 2); (2π, 2)
Maximum: ( π — 2 , 3 )
Minimum: ( 5π — 2 , 1 )
Step 4 Draw the graph through the key points.
x
y
2
3
1
π π2
−1
26. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = −4
Step 2 Draw the midline of the graph, y = −4.
Step 3 Find the fi ve key points.
On y = −4: ( π — 2 , −4 ) ; ( 3π —
2 , −4 )
Maximums: (0, −3); (2π, −3)
Minimum: (π, −5)
Step 4 Draw the graph through the key points.
x
y1
π π2−1
−2
−3
−4
−5
27. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = π — 2
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 0
Step 2 Draw the midline of the graph. Because k = 0, the
midline is the x-axis.
Step 3 Find the fi ve key points.
On y = k: (0, 0); (π, 0); (2π, 0)
Maximum: ( π — 2 , 1 )
Minimum: ( 3π — 2 , −1 )
Step 4 Draw the graph through the key points.
x
y1
−1
π2
Copyright © Big Ideas Learning, LLC Algebra 2 475All rights reserved. Worked-Out Solutions
Chapter 9
28. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = − π — 4
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 0
Step 2 Draw the midline of the graph. Because k = 0, the
midline is the x-axis.
Step 3 Find the fi ve key points.
On y = k: ( 3π — 4 , 0 ) ; ( 7π —
4 , 0 ) ; ( − π —
4 , 0 )
Maximum: ( π — 4 , 1 )
Minimum: ( 5π — 4 , 1 )
Step 4 Draw the graph through the key points.
x
y1
π
−1
π2
29. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 2 Horizontal shift: h = 0
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = −1
Step 2 Draw the midline of the graph, y = −1.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , −1 ) ; ( 3π —
2 , −1 )
Maximums: (0, 1); (2π, 1)
Minimums: (π, −3)
Step 4 Draw the graph through the key points.
x
y1
π π2
−1
−2
−3
30. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 3 Horizontal shift: h = 0
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 1
Step 2 Draw the midline of the graph, y = 1.
Step 3 Find the fi ve key points.
On y = k: (0, 1); (π, 1); (2π, 1)
Maximum: ( π — 2 , 4 )
Minimum: ( 3π — 2 , −2 )
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
π π2−1
−2
31. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = −2π
Period: 2π — b =
2π — 2 = π Vertical shift: k = 0
Step 2 Draw the midline of the graph. Because k = 0, the
midline is the x-axis.
Step 3 Find the fi ve key points.
On y = k: (0, 0); ( π — 2 , 0 ) ; (π, 0)
Maximum: ( π — 4 , 1 )
Minimum: ( 3π — 4 , −1 )
Step 4 Draw the graph through the key points.
x
y1
−1
π5 4
476 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
32. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = 2π
Period: 2π — b =
2π — 2 = π Vertical shift: k = 0
Step 2 Draw the midline of the graph. Because k = 0, the
midline is the x-axis.
Step 3 Find the fi ve key points.
On y = k: ( π — 4 , 0 ) ; ( 3π —
4 , 0 )
Maximums: (0, 1); (π, 1)
Minimum: ( π — 2 , −1 )
Step 4 Draw the graph through the key points.
x
y1
−1
π π2
33. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: k = −π
Period: 2π — b =
2π — 1 —
2
= 4π Vertical shift: k = 3
Step 2 Draw the midline of the graph, y = 3.
Step 3 Find the fi ve key points.
On y = k: (0, 3); (2π, 3); (4π, 3)
Maximum: (3π, 4)
Minimum: (π, 2)
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
π π4π3π2−1
34. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = 3π — 2
Period: 2π — b =
2π — 1 —
2
= 4π Vertical shift: k = −5
Step 2 Draw the midline of the graph, y = −5.
Step 3 Find the fi ve key points.
On y = k: (0, −5); (2π, −5); (4π, −5)
Maximum: (3π, −4)
Minimum: (π, −6)
Step 4 Draw the graph through the key points.
x
y1
π π4π3π2−1
−2
−3
−4
−5
−6
−7
35. To fi nd the period, use the expression 2π — ∣ b ∣
.
Period: 2π — ∣ b ∣
= 2π —
∣ 2 — 3 ∣ = 3π
36. π — 2 should be added to the x-coordinate.
Maximum: ( ( 1 — 4 ⋅ 2π ) +
π — 2 , 2 ) = ( π —
2 +
π — 2 , 2 ) = (π, 2)
37. The graph of g is a vertical stretch by a factor of 2 followed
by a translation π — 2 units right and 1 unit up of the graph of f.
38. The graph of g is a vertical stretch by a factor of 3 followed
by a translation π — 4 units left and 2 units down of the graph
of f.
39. The graph of g is a horizontal shrink by a factor of 1 —
3
followed by a translation 3π units left and 5 units down of
the graph of f.
40. The graph of g is a horizontal shrink by a factor of 1 —
6
followed by a translation π units right and 9 units up of the
graph of f.
Copyright © Big Ideas Learning, LLC Algebra 2 477All rights reserved. Worked-Out Solutions
Chapter 9
41. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 1
= 2π Vertical shift: k = 3
Step 2 Draw the midline of the graph, y = 3.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , 3 ) ; ( 3π —
2 , 3 )
Maximum: (π, 4)
Minimums: (0, 2); (2π , 2)
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
π π2−1
42. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 1
= 2π Vertical shift: k = −5
Step 2 Draw the midline of the graph, y = −5.
Step 3 Find the fi ve key points.
On y = k: (0, −5); (π, −5); (2π, −5)
Maximum: ( 3π — 2 , −4 )
Minimum: ( π — 2 , −6 )
Step 4 Draw the graph through the key points.
x
y1
π π2−1
−2
−3
−4
−5
−6
−7
43. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 1 —
2
= 4π Vertical shift: k = −2
Step 2 Draw the midline of the graph, y = −2.
Step 3 Find the fi ve key points.
On y = k: (0, −2); (2π, −2); (4π, −2)
Maximum: (3π, −1)
Minimum: (π, −3)
Step 4 Draw the graph through the key points.
x
y1
π π4π3π2−1
−2
−3
44. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 2
= π Vertical shift: k = 1
Step 2 Draw the midline of the graph, y = 1.
Step 3 Find the fi ve key points.
On y = k: ( π — 4 , 1 ) ; ( 5π —
4 , 1 )
Maximum: ( π — 2 , 2 )
Minimums: (0, 0); (π, 0)
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
π−1
45. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = π
Period: 2π — b =
2π — 1
= 2π Vertical shift: k = 4
Step 2 Draw the midline through the graph, y = 4.
Step 3 Find the fi ve key points.
On y = k: (0, 4); (π, 4); (2π, 0)
Maximum: ( π — 2 , 5 )
Minimum: ( 3π — 2 , 3 )
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
5
6
π π2−1
478 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
46. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = −π
Period: 2π — b =
2π — 1
= 2π Vertical shift: k = −2
Step 2 Draw the midline through the graph, y = −2.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , −2 ) ; ( 3π —
2 , −2 )
Maximums: (0, −1); (2π, −1)
Minimum: (π, −3)
Step 4 Draw the graph through the key points.
x
y1
π π2−1
−2
−3
−4
47. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −4 ∣ = 4 Horizontal shift: h = − π — 2
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = −1
Step 2 Draw the midline through the graph, y = −1.
Step 3 Find the fi ve key points.
On y = k: ( π — 4 , −1 ) ; ( 5π —
4 , −1 )
Maximums: ( 3π — 4 , 3 ) ; ( 11π —
4 , 3 )
Minimum: ( 7π — 4 −5 )
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
π π2−1
−2
−3
−4
−5
−6
48. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −5 ∣ = 5 Horizontal shift: h = π — 2
Period: 2π — b =
2π — 1
= 2π Vertical shift: k = 3
Step 2 Draw the midline through the graph, y = 3.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , 3 ) ; ( 3π —
2 , 3 )
Maximums: (0, 8); (2π, 8)
Minimum: (π, −2)
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
5
6
7
8
π π2−1
−2
49. A; The amplitude is ∣ a ∣ = ∣ −4 ∣ = 4. The period is
2π — b =
2π — 1
= 2π. The horizontal shift is h = π — 2 . Because the
graph is shifted π — 2 units horizontally, a maximum value
occurs at ( − π — 2 , 4 ) .
50. a. B; The graph of sine has been translated 3 units up.
b. C; The graph of cosine has been translated 3 units down.
c. A; The graph of sine has been shrunk horizontally by a
factor of 1 —
2 then translated
π — 2 units right.
d. D; The graph of cosine has been shrunk horizontally by a
factor of 1 —
2 then translated
π — 2 units right.
51. The transformation is g(x) = 3 sin(x − π) + 2.
52. The transformation is g(x) = cos 2π(x + 3) − 4.
53. The transformation is g(x) = – 1 — 3 cos π x − 1.
54. The transformation is g(x) =– 1 —
2 sin 6(x − 1) – 3 — 2 .
55.
0
1
2
3
4
5
6
7
8
0 10 20 30 40 50 60 70 80 90 θ
h
Angle (degrees)
Hei
gh
t (f
eet)
When θ = 45°, h = −8 cos 45° + 10 ≈ 4.3 feet.
Copyright © Big Ideas Learning, LLC Algebra 2 479All rights reserved. Worked-Out Solutions
Chapter 9
56. a.
t H L H — L
0 45 11.5 45
— 11.5
2.5 22.5 18 27.5
— 18
5 10 11.5 10
— 11.5
7.5 27.5 5 27.5
— 5
b. When the number of lynx is at the midline and increasing,
the number of hares decreases. Once the number of lynx
reaches a maximum and begins to decrease, the number of
hares decreases to its minimum. At this point, the number
of hares begins to increase to its midline and the number
of lynx starts to decrease until it reaches its minimum. The
number of lynx then starts to increase as the number of
hares reaches its maximum.
57. The days on which the average speed is 10 miles per hour are
days 205 and 328. When the function is graphed with the line
y = 10, the two points of intersection are about (205.5, 10)
and (328.7, 10).
Time (days)
Win
d s
pee
d (
mile
s p
er h
ou
r)
t
s
8
12
16
4
0200 3001000
58. The times when the depth is 7 feet are at midnight and about
12:24 p.m.. When the function is graphed with the line
y = 7, the two points of intersection are (0, 7) and (12.4, 7).
Time (hours)
Dep
th (
feet
)
t
d
30
45
60
15
012 1860
59. a. 2cos π − 2cos 0 ——
π − 0 =
−2 − 2 —
π
≈ −1.27
The average rate of change is about −1.27.
b. 1 − (−1) —
π − 0 =
2 — π
≈ 0.64
The average rate of change is about 0.64.
c. 1 − (−1) —
π − 0 ≈
2 —
π
≈ 0.64
The average rate of change is about 0.64.
60. a. x −2π − 3π —
2 −π − π —
2 0
y = sin(−x) 0 −1 0 1 0
y = cos(−x) 1 0 −1 0 1
x π — 2 π 3π —
2 2π
y = sin(−x) −1 0 1 0
y = cos(−x) 0 −1 0 1
b.
x
y1
π2
−1
y = sin(−x) y = cos(−x)
c. Both graphs are refl ections across the y-axis. Because
cosine is an even function, it is symmetrical across the
y-axis and the graph of y = cos(−x) is the same as the
graph of y = cos (x).
61. a.
0
20
40
60
80
100
120
140
160
180
0 40 80 120 160 t
h
Time (seconds)
Hei
gh
t (f
eet)
b. The Ferris wheel goes through 4.5 cycles in 180 seconds.
c. The maximum height is 175 feet and the minimum height
is 5 feet.
480 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
62. a. The graph represents a function of the form
f(x) = a cos bx because the y-coordinate is 5 when x = 0.
b. The maximum value is 5, the minimum value is −5, the
period is π, and the amplitude is 5.
63. Sample answer: The x-intercepts can be represented by
the expression (2n + 1) π — 4 , where n is an integer. So, the
x-intercepts occur when x = ± π — 4 , ±
3π — 4 , ± 5π —
4 , . . . .
64. Your friend is incorrect. The value of a indicates a vertical
stretch or a vertical shrink and changes the amplitude of the
graph. It does not affect the x-intercepts of the function. The
value of b indicates a horizontal stretch or a horizontal shrink
and changes the period of the graph, which is the horizontal
length of each cycle. So, only the value of b affects the
x-intercepts of the function.
65. The graph of g(x) = cos x is a translation π — 2 units to the right
of the graph of f(x) = sin x.
66. The form is y = sin x + cos 2x, where b1 = 1 and b2 = 2.
67. The period is 2π — b =
2π — 8π — 3
= 3 —
4 , so there is one heartbeat every
3 —
4 of a second. Thus, the pulse rate is
60 —
3 —
4
= 80 heartbeats
per minute.
68. a. The initial displacement is when t = 0. So, the initial
displacement is y = 0.2 cos (6.0) = 0.2 feet. The period
is 2π — b =
2π — 6 =
π — 3 .
t
y
0.1
0.2
−0.1
−0.2
3π
6π
b.
t
y
0.1
0.2
−0.1
−0.2
3π
4π
12π
The motion becomes imperceptible as the displacement
rapidly decreases toward 0.
Maintaining Mathematical Profi ciency
69. x2 + x − 6
— x + 3
= (x + 3)(x − 2)
—— x + 3
= x − 2 , x ≠ −3
70. x3 − 2x2 − 24x —— x2 – 2x − 24
= x (x + 4)(x − 6)
—— (x + 4)(x − 6)
= x , x ≠ −4, 6
71. x2 − 4x − 5
— x2 + 4x − 5
= (x − 5)(x + 1)
—— (x + 5)(x − 1)
72. x2 − 16
— x2 + x − 20
= (x + 4)(x − 4)
—— (x + 5)(x − 4)
= x + 4
— x + 5
, x ≠ 4
73. The least common multiple of 2x and 2(x − 5) is 2x (x − 5).
74. The least common multiple of (x2 − 4) = (x + 2)(x − 2) and
(x + 2) is (x + 2)(x − 2).
75. The least common multiple of x2 + 8x + 12 = (x + 6)(x + 2)
and x + 6 is (x + 6)(x + 2).
9.1−9.4 What Did You Learn? (p. 495)
1. Sample answer: The water from the sprinkler will travel the
maximum horizontal distance when the angle is 45°. If the
sprinkler is tilted up n degrees, the water will travel the same
horizontal distance when the sprinkler is tilted down n degrees.
2. Sample answer: The hare is the prey of the lynx. When the
hare population fl ourishes, there is more food for the lynx
and that population fl ourishes as well. As the lynx population
increases, the hare population will begin to decrease because
of the increase in predators and vice versa.
9.1–9.4 Quiz (p. 496)
1. Step 1 Draw a right triangle with acute angle θ such that
the leg opposite θ has length 2 and the hypotenuse
has length 7.
adj. = 3 5
27
θ
Step 2 Find the length of the adjacent side. By the
Pythagorean Theorem, the length of the other side is
adj. = √—
72 − 22 = 3 √—
5 .
Step 3 Find the values of the remaining fi ve trigonometric
functions. Because sin θ = 2 —
7 , csc θ =
hyp. —
opp. =
7 —
2 .
The other values are:
cos θ = adj. —
hyp. =
3 √—
5 —
7 tan θ =
opp. —
adj. =
2 √—
5 —
15
sec θ = hyp.
— adj.
= 7 √
— 5 —
15 cot θ =
adj. —
opp. =
3 √—
5 —
2
Copyright © Big Ideas Learning, LLC Algebra 2 481All rights reserved. Worked-Out Solutions
Chapter 9
2 . Write an equation using a trigonometric function that
involves the ratio of x and 8. Solve the equation for x.
sec 60° = hyp.
— adj.
2 = x —
8
16 = x
The length of the side is x = 16.
3. Write an equation using a trigonometric function that
involves the ratio of x and 12. Solve the equation for x.
tan 30° = opp.
— adj.
√
— 3 —
3 =
x —
12
4 √—
3 = x
The length of the side is x = 4 √—
3 .
4. Write an equation using a trigonometric function that
involves the ratio of x and 27. Solve the equation for x.
cos 49° = adj. —
hyp.
cos 49° = x —
27
27 cos 49° = x
17.7 ≈ x
The length of the side is x ≈ 17.7.
5. The terminal side is 40° counterclockwise from the positive
x-axis.
x
y
40°
40° + 360° = 400°
40° − 360° = −320°
6. Because 5π — 6 is
π — 3 more than
π — 2 , the terminal side is
π — 3
counterclockwise past the positive y-axis.
x
y
56π
5π — 6 + 2π =
17π — 6
5π — 6 − 2π = − 7π —
6
7. Because –960° is negative and 60° less than –900°, the
terminal side is 60° clockwise past the negative x-axis.
x
y
−960°
−960° + 3 ⋅ 360° = 120°
−960° + 2 ⋅ 360° = −240°
8. 3π — 10
= 3π — 10
radians ( 180 degrees —
π radians ) = 54°
9. −60° = −60 degrees ( π radians —
180 degrees )
= − π — 3
10. 72° = 72 degrees ( π radians —
180 degrees )
= 2π — 5
482 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
11. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x2 + y2
= √——
(−2)2 + (−6)2
= √—
40
= 2 √—
10
Using x = −2, y = −6, and r = 2 √—
10 , the values of the six
trigonometric functions of θ are:
sin θ = y —
r =
−6 —
2 √—
10 = − 6 √
— 10 —
20 = − 3 √
— 10 —
10
cos θ = x —
r =
−2 —
2 √—
10 = − 2 √
— 10 —
20 = − √
— 10 —
10
tan θ = y —
x = −6
— −2 = 3
csc θ = r —
y = 2 √
— 10 — −6
= − √—
10 —
3
sec θ = r —
x = 2 √
— 10 — −2
= − √—
10
cot θ = x —
y = −2
— −6 = 1 —
3
12. sin θ = 1 csc θ = 1
cos θ = 0 sec θ = undefi ned
tan θ = undefi ned cot θ = 0
13. sin θ = √
— 3 —
2 csc θ =
2 √—
3 —
2
cos θ = − 1 — 2 sec θ = −2
tan θ = − √—
3 cot θ = − √—
3 —
3
14. The function is of the form g(x) = a sin bx, where a = 3
and b = 1. So, the amplitude is a = 3 and the period
2π — b =
2π — 1
= 2π.
Intercepts: (0, 0); (π, 0); (2π, 0)
Maximum: ( π — 2 , 3 )
Minimum: ( 3π — 2 , −3 )
x
y
1
2
3
−1
−2
−3
3π2
π2
The graph of g is a vertical stretch by a factor of 3 of the
graph of f.
15. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 5π
= 2 —
5 Vertical shift: k = 3
Step 2 Draw the midline of the graph, y = 3.
Step 3 Find the fi ve key points.
On y = k: ( 1 — 10
, 3 ) ; ( 3 — 10
, 3 ) Maximums: (0, 4), ( 2 —
5 , 4 )
Minimum: ( 1 — 5 , 2 )
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
−115
25
The graph of g is a horizontal shrink by a factor
of 1 —
5π followed by a translation 3 units up of the
graph of f.
16. a. sin 70° = opp.
— 400
400 sin 70° = opp.
376 ≈ opp.
The total height of the kite is about 380 feet, including the
reel’s height above the ground.
b. Make the following diagram, using the height from part (a).
70°85°
376 ft
x y
tan 70° = 376
— x tan 85° =
376 —
y
x = 376 —
tan 70° y =
376 —
tan 85°
x ≈ 137 y ≈ 33
You and your friend are standing x + y feet apart, or about
137 + 33 = 170 feet.
Copyright © Big Ideas Learning, LLC Algebra 2 483All rights reserved. Worked-Out Solutions
Chapter 9
17. The time from 7:00 p.m. to 8:55 p.m. is 1 hour and
55 minutes or 23
— 12
hours. So, there are 23
— 12
revolutions,
which is 23
— 12
⋅ 2π = 23π —
6 radians. While you are seated, you
have rotated a distance of about (47.25) ( 23π — 6 ) ≈ 569 feet.
A person 5 feet from the window will have rotated a distance
of about (47.25 − 5) ( 23π — 6 ) ≈ 509 feet.
9.5 Explorations (p. 497)
1. a. x – π —
2 – π —
3 – π —
4 – π —
6 0
y = tan x undefi ned – √—
3 –1 – √—
3 —
3 0
x 2π — 3
3π — 4
5π — 6 π
7π — 6
y = tan x – √—
3 –1 – √—
3 —
3 0
√—
3 —
3
x π —
6 π —
4 π —
3 π —
2
y = tan x √
— 3 —
3 1 √
— 3 undefi ned
x 5π — 4
4π — 3
3π — 2
5π — 3
y = tan x 1 √—
3 undefi ned – √—
3
b.
x
y
4
2
1
3
5
6
−4
−3
−5
−6
−2
22−π π 74ππ
c. The asymptotes are at x = – π — 2 ,
π — 2 , –
3π — 2 . The x-intercepts
are x = 0, π. The function is increasing when – π — 2 < x <
π — 2
and π — 2 < x <
3π — 2 . The function is an odd function.
2. Sample answer: The tangent function does not have
maximum or minimum values and has a repeating pattern.
The asymptotes occur when x = ± π — 2 , ± 3π —
2 , ±
5π — 2 , . . . .
The x-intercepts occur when x = 0, ± π, ± 2π, ± 3π, . . . .
3. Because the function f (x) = cot x is undefi ned when x = 0
and x = π, the graph of y = cot x has asymptotes at x = 0
and x = π.
9.5 Monitoring Progress (pp. 500–501)
1. The function is of the form g(x) = a tan bx, where a = 1 and
b = 2. So, the period is π —
∣ b ∣ =
π — 2 .
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — 2(2)
, or x = π — 4 ; x = –
π — 2 ∣ b ∣
= – π —
2(2) ,
or x = – π — 4
Halfway points: ( π — 4b
, a ) = ( π — 4(2)
, 1 ) = ( π — 8 , 1 ) ;
( – π — 4b
, –a ) = ( – π — 4(2)
, –1 ) = ( – π — 8 , –1 )
x
y4
2
1
3
−4
−3
−2
4−π4π
8π
−1
The graph of g is a horizontal shrink by a factor of 1 —
2 of the
graph of f (x) = tan x.
2. The function is of the form g(x) = a cot bx, where a = 1 —
3 and
b = 1. So, the period is π —
∣ b ∣ =
π — 1 = π.
Intercept: ( π — 2b
, 0 ) = ( π — 2 , 0 )
Asymptotes: x = 0; x = π —
∣ b ∣ =
π — 1 = π
Halfway points: ( π — 4b
, a ) = ( π — 4(1)
, 1 —
3 ) = ( π —
4 ,
1 —
3 ) ; ( 3π —
4b , –a )
= ( 3π — 4(1)
, − 1 —
3 ) = ( 3π —
4 , – 1 —
3 )
x
y
1
2
3
4
π−1
−2
−3
−4
2π
The graph of g is a vertical shrink by a factor of 1 —
3 of the
graph of f (x) = cot x.
484 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
3. The function is of the form g(x) = a cot bx, where a = 2 and
b = 4. So, the period is π — b =
π — 4 .
Intercept: ( π — 2b
, 0 ) = ( π — 8 , 0 )
Asymptotes: x = 0; x = π —
∣ b ∣ =
π — 4
Halfway points: ( π — 4b
, a ) = ( π — 4(4)
, 2 ) = ( π — 16
, 2 ) ; ( 3π — 4b
, –a ) = ( 3π —
4(4) , –2 ) = 3π —
16 , –2
x
y
1
2
3
4
5
−2
−3
−4
−5
−6
4π
8π
−1
The graph of g is a vertical stretch by a factor of 2 and a
horizontal shrink by a factor of 1 —
4 of the graph of f (x) = cot x.
4. The function is of the form g(x) = a tan bx, where a = 5 and
b = π. So, the period is π —
∣ b ∣ = π — π
= 1.
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — (2π)
, or x = 1 —
2 ; x = –
π —
2 ∣ b ∣ = – π —
(2π) ,
or x = – 1 — 2
Halfway points: ( π — 4b
, a ) = ( π — 4(π)
, 5 ) = ( 1 — 4 , 5 ) ;
( – π — 4b
, –a ) = ( – π — 4(π)
, –5 ) = ( – 1 —
4 , –5 )
x
y
4
2
6
−6
−4
0.5−0.5
The graph of g is a vertical stretch by a factor of 5 and a
horizontal shrink by a factor of 1 — π of the graph of f (x) = tan x.
5. Step 1 Graph the function y = sin 3x. The period is 2π — 3 .
Step 2 Graph asymptotes of g. Because the asymptotes of g
occur when sin 3x = 0, graph x = 0, x = π — 3 , and
x = 2π — 3 .
Step 3 Plot points on g, such as ( π — 6 , 1 ) and ( π —
2 , –1 ) .
Then use the asymptotes to sketch the curve.
x
y
1
2
3
4
5
−1
−2
−3
−4
−5
−6
23π
3π
The graph of g is a horizontal shrink by a factor of 1 —
3 of the
graph of f (x) = csc x.
6. Step 1 Graph the function y = 1 —
2 cos x. The period is
2π — 1 = 2π.
Step 2 Graph asymptotes of g. Because the asymptotes of g
occur when 1 —
2 cos x = 0, graph x = – π —
2 , x = π —
2 , and
x = 3π — 2 .
Step 3 Plot points on g, such as ( 0, 1 —
2 ) and ( π,
1 —
2 ) .
Then use the asymptotes to sketch the curve.
x
y
1
2
3
4
−2
−3
−4
2−π 32π
2π
−1
The graph of g is a vertical shrink by a factor of 1 —
2 of the
graph of f (x) = sec x.
Copyright © Big Ideas Learning, LLC Algebra 2 485All rights reserved. Worked-Out Solutions
Chapter 9
7. Step 1 Graph the function y = 2 sin 2x. The period is 2π — 2 = π.
Step 2 Graph asymptotes of g. Because the asymptotes of g
occur when 2 sin 2x = 0, graph x = 0, x = π — 2 , and
x = π.
Step 3 Plot points on g, such as ( π — 4 , 2 ) and ( 3π —
4 , –2 ) .
Then use the asymptotes to sketch the curve.
2π x
y
2
4
π−2
−4
−6
The graph of g is a horizontal shrink by a factor of 1 —
2 and a
vertical stretch by a factor of 2 of the graph of f (x) = csc x.
8. Step 1 Graph the function y = 2 cos πx. The period is
2π — π = 2.
Step 2 Graph asymptotes of g. Because the asymptotes of g
occur when 2 cos πx = 0, graph x = – 1 — 2 , x =
1 —
2 , and
x = 3 —
2 .
Step 3 Plot points on g, such as (0, 2) and (1, −2). Then
use the asymptotes to sketch the curve.
x
y
4
6
8
−4
−6
−8
−20.5 1 1.5−.5
The graph of g is a horizontal shrink by a factor of 1 — π and a
vertical stretch by a factor of 2 of the graph of f (x) = sec x.
9.5 Exercises (pp. 502–504)
Vocabulary and Core Concept Check
1. The graphs of the tangent, cotangent, secant, and cosecant
functions have no amplitude because the graphs do not have
minimum and maximum values.
2. The cosecant and cotangent functions are undefi ned for
x-values at which sin x = 0.
3. The period of the function y = sec x is 2π, and the period of
y = cot x is π.
4. To graph y = a sec bx, fi rst graph y = a cos bx. Use the
asymptotes and several points of y = a sec bx to graph the
function.
Monitoring Progress and Modeling with Mathematics
5. The function is of the form g(x) = a tan bx, where a = 2 and
b = 1. So, the period is π —
∣ b ∣ =
π — 1 = π.
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — 2(1)
, or x = π — 2 ;
x = − π —
2 ∣ b ∣ = −
π — 2(1)
, or x = − π — 2
Halfway points: ( π — 4b
, a ) = ( π — 4(1)
, 2 ) = ( π — 4 , 2 ) ;
( − π — 4b
, −a ) = ( − π —
4(1) , −2 ) = ( −
π — 4 , −2 )
2π
4π
2−π4−π x
y4
2
1
3
−4
−3
−2
The graph of g is a vertical stretch by a factor of 2 of the
graph of f (x) = tan x.
6. The function is of the form g(x) = a tan bx, where a = 3 and
b = 1. So, the period is π —
∣ b ∣ =
π — 1 = π.
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — 2(1)
, or x = π — 2 ;
x = − π —
2 ∣ b ∣ = −
π — 2(1)
, or x = − π — 2
Halfway points: ( π — 4b
, a ) = ( π — 4(1)
, 3 ) = ( π — 4
, 3 ) ;
( − π — 4b
, −a ) = ( − π —
4(1) , −3 ) = ( −
π — 4 , −3 )
4π
4−π2π
2−π x
y4
2
1
3
−4
−3
−2
The graph of g is a vertical stretch by a factor of 3 of the
graph of f (x) = tan x.
486 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
7. The function is of the form g(x) = a cot bx, where a = 1 and
b = 3. So, the period is π —
∣ b ∣ =
π — 3 .
Intercept: ( π — 2b
, 0 ) = ( π — 2(3)
, 0 ) = ( π — 6 , 0 )
Asymptotes: x = 0; x = π —
∣ b ∣ =
π — 3
Halfway points: ( π — 4b
, a ) = ( π — 4(3)
, 1 ) = ( π — 12
, 1 ) ;
( 3π — 4b
, –a ) = ( 3π — 4(3)
, –1 ) = ( π — 4 , –1 )
3π x
y
1
2
3
−1
−2
−3
−4
The graph of g is a horizontal shrink by a factor of 1 —
3 of the
graph of f (x) = cot x.
8. The function is of the form g (x) = a cot bx, where a = 1 and
b = 2. So, the period is π —
∣ b ∣ =
π — 2 .
Intercept: ( π — 2b
, 0 ) = ( π — 2(2)
, 0 ) = ( π — 4 , 0 )
Asymptotes: x = 0; x = π — b =
π — 2
Halfway points: ( π — 4b
, a ) = ( π — 4(2)
, 1 ) = ( π — 8 , 1 ) ;
( 3π — 4b
, −a ) = ( 3π — 4(2)
, −1 ) = ( 3π — 8 , −1 )
4π
2π x
y
1
2
3
−1
−2
−3
−4
The graph of g is a horizontal shrink by a factor of 1 —
2 of the
graph of f (x) = cot x.
9. The function is of the form g(x) = a cot bx, where a = 3 and
b = 1 —
4 . So, the period is
π — ∣ b ∣
= π — 1 —
4
= 4π.
Intercept: ( π — 2b
, 0 ) = ( π —
2 ( 1 — 4 ) , 0 ) = (2π, 0)
Asymptotes: x = 0; x = π —
∣ b ∣ =
π — 1 —
4
= 4π
Halfway points: ( π — 4b
, a ) = ( π —
4 ( 1 — 4 ) , 3 ) = (π, 3);
( 3π — 4b
, −a ) = ( 3π —
4 ( 1 — 4 ) , −3 ) = (3π,−3)
32π
2π
x
y
2
4
6
8
10
−2
−4
−6
−8
−10
−12
π π2 π3 72π π4
The graph of g is a horizontal stretch by a factor of 4 and a
vertical stretch by a factor of 3 of the graph of f (x) = cot x.
10. The function is of the form g(x) = a cot bx, where a = 4
and b = 1 —
2 . So, the period is
π — ∣ b ∣
= π — 1 —
2
= 2π.
Intercept: ( π — 2b
, 0 ) = ( π —
2 ( 1 — 2 ) , 0 ) = ( π, 0 )
Asymptotes: x = 0; x = π —
∣ b ∣ =
π — 1 —
2
, or x = 2π
Halfway points: ( π — 4b
, a ) = ( π —
4 ( 1 — 2 ) , 4 ) = ( π —
2 , 4 ) ;
( 3π — 4b
, −a ) = ( 3π —
4 ( 1 — 2 ) , −4 ) = ( 3π —
2 , −4 )
32π
2π x
y
4
8
−4
−8
−12
π π2
The graph of g is a horizontal stretch by a factor of 2 and a
vertical stretch by a factor of 4 of the graph of f (x) = cot x.
Copyright © Big Ideas Learning, LLC Algebra 2 487All rights reserved. Worked-Out Solutions
Chapter 9
11. The function is of the form g(x) = a tan bx, where a = 1 —
2 and
b = π. So, the period is π —
∣ b ∣ =
π — π
= 1.
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — 2(π)
, or x = 1 —
2 ;
x = − π —
2 ∣ b ∣ = − π —
2(π) , or x = − 1 —
2
Halfway points: ( π — 4b
, a ) = ( π — 4(π)
, 1 —
2 ) = ( 1 —
4 ,
1 —
2 ) ;
( − π — 4b
, −a ) = ( − π — 4(π)
, − 1 — 2 ) = ( − 1 —
4 , − 1 —
2 )
4−π2π
2−π x
y4
2
1
3
The graph of g is a horizontal shrink by a factor of 1 — π and a
vertical shrink by a factor of 1 —
2 of the graph of f (x) = tan x.
12. The function is of the form g(x) = a tan bx, where a = 1 —
3 and
b = 2π. So, the period is π —
∣ b ∣ =
π — 2π
= 1 —
2 .
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — 2(2π)
, or x = 1 —
4 ;
x = − π —
2 ∣ b ∣ = − π —
2(2π) , or x = −
1 —
4
Halfway points: ( π — 4b
, a ) = ( π — 4(2π)
, 1 —
3 ) = ( 1 —
8 ,
1 —
3 )
( − π — 4b
, −a ) = ( − π — 4(2π)
, − 1 — 3 ) = ( − 1 —
8 , − 1 —
3 )
x
y
2
1
−2
−1
−0.25 0.25
The graph of g is a horizontal shrink by a factor of 1 —
2π and a
vertical shrink by a factor of 1 —
3 of the graph of f (x) = tan x.
13. To fi nd the period, use the expression π —
∣ b ∣ .
Period: π —
∣ b ∣ =
π — 3
14. The horizontal and vertical shrink factors are switched.
A vertical stretch by a factor of 2 and a horizontal shrink
by a factor of 1 —
5 .
15. a.
x
y
4
2
1
5
6
−4
−3
−5
−6
−2
4−π2π
4π
f(x) = 3 sec 2x
y = 3 cos 2x
b.
x
y
2
4
6
−2
−4
−6
−8
6π
3π
2π
f(x) = 4 csc 3x
y = 4 sin 3x
16. A; The asymptotes of the graph are x = π —
2 ∣ b ∣ =
π — 8 and
x = − π —
2 ∣ b ∣ = −
1 —
8 .
17. Step 1 Graph the function y = 3 sin x. The period is
2π — 1 = 2π.
Step 2 Graph asymptotes of g. Because the asymptotes of
g occur when 3 sin x = 0, graph x = 0, x = π, and
x = 2π.
Step 3 Plot points on g, such as ( π — 2 , 3 ) and ( 3π —
2 , −3 ) .
Then use the asymptotes to sketch the curve.
x
y
2
4
6
8
10
π π2−2
−4
−6
−8
−10
−12
32π
2π
The graph of g is a vertical stretch by a factor of 3 of the
graph of f (x) = csc x.
488 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
18. Step 1 Graph the function y = 2 sin x. The period is 2π — 1 = 2π.
Step 2 Graph asymptotes of g. Because the asymptotes of
g occur when 2 sin x = 0, graph x = 0, x = π, and
x = 2x.
Step 3 Plot points on g, such as ( π — 2 , 2 ) and ( 3π —
2 , −2 ) .
Then use the asymptotes to sketch the curve.
x
y
2
4
6
8
π π2−2
−4
−6
−8
−10
32π
2π
The graph of g is a vertical stretch by a factor of 2 of the
graph of f (x) = csc x.
19. Step 1 Graph the function y = cos 4x. The period is 2π — 4 =
π — 2 .
Step 2 Graph asymptotes of g. Because the asymptotes of
g occur when cos 4x = 0, graph x = − π — 8 , x =
π — 8 ,
and x = 3π — 8 .
Step 3 Plot points on g, such as (0, 1) and ( π — 9 , −1 ) .
Then use the asymptotes to sketch the curve.
8−π 38π
8π
4π x
y4
2
3
−4
−3
−2
−1
The graph of g is a horizontal shrink by a factor of 1 —
4 of the
graph of f (x) = sec x.
20. Step 1 Graph the function y = cos 3x. The period is 2π — 3 .
Step 2 Graph asymptotes of g. Because the asymptotes of g
occur when cos 3x = 0, graph x = − π — 6 , x =
π — 6 , and
x = π — 2 .
Step 3 Plot points on g, such as (0, 1) and ( π — 3 , −1 ) .
Then use the asymptotes to sketch the curve.
6−π6π
3π
2π x
y4
2
3
−4
−3
−2
−1
The graph of g is a horizontal shrink by a factor of 1 —
3 of the
graph of f (x) = sec x.
21. Step 1 Graph the function y = 1 —
2 cos π x. The period
is 2π — π
= 2.
Step 2 Graph asymptotes of g. Because the asymptotes of
g occur when 1 —
2 cos π x = 0, graph x = − 1 —
2 , x =
1 —
2 ,
and x = 3 —
2 .
Step 3 Plot points on g, such as ( 0, 1 —
2 ) and ( 1, − 1 —
2 ) .
Then use the asymptotes to sketch the curve.
x
y4
2
3
−4
−3
−2
−10.5 1.5−0.5
The graph of g is a horizontal shrink by a factor of 1 —
π and a
vertical shrink by a factor of 1 —
2 of the graph of f (x) = sec x.
Copyright © Big Ideas Learning, LLC Algebra 2 489All rights reserved. Worked-Out Solutions
Chapter 9
22. Step 1 Graph the function y = 1 —
4 cos 2π x. The period is
2π — 2π
= 1.
Step 2 Graph asymptotes of g. Because the asymptotes of g
occur when 1 —
4 cos 2 π x = 0, graph x = − 1 —
4 , x =
1 —
4 ,
and x = 3 —
4 .
Step 3 Plot points on g, such as ( 0, 1 —
4 ) and ( 1 —
2 , − 1 —
4 ) .
Then use the asymptotes to sketch the curve.
x
y
2
1
−2
−1
0.25 0.75−0.25
The graph of g is a horizontal shrink by a factor of 1 —
2π and a
vertical shrink by a factor of 1 —
4 of the graph of f (x) = sec x.
23. Step 1 Graph the function y = sin π — 2 x. The period is
2π — π — 2 = 4.
Step 2 Graph asymptotes of g. Because the asymptotes of
g occur when sin π — 2 x = 0, graph x = 0, x = 2, and
x = 4.
Step 3 Plot points on g, such as (1, 1) and (3, −1). Then
use the asymptotes to sketch the curve.
x
y
2
1
3
−4
−3
−2
−121 3 4
The graph of g is a horizontal stretch by a factor of 2 — π of the
graph of f (x) = csc x.
24. Step 1 Graph the function y = sin π — 4 x. The period is
2π — π — 4 = 8.
Step 2 Graph the asymptotes of g. Because the asymptotes
of g occur when sin π — 4 x = 0, graph x = 0, x = 4,
and x = 8.
Step 3 Plot points on g, such as (2, 1) and (6, −1). Then
use the asymptotes to sketch the curve.
x
y
2
1
3
−4
−3
−2
−1421 3 5 6 87
The graph of g is a horizontal stretch by a factor of 4 — π of the
graph of f (x) = csc x.
25. The period is π, so b = 1, and 6 = a tan π — 4 , so a = 6. Thus,
y = 6 tan x.
26. The period is π, so b = 1, and 1 —
2 = a tan
π — 4
, so a = 1 —
2 . Thus,
y = 1 —
2 tan x.
27. The period is 1, so b = π, and 2 = a tan ( π ⋅ 1 —
4 ) , so a = 2.
Thus, y = 2 tan π x.
28. The period is π — 2 , so b = 2, and 5 = a tan ( 2 ⋅
π — 8 ) , so a = 5.
Thus, y = 5 tan 2x.
29. B; The parent function is the tangent function and the graph
has an asymptote at x = π — 2 .
30. C; The parent function is the cotangent function and the
graph has an asymptote at x = 0.
31. D; The parent function is the cosecant function and the graph
has an asymptote at x = 1.
32. F; The parent function is the secant function and the graph
has an asymptote at x = 1 —
2 .
33. A; The parent function is the secant function and the graph
has an asymptote at x = π — 4 .
34. E; The parent function is the cosecant function and the graph
has an asymptote at x = π — 2 .
35. The tangent function that passes through the origin and has
asymptotes at x = π and x = −π can be stretched or shrunk
vertically to create more tangent functions with the same
characteristics.
490 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
36. a.
x
y
2
1
3
5
6
7
8
−4
−3
−2
4−π2−π
4π
2π 3
4π 5
4π 3
2ππ
The graph of g is a translation 3 units up of the graph of
f (x) = sec x.
b.
54π 7
4π3
2π x
y
1
2
3
π π2−1
−2
−3
−4
−5
−6
The graph of g is a translation 2 units down of the graph
of f (x) = csc x.
c.
x
y
1
2
3
4
π−1
−2
−3
−4
4π
2π
The graph of g is a translation π units right of the graph of
f (x) = cot x.
d.
x
y4
2
1
3
−4
−3
−2
−12−π4−π
2π
The graph of g is a refl ection across the x-axis of the
graph of f (x) = tan x.
37. The rule for the transformation is g (x) = cot ( 2x + π — 2 ) + 3.
38. The rule for the transformation is g(x) = 2 tan (3x − π).
39. The rule for the transformation is g(x) = −5 sec (x − π) + 2.
40. The rule for the transformation is g(x) = −8 csc x.
41. Function B has a local maximum value of −5 so Function
A’s local maximum value of − 1 — 4 is greater. Function A has a
local minimum of 1 —
4 so Function B’s local minimum value of
5 is greater.
42. Graph Average Rate of Change
A 1 − (−1)
— π — 4 − ( −
π — 4 )
= 4 —
π
B −1 −1
— π — 4 − ( −
π — 4 )
= − 4 — π
C
1 —
2 − ( −
1 —
2 ) —
π — 4 − ( −
π — 4 )
= 1 —
π
D −2 − 2 —
π — 4 − ( − π —
4 )
= − 8 — π
The order is D, B, C, and A.
43.
0
100
200
300
400
500
600
700
800
0 10 20 30 40 50 60 70 80 90 θ
d
Angle (degrees)
Dis
tan
ce (
ft)
As d increases, θ increases because as the car gets farther
away, the angle required to see the car gets larger.
44.
150
0
320
The viewing window is 0 < t < 15 and 0 < h < 320. The
Statue of Liberty is approximately 305 feet tall so it would
take almost 15 seconds to span the statue.
Copyright © Big Ideas Learning, LLC Algebra 2 491All rights reserved. Worked-Out Solutions
Chapter 9
45. a. tan θ = 260 − d
— 120
120 tan θ = 260 − d d = 260 − 120 tan θ b.
0
40
80
120
160
200
240
280
0 10 20 30 40 50 60 70 80 90 θ
d
The graph shows a negative correlation meaning that as
the angle gets larger, the distance from your friend to the
top of the building gets smaller. As the angle gets smaller,
the distance from your friend to the top of the building
gets larger.
46. a. tan θ = 200 − d
— 300
300 tan θ = 200 − d
d = 200 − 300 tan θ b.
0
50
100
150
200
250
300
0 10 20 30 40 θ
d
Angle (degrees)
Dis
tan
ce f
rom
to
p
of
clif
f (f
eet)
c. Using the trace feature, your friend is halfway down the
cliff, 100 feet from the top, at about 18.4°.
47. Your friend is incorrect. The graph of cosecant can be
translated π — 2 units right to create the same graph as y = sec x.
48. a. The period is 4.
b. The range is y > 2 and y < −2.
c. The function is of the form y = a csc bx because the
cosecant function has an asymptote at x = 0.
49. a sec bx = a —
cos bx
Because the cosine function is at most 1, y = a cos bx will
produce a maximum when cos bx = 1 and y = a sec bx will
produce a minimum. When cos bx = −1, y = a cos bx will
produce a minimum and y = a sec bx will produce
a maximum.
50.
2
−3
0
3
π
Graphing the function produces the same graph as the
cosecant function with asymptotes at 0, ±π, ±2π, . . .
So, csc x = 1 —
2 ( tan
x —
2 + cot
x —
2 ) .
51. Sample answer: y = 5 tan ( 1 — 2 x −
3π — 4 )
Maintaining Mathematical Profi ciency
52. Step 1 Use the three x-intercepts to write the function in
factored form. y = a (x + 1)(x − 1)(x − 3)
Step 2 Find the value of a by substituting the coordinates of
the point (0, 3).
3 = a (0 + 1)(x − 1)(x − 3)
3 = 3a
1 = a
The function is y = 3(x + 1)(x − 1)(x − 3)
= x 3 − 3x 2 − x + 3.
53. Step 1 Use the three x-intercepts to write the function in
factored form.
y = a (x + 2)(x − 1)(x − 3)
Step 2 Find the value of a by substituting the coordinates of
the point (0, −6).
−6 = a (0 + 2)(0 − 1)(0 − 3)
−6 = 6a
−1 = a
The function is y = −(x + 2)(x − 1)(x − 3)
= −x 3 + 2 x 2 + 5x − 6.
54. Step 1 Use the three x-intercepts to write the function in
factored form.
y = a (x + 1)(x − 2)(x − 3)
Step 2 Find the value of a by substituting the coordinates of
the point (1, −2).
−2 = a (1 + 1)(1 − 2)(1 − 3)
−2 = 4a
− 1 —
2 = a
The function is y = − 1 —
2 (x + 1)(x − 2)(x − 3)
= − 1 — 2 x 3 + 2 x 2 −
1 —
2 x − 3.
55. Step 1 Use the three x-intercepts to write the function in
factored form.
y = a (x + 3)(x + 1)(x − 3)
Step 2 Find the value of a by substituting the coordinates of
the point (−2, 1).
1 = a (−2 + 3)(−2 + 1)(−2 − 3)
1 = 5a
1 —
5 = a
The function is y = 1 —
5 (x + 3)(x + 1)(x − 3)
= 1 —
5 x 3 +
1 —
5 x 2 −
9 —
5 x −
9 —
5 .
56. The amplitude is 5 and the period is 2π.
57. The amplitude is 3 and the period is π.
492 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
58. The amplitude is 1 and the period is 8π.
9.6 Explorations (p. 505)
1. a. The function y = 5 sin 1.38π(x + 0.5) can model the
electric current shown. The amplitude is 5 and the period
is about 1.45.
b. The function y = 20 sin π(x + 0.4) can model the electric
current shown. The amplitude is 20 and the period is 2.
c. The function y = 15 sin 2π(x – 0.4) can model the electric
current shown. The amplitude is 15 and the period is 1.
d. The function y = 15 sin 1 —
2 π(x – 1.2) can model the
electric current shown. The amplitude is 15 and the period
is 4.
e. The function y = 12 sin 1.46π (x + 0.2) can model the
electric current shown. The amplitude is 12 and the period
is about 1.37.
f. The function y = 3 sin π(x + 3) can model the electric
current shown. The amplitude is 3 and the period is 2.
2. Sample answer: The problems involve oscillating motion or
patterns that repeat in cycles.
3. Answers will vary.
9.6 Monitoring Progress (pp. 507–509)
1. Because period and frequency are inversely related to each
other, the period would increase because the frequency is
decreased. The new function would be P = 2 sin 2000π t.
2. Sample answer:
Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 2 and the minimum
value is –2.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
——— 2 =
2 + (–2) —
2 =
0 —
2 = 0
Step 3 Decide whether the graph should be modeled by a
sine or cosine function. Because the graph crosses
the y-axis at a maximum value, the graph is a cosine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
2π — 3 =
2π — b → b = 3.
The amplitude is
∣ a ∣ = (maximum value) – (minimum value) ———
2 =
2 – (–2) —
2 =
4 —
2 = 2.
The graph is not a refl ection, so a > 0. Therefore,
a = 2. The function is y = 2 cos 3x.
3. Sample answer:
Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 1 and the minimum
value is –3.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
——— 2 =
1 + (–3) —
2 =
−2 —
2 = −1
Step 3 Decide whether the graph should be modeled by a
sine or cosine function. Because the graph crosses
the midline y = –1 on the y-axis, the graph is a sine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
2 = 2π — b → b = π.
The amplitude is
∣ a ∣ = (maximum value) – (minimum value)
——— 2 =
1 – (–3) —
2 =
4 —
2 = 2.
The graph is not a refl ection, so a > 0. Therefore,
a = 2. The function is y = 2 sin π x – 1.
4. The amplitude changes to 32.5 and the vertical shift becomes
37.5, but the period is not affected.
5. Enter the data in a graphing calculator and make a scatter
plot. The scatter plot appears sinusoidal. So, perform a
sinusoidal regression. Graph the data and the model in the
same viewing window.
120
0
80
The model appears to be a good fi t. So, a model for the data
is T = 21.8 sin (0.514m – 2.18) + 51.3. The period of the
graph represents the amount of time it takes for the weather
to repeat its cycle, which is about 12 months.
9.6 Exercises (pp. 510–512)
Vocabulary and Core Concept Check
1. Graphs of sine and cosine functions are called sinusoids.
2. From the graph, it can be determined that the period is 1 —
3 .
The frequency is the reciprocal of the period, so 3.
Copyright © Big Ideas Learning, LLC Algebra 2 493All rights reserved. Worked-Out Solutions
Chapter 9
Monitoring Progress and Modeling with Mathematics
3. The period is 2π.
frequency = 1 —
period
= 1 —
2π
4. The period is 2π — 3 .
frequency = 1 —
period
= 1 —
2π — 3
= 3 —
2π
5. The period is π — 2 .
frequency = 1 —
period
= 1 —
π — 2
= 2 —
π
6. The period is π.
frequency = 1 —
period
= 1 —
π
7. The period is 2 —
3 .
frequency = 1 —
period
= 1
— 2 —
3
= 3 —
2
8. The period is 8.
frequency = 1 —
period
= 1 —
8
9. The period is 8π — 3 .
frequency = 1 —
period
= 1
— 8π — 3
= 3 —
8π
10. The period is 10π.
frequency = 1 —
period
= 1 —
10π
11. Step 1 Find the values of a and b in the model P = a sin bt. The maximum pressure is 0.02, so a = 0.02. Use the
frequency to fi nd b.
frequency = 1 —
period
20 = b —
2π
40π = b
The pressure P as a function of time t is given by
P = 0.02 sin 40π t. Step 2 Graph the model. The amplitude is a = 0.02 and the
period is
1 —
f =
1 —
20 .
The key points are:
Intercepts: (0, 0); (0.025, 0); (0.05, 0)
Maximum: (0.0125, 0.02)
Minimum: (0.0375, –0.02)
x
y
0.01
0.02
−0.01
−0.02
0.01 0.02 0.03 0.04
494 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
12. Step 1 Find the values of a and b in the model P = a sin bt. The maximum pressure is 5, so a = 5. Use the
frequency to fi nd b.
frequency = 1 —
period
440 = b —
2π
880π = b
The pressure P as a function of time t is given by
P = 5 sin 880π t. Step 2 Graph the model. The amplitude is a = 5 and the
period is 1 —
f =
1 —
440 .
The key points are:
Intercepts: (0, 0); ( 1 —
880 , 0 ) ; ( 1
— 440
, 0 ) Maximum: ( 1
— 1760
, 5 ) Minimum: ( 3
— 1760
, −5 )
x
y
4
2
1
3
5
−4
−3
−5
−2
12200
11100
32200
1550
13. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 3 and the minimum
value is −3.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= 3 + (−3)
— 2 =
0 —
2 = 0
Step 3 Decide whether the graph should be modeled by a
sine or cosine function. Because the graph crosses
the midline y = 0 on the y-axis, the graph is a sine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
π = 2π — b → b = 2.
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 3 + (−3)
— 2 =
6 —
2 = 3.
The graph is not a refl ection, so a > 0. Therefore, a = 3.
The function is y = 3 sin 2x.
14. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 5 and the minimum
value is −5.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= 5 + (−5)
— 2 =
0 —
2 = 0
Step 3 Decide whether the graph should be modeled by a
sine or cosine function. Because the graph crosses
the y-axis at a maximum value, the graph is a cosine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
π — 2 =
2π — b → b = 4.
The amplitude is
∣ a ∣ = (maximum value) – (minimum value)
———— 2
= 5 – (−5)
— 2 =
10 —
2 = 5.
The graph is not a refl ection, so a > 0. Therefore, a = 5.
The function is y = 5 cos 4x.
15. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 2 and the minimum
value is −2.
Step 2 Identify the vertical shift, k. The value k is the mean
of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= 2 + (−2)
— 2 =
0 —
2 = 0.
Step 3 Decide whether the graph should be modeled by
a sine or cosine function. Because the graph has a
minimum value on the y-axis, the graph is a cosine
curve with a horizontal shift 4 units left.
So, h = −4.
Step 4 Find the amplitude and period. The period is
4 = 2π — b → b =
π — 2 .
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 2 − (−2)
— 2 =
4 —
2 = 2.
The graph is a refl ection, so a < 0. Therefore, a = −2.
The function is y = −2 cos π — 2 (x + 4).
Copyright © Big Ideas Learning, LLC Algebra 2 495All rights reserved. Worked-Out Solutions
Chapter 9
16. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is −1 and the minimum
value is −3.
Step 2 Identify the vertical shift, k. The value k is the mean
of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= −1 + (−3)
— 2 =
−4 —
2 = −2
Step 3 Decide whether the graph should be modeled by a
sine or cosine function. Because the graph crosses
the midline y = −2 on the y-axis, the graph is a sine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
2 = 2π — b → b = π.
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= −1 − (−3)
— 2 =
2 —
2 = 1.
The graph is a refl ection, so a < 0. Therefore, a = −1.
The function is y = −sinπ x − 2.
17. To fi nd the amplitude, take half of the difference between the
maximum and the minimum.
10 − (−6)
— 2 =
16 —
2 = 8
18. To fi nd the vertical shift, use the y-coordinates of the points.
−2 + (−8)
— 2 =
−10 —
2 = −5
19. Step 1 Identify the maximum and minimum values. The
maximum height is 9 feet. The minimum height is
4 feet.
Step 2 Identify the vertical shift, k.
k = (maximum value) + (minimum value)
———— 2
= 9 + 4
— 2 =
13 —
2 = 6.5
Step 3 Decide whether the height should be modeled by a
sine or cosine function. When t = 0, the height is at
its minimum. So, use a cosine function whose graph
is a refl ection in the x-axis with no horizontal shift
(h = 0).
Step 4 Find the amplitude and period.
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 9 − 4
— 2 = 2.5.
Because the graph is a refl ection in the x-axis, a < 0.
So, a = −2.5. Because the fl ywheel is rotating at a rate of
1 revolution per 2 seconds, one revolution is completed in
2 seconds. So, the period is 2π — b = 2, and b = π.
A model for the height of the handle is
h = −2.5 cos π t + 6.5.
20. Step 1 Identify the maximum and minimum values.
The maximum height of the bucket is 70.5 feet.
The minimum height is −2 feet.
Step 2 Identify the vertical shift, k.
k = (maximum value) + (minimum value)
———— 2
= 70.5 + (−2)
—— 2 = 34.25
Step 3 Decide whether the height should be modeled by a
sine or cosine function. When t = 0, the height is at its
minimum. So, use a cosine function whose graph is a
refl ection in the x-axis with no horizontal shift (h = 0).
Step 4 Find the amplitude and period.
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 70.5 – (−2)
— 2 = 36.25.
Because the graph is a refl ection in the x-axis, a < 0.
So, a = −36.25.
Because the bucket is rotating at a rate of 1 revolution per
24 seconds, one revolution is completed in 24 seconds.
So, the period is 2π — b = 24, and b =
π — 12
.
A model for the height of the bucket is
h = −36.25 cos π —
12t + 34.25.
21. Enter the data in a graphing calculator. Make a scatter plot.
The scatter plot appears sinusoidal. So, perform a sinusoidal
regression. Graph the data and the model in the same
viewing window.
130
0
110
The model appears to be a good fi t. So, a model for the data
is D = 19.81 sin (0.549t − 2.40) + 79.8. The period of the
graph represents the amount of time it takes for the weather
to repeat its cycle, which is about 11.4 months.
22. Enter the data in a graphing calculator. Make a scatter plot.
The scatter plot appears sinusoidal. So, perform a sinusoidal
regression. Graph the data and model in the same viewing
window.
130
0
100
The model appears to be a good fi t. So, a model for the data
is D = 7.38 sin (0.498t − 2.05) + 78.6. The period of the
graph represents the amount of time it takes for the weather
to repeat its cycle, which is about 12.6 months.
496 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
23. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 100 and the minimum
value is −100.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= 100 + (−100)
—— 2 =
0 —
2 = 0
Step 3 Decide whether the graph should be modeled by a
sine or cosine curve. Because the graph crosses the
midline y = 0 on the y-axis, the graph is a sine curve
with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
1 —
2 =
2π —
b → b = 4π.
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 100 − (−100)
—— 2 =
200 —
2 = 100.
The graph is not a refl ection, so a > 0. Therefore, a = 100.
The function is V = 100 sin 4π t.
24. Louisville has the greater average daily temperature. The
graph of the average daily temperature for Louisville is
always higher than the one for Lexington.
t
y
40
20
10
30
50
60
70
80
421 3 5 6 7 8 9 10 11
Louisville
Lexington
25. a. Using sinusoidal regression on a calculator, the data can
be modeled by N = 3.68 sin (0.776t − 0.70) + 20.4.
b. Using the model, when t = 12,
N = 3.68 sin (0.776(12) − 0.70) + 20.4 ≈ 23.1. So, there
are about 23,100 employees in the 12th year.
26.
x
y2
−2
−1
4 631−1−3−6
(−2π, 1)
(−π, −1) (π, −1)
(2π, 1)(0, 1)
( , 0)π2π(− , 0)π
2π
( , 0)32π(− , 0)3
2π
The slope of the graph of y = sin x is given by the function
y = cos x.
27. a. and b. Use a cosine function because it does not require
determining a horizontal shift.
c. Use a sine function because it does not require
determining a horizontal shift.
28. The frequency is 2 because the graph completes 2 full cycles
in 1 unit of x.
29. The vertical shift is k = 8 + 3
— 2 =
11 —
2 = 5.5. The period is
π — 2 =
2π — b ⇒ b = 4. The amplitude is ∣ a ∣ =
8 − 3 —
2 =
5 —
2 = 2.5.
The midline is 5.5. For the sine function, the graph is
shifted π — 8 units right, and for the cosine function, the graph is
a refl ection. So, the sine function is y = 2.5 sin 4 ( x − π — 8 ) + 5.5
and the cosine function is y = −2.5 cos 4x + 5.5.
30. Your friend is incorrect. The period is the reciprocal of
the frequency. The reciprocal of 1 —
2 , 2, is greater than the
reciprocal of 2, 1 —
2 .
31. a. The vertical shift is 16.5 + 3.5
— 2 = 10, so k = 10.
The period is 12, so b = π — 6 . The amplitude is
16.5 – 3.5
— 2 = 6.5, so a = −6.5. Therefore, the model is
d = −6.5 cos π — 6 t + 10.
b.
Time (hours)
Dep
th (
feet
)
t
d
8
12
16
4
012 1860
Low tide occurs at 12:00 a.m. (t = 0) and 12:00 p.m.
(t = 12), and high tide occurs at 6:00 a.m. (t = 6) and
6:00 p.m. (t = 18).
c. The graphs are related by a horizontal shift to the left by
3 units.
Maintaining Mathematical Profi ciency
32. 17
— √
— 2 =
17 —
√—
2 ⋅
√—
2 —
√—
2
= 17 √
— 2 —
( √—
2 ) 2
= 17 √
— 2 —
2
Copyright © Big Ideas Learning, LLC Algebra 2 497All rights reserved. Worked-Out Solutions
Chapter 9
33. 3 —
√—
6 − 2 =
3 —
√—
6 − 2 ⋅
√—
6 + 2 —
√—
6 + 2
= 3 ( √
— 6 + 2 ) —
√—
6 2 − 22
= 6 + 3 √
— 6 —
2
34. 8 —
√—
10 + 3 =
8 —
√—
10 + 3 ⋅
√—
10 − 3 —
√—
10 − 3
= 8 ( √
— 10 − 3 ) —
√—
10 2 − 32
= −24 + 8 √—
10
35. 13 —
√—
3 + √—
11 =
13 —
√—
3 + √—
11 ⋅
√—
3 − √—
11 —
√—
3 − √—
11
= 13 ( √
— 3 − √
— 11 ) ——
√—
3 2 − √—
11 2
= 13 √
— 11 − 13 √
— 3 ——
8
36. log8 x —
7 = log8 x − log8 7
37. ln 2x = ln 2 + ln x
38. log3 5x3 = log3 5 + log3 x3
= log3 5 + 3 log3 x
39. ln 4x6
— y = ln 4x6 − ln y
= ln 4 + ln x6 − ln y
= ln 4 + 6 ln x − ln y
9.7 Explorations (p. 513)
1. a. By the Pythagorean Theorem, a2 + b2 = c2.
b. The expressions are sin θ = opp.
— hyp.
= b —
c and
cos θ = adj.
— hyp.
= a —
c .
c. a2 + b2 = c2
a2
— c2
+ b2
— c2
= c2
— c2
sin2 θ + cos2 θ = 1
d. Sample answer:
θ sin2 θ cos2 θ sin2 θ + cos2 θ
QI π — 3
3 —
4
1 —
4 1
QII 3π — 4
1 —
2
1 —
2 1
QIII 7π — 6
1 —
4
3 —
4 1
QIV 7π — 4
1 —
2
1 —
2 1
2. a. sin2 θ + cos2 θ = 1
sin2 θ — cos2 θ
+ cos2 θ — cos2 θ =
1 —
cos2 θ
1 + tan2 θ = sec2 θ
b. sin2 θ + cos2 θ = 1
sin2 θ — sin2 θ
+ cos2 θ — sin2 θ
= 1 —
sin2 θ
cot2 θ + 1 = csc2 θ
3. Sample answer: To verify a trigonometric identity,
manipulate one side of an equation until both sides are equal.
4. no; sin θ = cos θ is not true for all values of θ. For example,
sin π — 3 =
√—
3 —
2 and cos
π — 3 =
1 —
2 .
5. Sample answer: From Lesson 9.1, you know that
csc θ = 1 —
sin θ and sec θ = 1 —
cos θ . These are examples of
trigonometric identities.
9.7 Monitoring Progress (pp. 515–516)
1. Find sin θ.
sin2 θ + csc2 θ = 1
sin2 θ + ( 1 — 6 ) 2 = 1
sin2 θ = 1 − ( 1 — 6 ) 2
sin2 θ = 35
— 36
sin θ = ± √—
35 —
6
sin θ = √
— 35 —
6
Find the values of the other four trigonometric functions of θ
using values of sin θ and cos θ.
tan θ = sin θ — csc θ
= √
— 35
—
6 —
1 —
6
= √
— 35 cot θ =
cos θ — sin θ
= 1 — 6 —
√
— 35 —
6 =
√—
35 —
35
csc θ = 1 —
sin θ =
1 — √
— 35
— 6 =
6 √—
35 —
35 sec θ =
1 —
cos θ = 1 —
1 — 6 = 6
2. sin x cot x sec x = ( sin x ) ( cos x —
sin x ) ( 1
— cos x
) = 1
3. cos θ − cos θ sin2 θ = cos θ (1 − sin2 θ)
= cos θ (cos2 θ)
= cos3 θ
498 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
4. tan x csc x — sec x
=
( sin x —
cos x ) ( 1
— sin x
) ——
1 —
cos x
=
1 —
cos x —
1 —
cos x
= 1
5. cot(−θ) = 1 —
tan(−θ)
= 1 —
−tan θ
= −cot θ
6. csc2 x(1 − sin2 x) = csc2 x(cos2 x)
= ( 1 —
sin2x ) cos2 x
= cos2 x
— sin2 x
= cot2 x
7. cos x csc x tan x = cos x ⋅ 1 —
sin x ⋅
sin x —
cos x
= 1
8. (tan2 x + 1)(cos2 x − 1) = (sec2 x)(−sin2 x)
= ( 1 —
cos2x ) −sin2x
= − sin2 x
— cos2 x
= −tan2 x
9.7 Exercises (pp. 517–518)
Vocabulary and Core Concept Check
1. A trigonometric equation is true for some values of a
variable but a trigonometric identity is true for all values of
the variable for which both sides of the equation are defi ned.
2. The reciprocal identity for secant can be used to write an
expression in terms of cosine. The negative angle identity can be
used to simplify the expression and then the reciprocal identity
can again be used to write the expression in terms of cosine.
Monitoring Progress and Modeling with Mathematics
3. Find cos θ.
sin2 θ + cos2 θ = 1
( 1 — 3 ) 2 + cos2 θ = 1
cos2 θ = 1 − ( 1 — 3 ) 2
cos2 θ = 8 —
9
cos θ = ± 2 √
— 2 —
3
cos θ = 2 √
— 2 —
3
Find the values of the other four trigonometric functions of θ
using the values of sine and cosine.
tan θ = sin θ — cos θ
=
1 —
3
—
2 √
— 2 —
3
= 2 —
√—
4 cot θ =
cos θ — sin θ
=
2 √
— 2 —
3
— 1 —
3
= 2 √—
2
csc θ = 1 —
sin θ = 1 —
1 —
3 = 3 sec θ =
1 —
cos θ =
1 —
2 √
— 2 —
3
= 3 √
— 2 —
4
4. Find cos θ.
sin2 θ + cos2 θ = 1
( −7 —
10 ) 2 + cos2 θ = 1
cos2 θ = 1 − ( −7 —
10 ) 2
cos2 θ = 51
— 100
cos θ = ± √
— 51 —
10
cos θ = − √—
51 — 10
Find the values of the other four trigonometric functions of θ
using the values of sine and cosine.
tan θ = sin θ — cos θ
=
− 7 —
10
—
− √
— 51 —
10
= 7 √
— 51 —
51
cot θ = cos θ — sin θ
=
− √
— 51 —
10
—
− 7 —
10
= √
— 51 —
7
csc θ = 1 — sin θ
= 1 —
− 7 —
10
= − 10
— 7
sec θ = 1 —
csc θ =
1 —
− √
— 51 —
10
= − 10 √
— 51 —
51
Copyright © Big Ideas Learning, LLC Algebra 2 499All rights reserved. Worked-Out Solutions
Chapter 9
5. Find sec θ.
1 + tan2 θ = sec2 θ
1 + ( − 3 —
7 ) 2 = sec2 θ
58 — 49
= sec2 θ
± √
— 58 —
7 = sec θ
− √—
58 — 7 = sec θ
Find the values of the other four trigonometric functions of θ
using the values of tangent and secant.
sin θ = tan θ — sec θ
=
− 3 —
7
—
− √
— 58 —
7
= 3 √
— 58 —
58
cos θ = 1 —
sec θ =
1 —
− √
— 58 —
7
= − 7 √—
58 —
58
cot θ = 1 —
tan θ =
1 —
− 3 —
7
= − 7 — 3
csc θ = sec θ — tan θ
=
− √
— 58 —
7
—
− 3 —
7
= √
— 58 —
3
6. Find csc θ.
1 + cot2 θ = csc2 θ
1 + ( − 2 —
5 ) 2 = csc2 θ
29 — 25
= csc2 θ
± √
— 29 —
5 = csc θ
√—
29 — 5 = csc θ
Find the values of the other four trigonometric functions of θ
using the values of cosecant and cotangent.
sin θ = 1 —
csc θ =
1 —
√
— 29 —
5
= 5 √
— 29 —
29
cos θ = cot θ — csc θ
=
− 2 —
5
—
√
— 29 —
5
= − 2 √
— 29 —
29
tan θ = 1 —
cot θ =
1 —
− 2 —
5
= − 5 —
2
sec θ = csc θ — cot θ
=
√
— 29 —
5
—
− 2 —
5
= − √
— 29 —
2
7. Find sin θ.
sin2 θ + cos2 θ = 1
sin2 θ + ( − 5 —
6 ) 2 = 1
sin2 θ = 1 − ( − 5 —
6 ) 2
sin2 θ = 11
— 36
sin θ = ± √
— 11 —
6
sin θ = − √
— 11 —
6
Find the values of the other four trigonometric functions of θ
using the values of sine and cosine.
tan θ = sin θ — cos θ
=
− √
— 11 —
6
—
− 5 —
6
= √
— 11 —
5
cot θ = cos θ — sin θ
=
− 5 —
6
—
− √
— 11 —
6
= 5 √
— 11 —
11
sec θ = 1 —
cos θ =
1 —
− 5 —
6
= − 6 —
5
csc θ = 1 —
sin θ =
1 —
− √
— 11 —
6
= − 6 √
— 11 —
11
8. Find tan θ.
1 + tan2 θ = sec2 θ
1 + tan2 θ = ( 9 — 4 ) 2
tan2 θ = ( 9 — 4 ) 2 −1
tan2 θ = 65
— 16
tan θ = ± √
— 65 —
4
tan θ = − √—
65 —
4
Find the values of the other four trigonometric functions of θ
using the values of tangent and secant.
sin θ = tan θ — sec θ
=
− √
— 65 —
4
— 9 —
4
= − √
— 65 —
9
cos θ = 1 —
sec θ =
1 —
9 —
4
= 4 —
9
cot θ = 1 —
tan θ =
1 —
− √
— 65 —
4
= − 4 √
— 65 —
65
csc θ = sec θ — tan θ
=
9 —
4
—
− √
— 65 —
4
= − 9 √—
65 —
65
500 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
9. Find csc θ.
1 + cot2 θ = csc2 θ 1 + (−3)2 = csc2 θ 10 = csc2 θ ± √
— 10 = csc θ
− √—
10 = csc θ
Find the values of the other four trigonometric functions of θ
using the values of cosecant and cotangent.
sin θ = 1 —
csc θ =
1 —
− √—
10 = −
√—
10 —
10
cos θ = cot θ — csc θ
= −3
— − √
— 10 =
3 √—
10 —
10
tan θ = 1 —
cot θ =
1 —
−3 = −
1 —
3
sec θ = csc θ — cot θ
= − √
— 10 —
−3 =
√—
10 —
3
10. Find cot θ.
1 + cot2 θ = csc2 θ
1 + cot2 θ = ( − 5 —
3 ) 2
cot2 θ = ( − 5 —
3 ) 2 −1
cot2 θ = 16
— 9
cot θ = 4 —
3
Find the values of the other four trigonometric functions of θ
using the values of cosecant and tangent.
sin θ = 1 —
csc θ =
1 —
− 5 —
3
= − 3 — 5 cos θ =
cot θ — csc θ
=
4 —
3
— −
5 —
3
= − 4 — 5
tan θ = 1 —
cot θ =
1 —
4 —
3
= 3 —
4 sec θ =
csc θ — cot θ
=
− 5 —
3
— 4 —
3
= − 5 — 4
11. sin x cot x = (sin x) ( cos x —
sin x )
= cos x
12. cos θ (1 + tan2 θ) = cos θ(sec2 θ)
= cos θ ( 1 —
cos2 θ )
= 1 —
cos θ
= sec θ
13. sin(−θ) —
cos(−θ) =
−sin θ — cos θ
= −tan θ
14. cos2 x — cot2 x
= cos2 x
—
( cos2 x —
sin2 x )
= sin2 x
15. cos ( π —
2 − x ) —
csc x =
sin x —
1 —
sin x
= sin2 x
16. sin ( π — 2 − θ ) sec θ = cos θ ( 1
— cos θ )
= 1
17. csc2 x − cot2 x —— sin(−x) cot(x)
= 1 ——
sin(−x) cot x
= 1 —
−sin x cot x
= 1 ——
−sin x ( cos x —
sin x )
= − 1 —
cos x
= −sec x
18. cos2 x tan2(−x) − 1 ——
cos2 x =
cos2 x tan2 x − 1 ——
cos2 x
=
cos2 x ( sin2 x —
cos2 x ) − 1 ——
cos2 x
= sin2 x − 1
— cos2 x
= −cos2 x
— cos2 x
= −1
Copyright © Big Ideas Learning, LLC Algebra 2 501All rights reserved. Worked-Out Solutions
Chapter 9
19. cos ( π —
2 − θ ) —
csc θ + cos2 θ =
sin θ —
1 —
sin θ + cos2 θ
= sin2 θ + cos2 θ
= 1
20. sec x sin x + cos ( π —
2 − x ) ———
1 + sec x =
sec x sin x + sin x ——
1 + sec x
= sin x(sec x + 1)
—— 1 + sec x
= sin x
21. (1 + cos2 θ) was substituted for sin2 θ instead of (1 − cos2 θ).
1 − sin2 θ = 1 − (1 − cos2 θ)
= 1 − 1 + cos2 θ = cos2 θ
22. The identity for tan x involving sin x and cos x is
tan x = sin x
— cos x
.
tan x csc x = sin x
— cos x
⋅ 1 —
sin x
= 1 —
cos x
= sec x
23. sin x csc x = sin x ⋅ 1 —
sin x
= 1
24. tan θ csc θ cos θ = sin θ — cos θ
⋅ 1 —
sin θ ⋅ cos θ
= 1
25. cos ( π — 2 − x ) cot x = sin x ⋅
cos x —
sin x
= cos x
26. sin ( π — 2 − x ) tan x = cos x ⋅
sin x —
cos x
= sin x
27. cos ( π —
2 − θ ) + 1
—— 1 − sin(−θ)
= sin θ + 1
—— 1 − sin(− θ)
= sin θ + 1
—— 1 − (−sin θ)
= sin θ + 1
— 1 + sin θ
= 1
28. sin2(−x) —
tan2 x =
(−sin x)2
— tan2 x
= sin2 x
— tan2 x
= sin2 x ⋅ cot2 x
= sin2 x ⋅ cos2 x
— sin2 x
= cos2 x
29. 1 + cos x — sin x
+ sin x —
1 + cos x =
1 + cos x —
sin x +
sin x(1 − cos x) ——
(1 + cos x)(1 − cos x)
= 1 + cos x
— sin x
+ sin x(1 − cos x)
—— 1 − cos2 x
= 1 + cos x
— sin x
+ sin x(1 − cos x)
—— sin2 x
= sin x(1 + cos x)
—— sin2 x
+ sin x(1 − cos x)
—— sin2 x
= sin x(1 + cos x) + sin x(1 − cos x)
——— sin2 x
= sin x(1 + cos x + 1 − cos x)
——— sin2 x
= sin x(2)
— sin2 x
= 2 —
sin x
= 2 csc x
30. sin x —— 1 − cos(−x)
= sin x —
1 − cos x
= sin x(1 + cos x)
—— (1 − cos x)(1 + cos x)
= sin x(1 + cos x)
—— 1 − cos2 x
= sin x(1 + cos x)
—— sin2 x
= 1 + cos x
— sin x
= 1 —
sin x +
cos x —
sin x
= csc x + cot x
502 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
31.
x
y
2
1
3
−2π
( , 1)π2π
(− , −1)π2π
y = csc x
y = sin x
x
y
2
1
3
( , 1)π4π
(− , −1)π4π
y = tan x
y = cot x
−2π
x
y
3
−1
−3
−2
2π−2π
(− , −1)π ( , −1)π
y = sec xy = cos x
The functions sin x, csc x, tan x, and cot x are odd functions,
and cos x and sec x are even functions.
sin(−θ) = −sin θ
csc(−θ) = 1 —
sin(− θ) =
1 —
−sin θ = −csc θ
tan(−θ) = −tan θ cot(−θ) =
1 —
tan(− θ) =
1 —
−tan θ = −cot θ
cos(−θ) = cos θ
sec(−θ) = 1 —
cos(− θ) =
1 —
cos θ = sec θ
32. Just as 1 —
x decreases as x increases, sec θ =
1 —
cos θ decreases as
cos θ increases. This happens on the intervals
π + 2nπ < θ < 3π — 2 + 2nπ and
3π — 2 + 2nπ < θ < 2π + 2nπ,
where n is an integer.
33. Your answers are equivalent.
sec x tan x − sin x = 1 —
cos x ⋅
sin x —
cos x − sin x
= sin x
— cos2 x
− sin x
= sec2 x sin x − sin x
= sin x(sec2 x −1)
= sin x tan x
34. a. The function sin θ is positive because y is positive and
both cos θ and tan θ are negative because x is negative.
b. The angle −θ lies in Quadrant III.
c. The function sin(−θ) is negative because y is negative,
cos (−θ) is negative because x is negative, and tan (−θ) is
positive because x and y are negative.
35. s = h sin (90° − θ)
—— sin θ
s = h cos θ — sin θ
s = h cot θ
36. Sample answer: Multiply both sides by 1 —
cos x , so
sin x —
cos x = 1.
Then tan x = 1 because sin x
— cos x
= tan x.
37. a. u Wcos θ = Wsin θ u cos θ = sin θ
u = sin θ — cos θ
u = tan θ b. As θ increases from 0° to 90°, u starts at 0 and increases
without bound.
38. a. n1 ——
√— cot2 θ 1 + 1 =
n2 ——
√— cot2 θ 2 + 1
n1 —
√— csc2 θ 1 =
n2 —
√—
csc2 θ 2
n1 —
csc θ 1 =
n2 — csc θ 2
n1 sin θ1 = n2 sin θ2
b. When θ1 = 55°, θ2 = 35°, and n2 = 2:
n1 sin 55° = 2 sin 35°
n1 = 2 sin 35°
— sin 55°
n1 ≈ 1.4
c. It would be the case that n1 = n2. This situation could
occur when the mediums have the same composition.
39. You can obtain the graph of y = cos x by refl ecting the
graph of f (x) = sin x in the y-axis and translating it
π — 2 units right.
40. a. ln ∣ sec θ ∣ = ln 1 —
∣ cos θ ∣
= ln ∣ (cos θ)−1 ∣ = −ln ∣ cos θ ∣ b. ln ∣ tan θ ∣ = ln ∣ sin θ —
cos θ ∣
= ln ∣ sin θ ∣ − ln ∣ cos θ ∣
Copyright © Big Ideas Learning, LLC Algebra 2 503All rights reserved. Worked-Out Solutions
Chapter 9
Maintaining Mathematical Profi ciency
41. Write an equation using a trigonometric function that
involves the ratio of x and 11.
cot 45° = adj.
— opp.
1 = x —
11
11 = x
The length of the side is x = 11.
42. Write an equation using a trigonometric function that
involves the ratio of x and 13
sin 60° = opp.
— hyp.
√
— 3 —
2 =
x —
13
13 √
— 3 —
2 = x
The length of the side is x = 13 √
— 3 —
2 .
43. Write an equation using a trigonometric function that
involves the ratio of x and 7.
cos 30° = adj.
— hyp.
√
— 3 —
2 =
7 —
x
x √—
3 = 14
x = 14
— √
— 3
x = 14 √
— 3 —
3
The length of the side is x = 14 √
— 3 —
3 .
9.8 Explorations (p. 519)
1. a. Sample answer: For each triangle shown, d is a side that is
opposite the angle a − b and the other two sides are 1 unit
in length. So, the triangles are congruent by the SAS
Congruence Theorem.
b. d = √———
(cos b − cos a) 2 + (sin b − sin a) 2
c. d = √————
(cos(a − b) − 1) 2 + (sin(a − b) − 0) 2
d. √———
(cos b − cos a) 2 + (sin b − sin a) 2
= √————
(cos(a − b) − 1) 2 + (sin(a − b) − 0) 2
(cos b − cos a) 2 + (sin b − sin a) 2
= (cos(a − b) − 1) 2 + (sin(a − b) − 0) 2
cos2 b − 2 cos a cos b + cos2 a + sin2 b
− 2 sin a sin b + sin2 a = cos2(a − b) − 2 cos(a − b) + 1 + sin2(a − b)
−2 cos a cos b − 2 sin a sin b + 2
= −2 cos(a − b) + 2
cos a cos b + sin a sin b
= cos(a − b)
2. cos(a + b) = cos[a − (−b)]
= cos a cos(−b) + sin a sin(−b)
= cos a cos b + sin a(−sin b)
= cos a cos b − sin a sin b
3. sin(a − b) = cos [ π — 2 − (a − b) ]
= cos [ ( π — 2 − a ) + b ]
= cos ( π — 2 − a ) cos b − sin ( π —
2 − a ) sin b
= sin a cos b − cos a sin b
sin(a + b) = cos [ π — 2 − (a + b) ]
= cos [ ( π — 2 − a ) − b ]
= cos ( π — 2 − a ) cos b + sin ( π —
2 − a ) sin b
= sin a cos b + cos a sin b
4. Sample answer: Rewrite the formula using a sum or
difference formula and simplify.
5. a. sin 75° can be written as sin(45° + 30°) and cos 75° can
be written as cos(45° + 30°) so that the sum formulas can
be applied.
sin 75° = sin(45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
= √
— 6 + √
— 2 —
4
cos 75° = cos(45° + 30°)
= cos 45° cos 30° − sin 45° sin 30°
= √—
6 − √—
2 —
4
b. sin 75° = sin(120° − 45°)
= sin 120° cos 45° − cos 120° sin 45°
= √
— 6 + √
— 2 —
4
cos 75° = cos(120° − 45°)
= cos 120° cos 45° + sin 120° sin 45°
= √
— 6 − √
— 2 —
4
The values use the same.
504 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
9.8 Monitoring Progress (pp. 521–522) 1. sin 105° = sin(60° + 45°)
= sin 60° cos 45° + cos 60° sin 45°
= √
— 3 —
2 ( √
— 2 —
2 ) + 1 —
2 ( √
— 2 —
2 )
= √
— 6 + √
— 2 —
4
2. cos 15° = cos(45° − 30°)
= cos 45° cos 30° + sin 45° sin 30°
= √
— 2 —
2 ( √
— 3 —
2 ) +
√—
2 —
2 ( 1 —
2 )
= √
— 6 + √
— 2 —
4
3. tan 5π — 12
= tan ( π — 6 +
π — 4 )
=
tan π — 6 + tan
π — 4
—— 1 − tan
π — 6 tan
π — 4
=
√
— 3 —
3 + 1
—
1 − √
— 3 —
3 (1)
= 2 + √—
3
4. cos π — 12
= cos ( π — 3 −
π — 4 )
= cos π — 3 cos
π — 4 + sin
π — 3 sin
π — 4
= 1 —
2 ( √
— 2 —
2 ) +
√—
3 —
2 ( √
— 2 —
2 )
= √
— 6 + √
— 2 —
4
5. Find cos a and sin b. Because sin a = 8 —
17 and a is in
Quadrant I, cos a = 15
— 17
, as shown in the fi gure.
178
172 − 82 = 15
ax
y
Because cos b = − 24
— 25 and b is in Quadrant III, sin b = − 7 — 25 ,
as shown in the fi gure.
24
25252 − 242 = 7 b x
y
Use the difference formula for sine to fi nd sin(a − b).
sin(a − b) = sin a cos b − cos a sin b
= 8 — 17
( − 24
— 25 ) − 15
— 17
( − 7 — 25 )
= − 87
— 425
6. sin(x + π) = sin x cos π + cos x sin π = (sin x)(−1) + (cos x)(0)
= −sin x
7. cos(x − 2π) = cos x cos 2π + sin x sin 2π = (cos x)(1) + (sin x)(0)
= cos x
8. tan(x − π) = tan x − tan π ——
1 + tan x tan π
= tan x − 0 ——
1 + (tan x)(0)
= tan x
9. sin ( π — 4 − x ) − sin ( x +
π — 4 ) = 1
sin π — 4 cos x − cos
π — 4 sin x − ( sin x cos
π — 4 + cos x sin
π — 4 ) = 1
√
— 2 —
2 cos x −
√—
2 —
2 sin x −
√—
2 —
2 sin x −
√—
2 —
2 cos x = 1
− √—
2 sin x = 1
sin x = − √
— 2 —
2
In the interval 0 ≤ x < 2π, the solutions are x = 5π — 4 and x =
7π — 4 .
9.8 Exercises (pp. 523–524)
Vocabulary and Core Concept Check
1. The expression is cos 170°.
2. First break 75° into the sum or difference of two angles
whose tangent values are known such as 45° + 30°. Rewrite
the expression using the corresponding sum or difference
formula and evaluate.
Monitoring Progress and Modeling with Mathematics
3. tan(−15°) = tan(30° − 45°)
= tan 30° − tan 45° ——
1 + tan 30° tan 45°
=
√
— 3 —
3 − 1
—
1 + √
— 3 —
3 (1)
= √—
3 − 2
4. tan 195° = tan(150° + 45°)
= tan 150° + tan 45°
—— 1 − tan 150° tan 45°
= −
√—
3 —
3 + 1
——
1 − ( − √
— 3 —
3 ) (1)
= 2 − √—
3
Copyright © Big Ideas Learning, LLC Algebra 2 505All rights reserved. Worked-Out Solutions
Chapter 9
5. sin 23π — 12
= sin ( π — 4 +
5π — 3 )
= sin π — 4 cos
5π — 3 + cos
π — 4 sin
5π — 3
= √
— 2 —
2 ( 1 —
2 ) +
√—
2 —
2 ( −
√—
3 —
2 )
= √
— 2 − √
— 6 —
4
6. sin(−165°) = sin(−120° + (−45°))
= sin(−120°) cos(−45°) + cos(−120°) sin(−45°)
= − √
— 3 —
2 ( √
— 2 —
2 ) + ( −
1 —
2 ) ( −
√—
2 —
2 )
= √
— 2 − √
— 6 —
4
7. cos 105° = cos(60° + 45°)
= cos 60° cos 45° − sin 60° sin 45°
= 1 —
2 ( √
— 2 —
2 ) −
√—
3 —
2 ( √
— 2 —
2 )
= √
— 2 − √
— 6 —
4
8. cos 11π — 12
= cos ( π — 4 +
2π — 3 )
= cos π — 4 cos
2π — 3 − sin
π — 4 sin
2π — 3
= √
— 2 —
2 ( −
1 —
2 ) −
√—
2 —
2 ( √
— 3 —
2 )
= − √—
2 − √—
6 —
4
9. tan 17π — 12
= tan ( 3π — 4 +
2π — 3 )
=
tan 3π — 4 + tan
2π — 3
—— 1 − tan
3π — 4 tan
2π — 3
= −1 + ( − √
— 3 ) ——
1 − (−1) ( − √—
3 )
= √—
3 + 2
10. sin ( − 7π — 12
) = sin ( − π — 4 −
π — 3 )
= sin ( − π — 4 ) cos
π — 3 − cos ( −
π — 4 ) sin
π — 3
= ( − √—
2 —
2 ) ( 1 —
2 ) − ( √
— 2 —
2 ) ( √
— 3 —
2 )
= − √—
2 − √—
6 —
4
11. Find sin a and cos b.
Because cos a = 4 —
5 and a is in Quadrant I, sin a =
3 —
5 as
shown in the fi gure.
5
4
52 − 42 = 3
ax
y
Because sin b = − 15
— 17
and b is in Quadrant IV, cos b = 8 —
17 as
shown in the fi gure.
1715
172 − 152 = 8b x
y
Use the sum formula for sine to fi nd sin(a + b).
sin(a + b) = sin a cos b + cos a + sin b
= 3 —
5 ( 8 —
17 ) +
4 —
5 ( −
15 —
17 )
= −36
— 85
12. Find sin a and cos b.
Because cos a = 4 —
5 and a is in Quadrant I, sin a = 3 — 5 as shown
in the fi gure.
5
4
52 − 42 = 3
ax
y
Because sin b = − 15
— 17 and b is in Quadrant IV, cos b = 8 —
17 as
shown in the fi gure.
1715
172 − 152 = 8b x
y
Use the difference formula for sine to fi nd sin(a – b).
sin(a − b) = sin a cos b − cos a sin b
= 3 —
5 ( 8 —
17 ) − ( 4 —
5 ) ( −
15 — 17 )
= 84
— 85
506 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
13. Find sin a and cos b.
Because cos a = 4 —
5 and a is in Quadrant I, sin a =
3 —
5 as shown
in the fi gure.
5
4
52 − 42 = 3
ax
y
Because sin b = − 15
— 17 and b is in Quadrant IV, cos b = 8 — 17 as
shown in the fi gure.
1715
172 − 152 = 8b x
y
Use the difference formula for cosine to fi nd cos(a − b).
cos(a − b) = cos a cos b + sin a sin b
= 4 — 5 ( 8 —
17 ) +
3 —
5 ( −
15 — 17 )
= − 13
— 85
14. Find sin a and cos b.
Because cos a = 4 —
5 and a is in Quadrant I, sin a =
3 —
5 as shown
in the fi gure.
5
4
52 − 42 = 3
ax
y
Because sin b = − 15
— 17 and b is in Quadrant IV, cos b = 8 —
17 as
shown in the fi gure.
1715
172 − 152 = 8b x
y
Use the sum formula for cosine to fi nd cos(a + b).
cos(a + b) = cos a cos b − sin a sin b
= ( 4 — 5 ) ( 8 —
17 ) − ( 3 —
5 ) ( −
15 — 17 )
= 77 — 85
15. Find tan a and tan b.
Because cos a = 4 —
5 and a is in Quadrant I, tan a =
3 —
4 as shown
in the fi gure.
5
4
52 − 42 = 3
ax
y
Because sin b = − 15
— 17 and b is in Quadrant IV, tan b = − 15
— 8 as
shown in the fi gure.
1715
172 − 152 = 8b x
y
Use the sum formula for tangent to fi nd tan(a + b).
tan(a + b) = tan a + tan b
—— 1 − tan a tan b
= 3 —
4 + ( −
15 —
8 ) ——
1 − ( 3 — 4 ) ( − 15
— 8 )
= − 36
— 77
Copyright © Big Ideas Learning, LLC Algebra 2 507All rights reserved. Worked-Out Solutions
Chapter 9
16. Find tan a and tan b.
Because cos a = 4 —
5 and a is in Quadrant I, tan a =
3 —
4 as
shown in the fi gure.
5
4
52 − 42 = 3
ax
y
Because sin b = − 15
— 17
and b is in Quadrant IV, tan b = − 15
— 8
as shown in the fi gure.
1715
172 − 152 = 8b x
y
Use the difference formula for tangent to fi nd tan(a − b).
tan(a − b) = tan a − tan b
—— 1 + tan a tan b
=
3 —
4 − ( −
15 —
8 ) ——
1 + ( 3 — 4 ) ( −
15 —
8 )
= − 84
— 13
17. tan(x + π) = tan x + tan π ——
1 − tan x tan π
= tan x + 0
—— 1 − (tan x)(0)
= tan x
18. cos ( x − π — 2 ) = cos x cos
π — 2 + sin x sin
π — 2
= (cos x)(0) + (sin x)(1)
= sin x
19. cos(x + 2π) = cos x cos 2π − sin x sin 2π = (cos x)(1) − (sin x)(0)
= cos x
20. tan(x − 2π) = tan x − tan 2π ——
1 + tan x tan 2π
= tan x − 0
—— 1 + (tan x)(0)
= tan x
21. sin ( x − 3π — 2
) = sin x cos 3π — 2 − cos x sin
3π — 2
= (sin x)(0) − (cos x)(−1)
= cos x
22. tan ( x + π — 2 ) =
sin ( x + π — 2 ) —
cos ( x + π — 2 )
=
sin x cos π — 2 + cos x sin
π — 2
——— cos x cos
π — 2 − sin x sin
π — 2
= sin x(0) + cos x (1)
—— cos x (0) − sin x (1)
= cos x
— −sin x
= −cot x
23. The sign in the denominator should be negative when using
the sum formula.
tan ( x + π — 4 ) =
tan x + tan π — 4
—— 1 − tan x tan
π — 4
= tan x + 1 —
1 − tan x
24. The a and b were reversed when the difference formula
was used.
sin ( x − π — 4 ) = sin x cos
π — 4 − cos x sin
π — 4
= (sin x) ( √—
2 —
2 ) − (cos x) ( √
— 2 —
2 )
= √
— 2 —
2 (sin x − cos x)
25. B, D; 2 sin π — 6 − 1 = 0 and 2 sin
5π — 6 − 1 = 0
26. B, D; tan 3π — 4 + 1 = 0 and tan
7π — 4 + 1 = 0
27. sin ( x + π — 2 ) =
1 —
2
sin x cos π — 2 + cos x sin
π — 2 =
1 —
2
0 sin x + 1 cos x = 1 —
2
cos x = 1 —
2
In the interval 0 ≤ x < 2π, the solutions are x = π — 3 and x = 5π —
3 .
28. tan ( x − π — 4 ) = 0
tan x − tan π — 4
—— 1 + tan x tan
π — 4 = 0
tan x − 1
— 1 + tan x
= 0
tan x − 1 = 0 tan x = 1
In the interval 0 ≤ x < 2π, the solutions are x = π — 4 and x = 5π —
4 .
508 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
29. cos ( x + π — 6 ) − cos ( x − π —
6 ) = 1
cos x cos π — 6 − sin x sin
π — 6 − ( cos x cos
π — 6 + sin x sin
π — 6 ) = 1
√
— 3 —
2 cos x − 1 —
2 sin x − √
— 3 —
2 cos x − 1 —
2 sin x = 1
− sin x = 1 sin x = −1
In the interval 0 ≤ x < 2π, the solution is x = 3π — 2 .
30. sin ( x + π — 4 ) + sin ( x − π —
4 ) = 0
sin x cos π — 4 + cos x sin
π — 4 + sin x cos
π — 4 − cos x sin
π — 4 = 0
√
— 2 —
2 sin x +
√—
2 —
2 cos x +
√—
2 —
2 sin x −
√—
2 —
2 cos x = 0
√—
2 sin x = 0 sin x = 0
In the interval 0 ≤ x < 2π, the solutions are x = 0 and x = π.
31. tan(x + π) − tan(π − x) = 0
tan x + tan π —— 1− tan x tan π
− tan π − tan x ——
1+ tan π tan x = 0
tan x − (− tan x) = 0 2 tan x = 0 tan x = 0
In the interval 0 ≤ x < 2π, the solutions are x = 0 and x = π.
32. sin(x + π) + cos(x + π) = 0
sin x cos π + cos x sin π + cos x cos π − sin x sin π = 0 −sin x − cos x = 0 −sin x = cos x
In the interval 0 ≤ x < 2π, the solutions are x = 3π — 4 and x =
7π — 4 .
33. sin ( π — 2 − θ ) = sin
π — 2 cos θ − cos
π — 2 sin θ
= (1) cos θ − (0) sin θ = cos θ
34. Your friend is incorrect. The difference formula for tan ( π — 2 − θ )
would require fi nding tan π — 2 , which is undefi ned.
35. When t = 45°:
WQ — NA
= 35 tan(θ − 45°) + 35 tan 45° ———
h tan θ
= 35 ( tan θ − tan 45°
—— 1 + tan θ tan 45°
) + 35
——— h tan θ
= 35 ( tan θ − 1
— 1 + tan θ
) + 35
—— h tan θ
= 35(tan θ − 1) + 35(1 + tan θ) ———
h tan θ(1 + tan θ)
= 35 tan θ − 35 + 35 + 35 tan θ ——— h tan θ(1 + tan θ)
= 70 tan θ —— h tan θ(1 + tan θ)
= 70 ——
h(1 + tan θ)
36. When t = 1:
y = A cos ( 2π — 3 − 2π x
— 5 ) + A cos ( 2π —
3 + 2π x
— 5 )
= A cos 2π — 3 cos 2π x
— 5 + A sin
2π — 3 sin 2π x
— 5
+ A cos 2π — 3 cos
2π x —
5 − A sin
2π — 3 sin 2π x
— 5
= A cos 2π — 3 cos 2π x
— 5 + A cos
2π — 3 cos 2π x
— 5
= 2A cos 2π — 3 cos 2π x
— 5
= 2A ( − 1 —
2 ) cos 2π x
— 5
= −A cos 2π x —
5
37. y1 + y2 = cos 960πt + cos 1240πt
= cos(1100πt − 140πt) + cos(1100πt + 140πt)
= cos 1100πt cos 140πt + sin 1100πt sin 140πt
+ cos 1100πt cos 140πt − sin 1100πt sin 140πt
= cos 1100πt cos 140πt + cos 1100πt cos 140πt
= 2 cos 1100πt cos 140πt
38. Any point where the two graphs intersect is a solution
because if f(x) = g(x), then f(x) − g(x) = 0.
Copyright © Big Ideas Learning, LLC Algebra 2 509All rights reserved. Worked-Out Solutions
Chapter 9
39. a. First note that tan θ1 = m1 and tan θ2 = m2.
tan(θ2 − θ1) = tan θ2 − tan θ1
—— 1 + tan θ2 tan θ1
= m2 – m1
— 1 + m2 m1
b. The slopes of the lines are m1 = 1 and m2 = 1 —
√—
3 − 2 .
tan(θ2 − θ1) = m2 − m1 — 1 + m2 m1
tan(θ2 − θ1) =
1 —
√—
3 −2 − 1
——
1 + ( 1 —
√—
3 −2 ) (1)
tan(θ2 − θ1) = 1 − ( √
— 3 −2 ) ——
( √—
3 −2 ) + 1
tan(θ2 − θ1) = 3 − √
— 3 —
√—
3 − 1
tan(θ2 − θ1) = ( 3 − √
— 3 ) ( √
— 3 + 1 ) ——
√—
3 2 − 12
tan(θ2 − θ1) = 2 √—
3 —
2
tan(θ2 − θ1) = √—
3
θ2 − θ1 = 60°
So, the acute angle is 60°.
40. a. sin 3x = sin(2x + x)
= sin 2x cos x + cos 2x sin x
= sin(x + x) cos x + cos(x + x) sin x
= (sin x cos x + cos x sin x) cos x + (cos x cos x −
sin x sin x) sin x
= sin x cos2 x + cos2 x sin x + cos2 x sin x − sin3 x
= 3 sin x cos2 x − sin3 x
= 3 sin x(1 − sin2 x) − sin3 x
= 3 sin x − 3 sin3 x − sin3 x
= 3 sin x − 4 sin3 x
b. cos 3x = cos(2x + x)
= cos 2x cos x − sin 2x sin x
= cos(x + x) cos x − sin(x + x) sin x
= (cos x cos x − sin x sin x) cos x −
(sin x cos x + cos x sin x) sin x
= (cos2 x − sin2 x) cos x − 2 sin x cos x sin x
= (cos2 x − sin2 x) cos x − 2 sin2 x cos x
= (cos2 x − (1 − cos2 x)) cos x − 2(1 − cos2 x)
cos x
= (2 cos2 x − 1) cos x − 2 (cos x − cos3 x)
= 2 cos3 x − cos x − 2 cos x + 2 cos3 x
= 4 cos3 x − 3 cos x
c. tan 3x = sin 3x —
cos 3x
= 3 sin x − 4 sin3x ——
4 cos3x − 3 cos x
= 3 sin x − 4 sin3x ——
4 cos3x − 3 cos x ⋅
1 —
cos3 x —
1 —
cos3 x
= 3 sin x
— cos3x
− 4 sin3x
— cos3x
——
4cos3x
— cos3x
− 3cos x
— cos3x
= 3 tan x
— cos2x
− 4 tan3x ——
4 − 3 —
cos2x
= 3 sec2x tan x − 4 tan3x
—— 4 − 3 sec2x
= 3(1 + tan2x) tan x − 4 tan3x ———
4 − 3(1 + tan2x)
= 3 tan x + 3 tan3x − 4 tan3x ———
4 − 3 − 3 tan2x
= 3 tan x − tan3x
—— 1 − 3 tan2x
Maintaining Mathematical Profi ciency
41. 1 − 9 —
x − 2 = −
7 —
2 Check x = 4:
− 9 —
x − 2 = −
9 —
2 1 −
9 —
4 − 2 =
? −
7 —
2
9 —
x − 2 =
9 —
2 1 −
9 —
2 =
? −
7 —
2
2 ⋅ 9 = 9(x − 2) − 7 —
2 = −
7 —
2 ✓
2 = x − 2
4 = x
The solution is x = 4.
510 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
42. 12
— x +
3 —
4 =
8 —
x
4x ( 12 —
x +
3 —
4 ) = 4x ( 8 —
x )
4 ⋅ 12 + 3x = 4 ⋅ 8
48 + 3x = 32
3x = −16
x = − 16
— 3
The solution is x = − 16
— 3 .
43. 2x − 3
— x + 1
= 10 —
x2 − 1 + 5
2x − 3
— x + 1
= 10 ——
( x + 1 ) ( x − 1 ) + 5
(x + 1)(x − 1) ( 2x − 3 —
x + 1 ) = (x + 1)(x − 1) ( 10
—— (x + 1)(x − 1)
+ 5 ) (x − 1)(2x − 3) = 10 + 5 (x + 1)(x − 1)
2x2 − 5x + 3 = 10 + 5x2 − 5
2x2 − 5x + 3 = 5x2 + 5
0 = 3x2 + 5x + 2
0 = (x + 1)(3x + 2)
x = −1, − 2 —
3
Check x = −1:
2(−1) − 3
— −1 + 1
=?
10 —
(−1)2 − 1 + 5
−5
— 0 =
? 10
— 0 + 5
Division by 0 is undefi ned.
Check x = − 2 —
3 :
2 ( −
2 —
3 ) − 3 —
− 2 —
3 + 1
=?
10 —
( − 2 —
3 ) 2 − 1
+ 5
−
4 —
3 − 3 —
1 —
3
=?
10 —
4 —
9 − 1
+ 5
−
13 —
3 —
1 —
3
=? 10
—
− 5 —
9
+ 5
−13 =? −18 + 5
−13 = −13 ✓
The solution is x = − 2 —
3 .
9.5–9.8 What Did You Learn? (p. 525)
1. As the angle at which you are looking down at the car
increases, the distance between you and the car increases.
2. Sample answer: The slope is the ratio of rise to run, and these
form the legs of a right triangle when drawn on the graph.
The tangent is the ratio of the lengths of the legs, so
m = rise
— run
= tan θ.
Chapter 9 Review (pp. 526–530)
1. Step 1 Draw a right triangle with acute angle θ such that
the leg adjacent θ has length 6 and the hypotenuse
has length 11.
Step 2 Find the length of the opposite side. By the
Pythagorean Theorem, the length of the other leg is
opp. = √—
112 − 62 = √—
85 .
opp. = 85
6
11
θ
Step 3 Find the values of the remaining fi ve trigonometric
functions. Because cos θ = 6 —
11 , sec θ =
hyp. —
adj. =
11 —
6 .
The other values are:
sin θ = opp.
— hyp.
= √
— 85 —
11
tan θ = opp.
— adj.
= √
— 85 —
6
csc θ = hyp.
— opp.
= 11
— √
— 85 =
11 √—
85 —
85
cot θ = adj.
— opp.
= 6 —
√—
85 =
6 √—
85 —
85
2. tan 31° = h —
25
25(tan 31°) = h
15 = h
The height is about 15 feet.
3. 382° − 360° = 22° 382° − 2 ⋅ 360° = −338°
4. 30° = 30 degrees ( π radians —
180 degrees )
= π — 6
5. 225° = 225 degrees ( π radians —
180 degrees )
= 5π — 4
Check x = − 16
— 3 :
12
—
− 16
— 3
+ 3 —
4 =
?
8 —
− 16
— 3
−9
— 4 +
3 —
4 =
? −
3 —
2
− 3 —
2 = −
3 —
2 ✓
Copyright © Big Ideas Learning, LLC Algebra 2 511All rights reserved. Worked-Out Solutions
Chapter 9
6. 3π — 4 =
3π — 4 radians ( 180° —
π radians )
= 135°
7. 5π — 3 =
5π — 3 radians ( 180° —
π radians )
= 300°
8.
140°
x
y
Convert 140° to radians.
140° = 140 degrees ( π radians —
180 degrees )
= 7π — 9
Find the area of the sector.
A = 1 —
2 r 2 θ
= 1 —
2 ( 35 ) 2 ( 7π —
9 )
= 8525
— 18
π
≈ 1497
The area is about 1497 square meters.
9. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x2 + y2
= √—
02 + 12
= √—
1
= 1
Using x = 0, y = 1, and r = 1, the values of the six
trigonometric functions of θ are:
sin θ = y —
r =
1 —
1 = 1 csc θ =
r —
y =
1 —
1 = 1
cos θ = x —
r =
0 —
1 = 0 sec θ =
r —
x =
1 —
0 = undefi ned
tan θ = y —
x =
1 —
0 = undefi ned cot θ =
x —
y =
0 —
1 = 0
10. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x2 + y2
= √——
242 + (−7)2
= √—
625
= 25
Using x = 24, y = −7, and r = 25, the values of the six
trigonometric functions of θ are:
sin θ = y —
r = −
7 — 25 csc θ =
r —
y = −
25 — 7
cos θ = x —
r = 24
— 25
sec θ = r —
x =
25 —
24
tan θ = y —
x = −
7 — 24 cot θ =
x —
y = −
24 — 7
11. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x2 + y2
= √—
( −4 ) 2 + 62
= √—
52
= 2 √—
13
Using x = −4, y = 6, and r = 2 √—
13 , the values of the six
trigonometric functions of θ are:
sin θ = y —
r =
3 √—
13 —
13 csc θ =
r —
y =
√—
13 —
3
cos θ = x —
r = −
2 √—
13 —
13 sec θ =
r —
x = −
√—
13 —
2
tan θ = y —
x = −
3 —
2 cot θ =
x —
y = −
2 —
3
12. The reference angle is 360° − 330° = 30°. The tangent
function is negative in Quadrant IV, so
tan 330° = −tan 30° = − √
— 3 —
3 .
13. The angle −405° is coterminal with 315°. The reference
angle is 360° − 315° = 45. The secant function is positive in
Quadrant IV, so
sec(−405°) = sec 45° = √—
2 .
14. The angle 13π —
6 is coterminal with
π — 6 . The sine function is
positive in Quadrant I, so
sin 13π —
6 = sin
π — 6 =
1 —
2 .
15. The angle 11π —
3 is coterminal with
5π — 3 . The reference angle is
2π − 5π — 3 =
π — 3 . The secant function is positive in Quadrant IV,
so sec 11π —
3 = sec
π — 3 = 2.
512 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
16. The function is of the form g(x) = a cos bx, where a = 8 and
b = 1.
So, the amplitude is a = 8 and the period is 2π — b =
2π — 1 = 2π.
Intercepts: ( − π — 2 , 0 ) ; ( π —
2 , 0 ) ; ( 3π —
2 , 0 )
Maximum: (0, 8)
Minimum: (π, −8)
x
y8
2
−8
−6
−4
−π 2ππ
The graph of g is a vertical stretch by a factor of 8 of the
graph of f(x) = cos x.
17. The function is of the form g(x) = a sin bx, where a = 6 and
b = π.
So, the amplitude is a = 6 and the period is 2π — b =
2π — π
= 2.
Intercepts: (0, 0); (1, 0); (2, 0)
Maximum: ( 1 — 2 , 6 )
Minimum: ( 3 — 2 , −6 )
x
y
4
2
6
−6
−4
−21.50.5
The graph of g is a horizontal shrink by a factor of 1 —
π and a
vertical stretch by a factor of 6 of the graph of f(x) = sin x.
18. The function is of the form g(x) = a cos bx, where a = 1 —
4 and
b = 4. So, the amplitude is a = 1 —
4 and the period is
2π — b =
2π — 4 =
π — 2 .
Intercepts: ( π — 8 , 0 ) ; ( 3π —
8 , 0 )
Maximums: ( 0, 1 —
4 ) ; ( π —
2 ,
1 —
4 )
Minimum: ( π — 4 , −
1 —
4 )
x
y
0.25
0.5
−0.5
−0.25
8π
4π
2π
The graph of g is a horizontal shrink by a factor of 1 —
4 and a
vertical shrink by a factor of 1 —
4 of the graph of f(x) = cos x.
19. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 1 Horizontal shift: h = −π
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 2
Step 2 Draw the midline of the graph, y = 2.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , 2 ) ; ( 3π —
2 , 2 )
Maximum: (π, 3)
Minimums: (0, 2); (2π, 2)
Step 4 Draw the graph through the key points.
x
y
2
3
4
−π 2ππ−2π−1
Copyright © Big Ideas Learning, LLC Algebra 2 513All rights reserved. Worked-Out Solutions
Chapter 9
20. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = ∣ −1 ∣ = 1 Horizontal shift: h = 0
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = −4
Step 2 Draw the midline of the graph, y = −4.
Step 3 Find the fi ve key points.
On y = k: (0, − 4); (π, − 4); (2π, −4)
Maximum: ( 3π — 2 , −3 )
Minimum: ( π — 2 , −5 )
Step 4 Draw the graph through the key points.
x
y1
−π 2ππ−2π−1
−2
−4
−5
−6
21. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: a = 2 Horizontal shift: h = − π — 2
Period: 2π — b =
2π — 1 = 2π Vertical shift: k = 0
Step 2 Draw the midline of the graph, y = 0.
Step 3 Find the fi ve key points.
On y = k: ( π — 2 , 0 ) ; ( 3π —
2 , 0 )
Maximums: (0, 2); (2π, 2)
Minimum: (π, –2)
Step 4 Draw the graph through the key points.
x
y3
−π 2ππ−2π−1
−2
−3
22. The function is of the form g(x) = a tan bx, where a = 1 and
b = 1 —
2 . So, the period is
π — ∣ b ∣
= π — 1 —
2
= 2π.
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π —
2 ( 1 — 2 )
= π ;
x = − π —
2 ∣ b ∣ = −
π —
2 ( 1 — 2 ) = −π
Halfway points: ( π — 4b
, a ) = ( π —
4 ( 1 — 2 ) , 1 ) = ( π —
2 , 1 )
( − π — 4b
, −a ) = ( − π —
4 ( 1 — 2 ) , −1 ) = ( −
π — 2 , −1 )
2π x
y
1
3
2
4
−π π−1
−2
−3
−4
The graph of g is a horizontal stretch by a factor of 2 of the
graph of f (x) = tan x.
23. The function is of the form g(x) = a tan bx, where a = 2 and
b = 1. So, the period is π —
∣ b ∣ =
π — 1 = π.
Intercepts: ( π — 2b
, 0 ) = ( π — 2 , 0 )
Asymptotes: x = 0; x = π —
∣ b ∣ = π — 1 = π
Halfway points: ( π — 4b
, a ) = ( π — 4(1)
, 2 ) = ( π — 4 , 2 ) ;
( 3π — 4b
, −a ) = ( 3π — 4(1)
, −2 ) = ( 3π — 4 , −2 )
x
y
2
3
1
−4
−2
−1
−3
4π
2π 3
4π π
The graph of g is a vertical stretch by a factor of 2 of the
graph of f (x) = cot x.
514 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
24. The function is of the form g(x) = a tan bx, where a = 4 and
b = 3π. So, the period is π —
∣ b ∣ =
π — 3π
= 1 —
3 .
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — 6π
= 1 —
6 ;
x = − π —
2 ∣ b ∣ = −
π — 6π = −
1 —
6
Halfway points: ( π — 4b
, a ) = ( π — 4 ( 3π )
, 4 ) = ( 1 — 12
, 4 ) ; ( −
π — 4b
, −a ) = ( − π —
4(3π) , −4 ) = ( − 1 —
12 , −4 )
x
y
16
20
8
4
12
−12
−16
−20
−8
−416
16−
The graph of g is a horizontal shrink by a factor of 1 —
3π and a
vertical stretch by a factor of 4 of the graph of f (x) = tan x.
25. Step 1 Graph the function y = 5 sin x. The period is
2π — 1 = 2π.
Step 2 Graph the asymptotes of g. Because the asymptotes
of g occur when 5 sin x = 0, graph x = 0, x = π,
and x = 2π.
Step 3 Plot points on g, such as ( π — 2 , 6 ) and ( 3π —
2 , −6 ) .
Then use the asymptotes to sketch the curve.
x
y
2
4
6
8
π−2
−4
−6
−8
−10
2π 2π3
2π
26. Step 1 Graph the function y = cos 1 —
2 x.
The period is 2π — 1 —
2
= 4π.
Step 2 Graph the asymptotes of g. Because the asymptotes
of g occur when cos 1 —
2 x = 0, graph x = –π, x = π,
and x = 3π.
Step 3 Plot points on g, such as (0, 1) and (2π, −1). Then
use the asymptotes to sketch the curve.
x
y8
4
2
6
−8
−6
−4
−23ππ−π
27. Step 1 Graph the function y = 5 cos π x.
The period is 2π — π
= 2.
Step 2 Graph the asymptotes of g. Because the asymptotes
of g occur when 5 cosπ x = 0, graph x = − 1 —
2 ,
x = 1 —
2 , and x =
3 —
2 .
Step 3 Plot points on g, such as (0, 5) and (1, −5).
Then use the asymptotes to sketch the curve.
x
y
16
20
8
4
12
−16
−20
−12
−8
−40.5 1.0 1.5−0.5
28. Step 1 Graph the function y = 1 —
2 sin
π — 4 x.
The period is 2π — π — 4 = 8.
Step 2 Graph the asymptotes of g. Because the asymptotes
of g occur when 1 —
2 sin
π — 4 x = 0, graph x = 0, x = 4,
and x = 8.
Step 3 Plot points on g, such as (2, 1) and (6, –1). Then use
the asymptotes to sketch the curve.
x
y
2
1
3
−4
−3
−2
−1421 3 8
Copyright © Big Ideas Learning, LLC Algebra 2 515All rights reserved. Worked-Out Solutions
Chapter 9
29. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 1 and the minimum
value is −1.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= 1 + (−1)
— 2 =
0 —
2 = 0
Step 3 Decide whether the graph should be modeled by a
sine or cosine function. Because the graph crosses
the midline y = 0 on the y−axis, the graph is a sine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period.
The period is 4π = 2π — b . So, b =
1 —
2 . The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 1 − (−1)
— 2 =
2 —
2 = 1.
The graph is a refl ection, so a < 0. So, a = −1. The function
is y = −sin 1 —
2 x.
30. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is −1 and the minimum
value is –3.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= −1 + (−3)
— 2 =
−4 —
2 = −2
Step 3 Decide whether the graph should be modeled by
a sine or cosine function. Because the graph has a
maximum value on the y-axis, the graph is a cosine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period.
The period is 2 = 2π — b . So, b = π. The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= −1 − (−3)
— 2 =
2 —
2 = 1
The graph is not a refl ection, so a > 0. So, a = 1.
The function is y = cosπ x −2.
31. The refl ector oscillates between 25 inches and 2 inches
above the ground. So, fi nd a sine or cosine function that may
be an appropriate model for the height over time.
Step 1 Identify the maximum and minimum values. The
maximum height is 25 inches. The minimum height
is 2 inches.
Step 2 Identify the vertical shift, k.
k = (maximum value) + (minimum value)
———— 2
= 25 + 2
— 2 =
27 —
2 = 13.5
Step 3 Decide whether the height should be modeled by a
sine or cosine function. When t = 0, the height is at
its minimum. So, use a cosine function whose graph
is a refl ection in the x-axis with no horizontal shift
(h = 0).
Step 4 Find the amplitude and period. The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 25 − 2
— 2 = 11.5.
Because the graph is a refl ection in the x−axis,
a < 0. So, a = −11.5. Because the wheel is rotating
at a rate of 1 revolution per second, one revolution
is completed in 1 second. So, the period is 2π — b = 1,
and b = 2π.
A model for the height of the refl ector is
h = −11.5 cos 2π t + 13.5.
32. Enter the data in a graphing calculator. Make a scatter plot.
The scatter plot appears sinusoidal. So, perform a sinusoidal
regression. Graph the data and the model in the same
viewing window.
130
0
3
The model appears to be a good fi t. So, a model for the data
is P = 1.08 sin(0.585t − 2.33). The period represents the
amount of time it takes for the precipitation level to complete
one cycle, which is about 10.7 months.
33. cot2x − cot2x cos2x = cot2x (1 − cos2x)
= cot2x (sin2x)
= cos2x
— sin2x
∙ sin2x
= cos2x
516 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
34. (sec x + 1)(sec x − 1) ——
tan x =
sec2x − 12
— tan x
= sec2x − 1
— tan x
= tan2x
— tan x
= tan x
35. sin ( π — 2 − x ) tan x = cos x ⋅ tan x
= cos x ⋅ sin x
— cos x
= sin x
36. cos x sec x — 1 + tan2x
= cos x ⋅ sec x
— sec2 x
= cos x
— sec x
= cos x cos x
= cos2 x
37. tan ( π — 2 − x ) cot x = cot x ⋅ cot x
= cot2 x
= csc2 x − 1
38. sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
= √
— 2 —
2 ( √
— 3 —
2 ) +
√—
2 —
2 ( 1 —
2 )
= √
— 2 + √
— 6 —
4
39. tan ( −15° ) = tan(30° − 45°)
= tan 30° − tan 45° ——
1 + tan 30° tan 45°
=
√
— 3 —
3 − 1
——
1 + ( √—
3 —
3 ) (1)
= √—
3 − 2
40. cos π — 12
= cos ( π — 3 −
π — 4 )
= cos π — 3 cos
π — 4 + sin
π — 3 sin
π — 4
= 1 —
2 ( √
— 2 —
2 ) +
√—
3 —
2 ( √
— 2 —
2 )
= √
— 2 + √
— 6 —
4
41. tan(a + b) = tan a + tan b
—— 1 − tan a tan b
= 1 — 4 +
3 —
7
—
1 − ( 1 — 4 ) ( 3 —
7 )
= 19
— 25
42. cos ( x + 3π — 4 ) + cos ( x −
3π — 4 ) = 1
cos x cos 3π — 4 − sin x sin
3π — 4 + cos x cos
3π — 4 + sin x sin
3π — 4 = 1
− √
— 2 —
2 cos x −
√—
2 —
2 sin x −
√—
2 —
2 cos x +
√—
2 —
2 sin x = 1
− √—
2 cos x = 1
cos x = − √
— 2 —
2
In the interval 0 ≤ x < 2π, the solutions are x = 3π — 4 and
x = 5π — 4 .
43. tan ( x + π ) + cos ( x + π — 2 ) = 0
tan x + tan π ——
1 − tan x tan π + cos x cos
π — 2 − sin x sin
π — 2 = 0
tan x − sin x = 0
tan x = sin x
In the interval 0 ≤ x < 2π, the solutions are x = 0 and x = π.
Chapter 9 Test (p. 531)
1. cos2 x + sin2 x —— 1 + tan2 x
= 1 —
sec2 x
= cos2 x
2. 1 + sin x — cos x
+ cos x
— 1 + sin x
= (1 + sin x)(1 + sin x) + cos2 x
——— cos x(1 + sin x)
= 1 + 2 sin x + sin2 x + cos2 x
——— cos x (1 + sin x)
= 2 + 2 sin x
—— cos x(1 + sin x)
= 2 —
cos x
= 2 sec x
3. cos ( x + 3π — 2 ) = cos x cos
3π — 2 − sin x sin
3π — 2
= (cos x)(0) − (sin x)(−1)
= sin x
4. An angle coterminal to −300° is 60°. The secant function is
positive in Quadrant I, so
sec(−300°) = sec 60° = 2.
Copyright © Big Ideas Learning, LLC Algebra 2 517All rights reserved. Worked-Out Solutions
Chapter 9
5. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 5 and the minimum
value is −1.
Step 2 Identify the vertical shift k. The value of k is the
mean of the maximum value and minimum value.
k = (maximum value) − (mininum value)
———— 2
= 5 + ( −1 )
— 2 =
4 —
2 = 2
Step 3 Decide whether the graph should be modeled by
a sine or cosine function. Because the graph has a
maximum value on the y-axis. The graph is a cosine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
2 = 2π — b → b = π.
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 5 − ( −1 )
— 2 =
6 —
2 = 3.
The graph is not a refl ection, so a > 0. Therefore, a = 3.
The function is y = 3 cosπ x + 2.
6. Step 1 Find the maximum and minimum values. From the
graph, the maximum value is 1 and the minimum
value is −5.
Step 2 Identify the vertical shift, k. The value of k is the
mean of the maximum and minimum values.
k = (maximum value) + (minimum value)
———— 2
= 1 + ( −5 )
— 2 =
−4 —
2 = −2
Step 3 Decide whether the graph should be modeled by a
sine or cosine function. Because the graph crosses
the midline y = −2 on the y-axis, the graph is a sine
curve with no horizontal shift. So, h = 0.
Step 4 Find the amplitude and period. The period is
3π — 2 =
2π — b → b =
4 —
3 .
The amplitude is
∣ a ∣ = (maximum value) − (minimum value)
———— 2
= 1 − ( −5 )
— 2 =
4 —
2 = 2.
The graph is a refl ection, so a < 0. Therefore, a = −2.
The function is y = −3 sin 4 —
3 x −2.
7. The function is of the form g(x) = a tan bx, where a = −4
and b = 2.
So, the period is π —
∣ b ∣ =
π — 2 .
Intercept: (0, 0)
Asymptotes: x = π —
2 ∣ b ∣ =
π — 2 ( 2 )
= π — 4 ; x = −
π —
2 ∣ b ∣
= − π —
2 ( 2 ) = −
π — 4
Halfway points: ( π — 4b
, a ) = ( π — 4(2)
, −4 ) = ( π — 8 , −4 ) ;
( − π — 4b
, − a ) = ( − π —
4(2) , −(−4) ) = ( −
π — 8 , 4 )
x
y16
8
4
12
−16
−12
−8
−44−π8−π
4π
The graph of g is a horizontal shrink by a factor of 1 —
2 and
a vertical stretch by a factor of 4 followed by a refl ection
across the x-axis of the graph of f (x) = tan x.
8. Step 1 Identify the amplitude, period, horizontal shift, and
vertical shift.
Amplitude: ∣ a ∣ = 2 Horizontal shift: h = 0
Period: 2π — b =
3π — 1 —
3
= 6π Vertical shift: k = 3
Step 2 Draw the midline of the graph, y = 3.
Step 3 Find the fi ve key points.
On y = k: ( 3π — 2 , 3 ) ; ( 9π —
2 , 3 )
Maximum: (3π, 5)
Minimums: (0, 1); (6π, 1)
Step 4 Draw the graph through the key points.
x
y
1
2
3
4
5
6
π−1
3π 5π
The graph of g is a horizontal stretch by a factor of 3 and
a vertical stretch by a factor of 2 followed by a refl ection
across the x-axis and a translation 3 units up of the graph of
f (x) = cos x.
518 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
9. Step 1 Graph the function y = 3 sinπ x. The period is
2π — π
= 2.
Step 2 Graph asymptotes of g. Because the asymptotes of g
occur when 3 sin 2π = 0, graph x = 0, x = 1,
and x = 2.
Step 3 Plot points on g, such as ( 1 — 2 , 3 ) and ( 3 —
2 , −3 ) .
Then use the asymptotes to sketch the curve.
x
y
2
4
6
8
−2
−4
−6
−8
−10
0.5 1 1.5 2
The graph of g is a horizontal shrink by a factor of 1 — π and a
vertical stretch by a factor of 3 of the graph of f (x) = csc x.
10. −50° = −50 degrees ( π radians —
180 degrees )
= − 5π — 18
−50° + 360° = 310° −50° − 360° = −410°
11. 4π — 5 =
4π — 5 radians ( 180 degrees
— π radians
) = 144°
4π — 5 + 2π =
14π — 5
4π — 5 − 2π = −
6π — 5
12. 8π — 3 =
8π — 3 radians ( 180 degrees
— π radians
) = 480°
8π — 3 − 2π =
2π — 3
8π — 3 − 2 ⋅ 2π = −
4π — 3
13. Convert 40° to radians: 40° = 2π — 9 .
Arc length: s = θr Area of sector: A = 1 —
2 θr2
= ( 2π — 9 ) (13) =
1 —
2 ( 2π —
9 ) (13)2
= 26π —
9 =
169π — 9
≈ 9.08 ≈ 59.00
The arc length is about 9.08 inches and the area is about
59.00 square inches.
14. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x2 + y2
= √—
22 + (−9)2
= √—
85
Using x = 2, y = −9, and r = √—
85 , the values of the six
trigonometric functions of θ are:
sin θ = y —
r = −
9 √—
85 —
85 csc θ =
r —
y = −
√—
85 —
9
cos θ = x —
r =
2 √—
85 —
85 sec θ =
r —
x =
√—
85 —
2
tan θ = y —
x = −
9 —
2 cot θ =
x —
y = −
2 —
9
15. Use the Pythagorean Theorem to fi nd the length of r.
r = √—
x2 + y2
= √—
(−1)2 + 02
= √—
1
= 1
Using x = −1, y = 0, and r = 1, the values of the six
trigonometric functions of θ are:
sin θ = y —
r = 0 csc θ =
r —
y =
0 —
0 = undefi ned
cos θ = x —
r = −1 sec θ =
r —
x = −1
tan θ = y —
x = 0 cot θ =
x —
y = −
1 —
0 = undefi ned
16. The quadrant is Quadrant III because cos θ < 0, then θ must
be in Quadrant II or Quadrant III and because tan θ > 0, then
θ must be in Quadrant III.
17. sin 60° = d —
200 , where d is the distance from the base of the
crane to the top. So, d = 200 sin 60°. So, the height of the
building is h = 200 sin 60° + 5 ≈ 178 feet.
18. Enter the data in a graphing calculator. Make a scatter plot.
The scatter plot appears sinusoidal. So, perform a sinusional
regression. Graph the data and the model in the same
viewing window.
120
0
100
The model appears to be a good fi t. So, a model for the
data is T = 23.14 sin (0.495 m − 1.95) + 63.7. The period
represents the amount of time it takes for the weather to
complete one cycle, which is about 12.7 months.
Copyright © Big Ideas Learning, LLC Algebra 2 519All rights reserved. Worked-Out Solutions
Chapter 9
Chapter 9 Standards Assessment (pp. 532–533)
1. tan x sec x cos x = sin x
— cos x
⋅ 1 —
cos x ⋅ cos x
= sin x
— cos x
= tan x
sin2 x + cos2 x = 1
cos2(−x) tan2 x ——
sin2(−x) =
cos2 x tan2 x —
sin2 x
=
cos2 x ⋅ sin2 x
— cos2 x
——
sin2 x
= 1
cos ( π — 2 − x ) csc x = sin x csc x
= sin x ⋅ 1 —
sin x
= 1
The expressions are sin2x + cos2x, cos2 (−x) tan2x
—— sin2(−x)
,
cos ( π — 2 − x ) csc x.
2. A; Perimeter of the playground
= 2x + 6x + (2x + x) + 2x + x + (6x − 2x)
= 18x
Area of the playground = (2x)(6x) + (x)(2x)
= 12x2 + 2x2
= 14x2
So, the ratio of the perimeter to the area is 18x
— 14x2
= 9 —
7x .
3. a. Using the sinusoidal regression feature, the functions that
model the data are
y1 = 28.53 sin(0.548t + 3.12) + 45.7 and
y2 = 7.89 sin(0.610t − 0.09) + 17.1
b.
130
0
80
Sample answer: The two graphs increase and decrease in
about the same intervals.
4. a. log 1000 = log 103
= 3 log 10
= 3
b. log2 15 = log2 (3 ⋅ 5)
= log2 3 + log2 5
≈ 1.585 + 2.322
= 3.907
c. ln e = 1
d. log2 9 = log2 (32)
= 2 log2 3
≈ 2(1.585)
= 3.17
e. log2 5 —
3 = log2 5 − log2 3
≈ 2.322 − 1.585
= 0.737
f. log2 1 = 0
The order is f, e, c, a, d, and b.
5. C; A. y = 5 sin x
B. y = 5 cos ( π — 2 − x )
= 5 sin x
C. y = 5 cos ( x + π — 2 )
= 5 [ cos x cos π — 2 − sin x sin
π — 2 ]
= −5 sin x
D. y = −5 sin ( π + x)
= −5 [sin π cos x + cos π sin x]
= −5 (−sin x)
= 5 sin x
6. a. s = rθ 4π = 6θ
4π — 6 = θ
2.09 ≈ θ θ < 3 radians
b. The tangent function is negative in Quadrant II.
tan θ < 0
c. θ′ = π − θ = π −
4π — 6
= π — 3
= π — 3 ⋅
180° — π
= 60°
θ′ > 45°
7. The factors of −6 are ±1, ±2, ±3, ±6.
The factors of 2 are ±1, ±2.
The possible rational roots are ±1, ± 1 — 2 , ±2, ±3,
± 3 — 2 , ±6.
From the graph, you can see that the real zeros of the
functions are −2, − 1 — 2 , and 3.
520 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.
Chapter 9
8. −210° + 360° = 150°
150° = 150° ( π — 180°
) =
5π — 6
yes; Both angles lie in Quadrant II and have a reference
angle of 180° − 150° = 30° or π − 5π — 6 =
π — 6 and, therefore,
–210° is coterminal with 5π — 6 .
9. a. For Company A, the fi rst term is a1 = 20,000 and the
common difference is d = 1000.
an = a1 + (n − 1) d
= 20000 + (n − 1) 1000
= 1000n + 19,000
The sequence is represented by an = 1000n + 19,000, and
it is an arithmetic sequence.
For Company B, the fi rst term is a1 = 20,000 and the
common ratio is r = 0.04 + 1 = 1.04.
an = a1 r n−1
= 20,000 (1.04)n−1
The sequence is represented by an = 20,000 (1.04)n−1,
and it is a geometric sequence.
b.
0
20,000
30,000
40,000
0 5 10 15 20 n
an
anbn
Years
Sala
ry (
do
llars
)
c. Sample answer: You would choose to work for Company B
if you would be working for longer than 12 years.
d. ∑ n=1
20
an = ∑ n=1
20
(1000n + 19,000)
= 1000 [ 20 (21) —
2 ] + 19,000(20)
= 590,000
∑ n=1
20
bn = ∑ n=1
20
20,000 (1.04)n−1
= −20,000 [ 1 + 1.04 + 1.042 + 1.043 + . . . + 1.0419 ]
≈ 595, 562
The total earnings for Company A is $590,000 and for
Company B is about $595,562.