Download Solutions to Trigonometry Practice

MATH & PHYSICS RESOURCE CENTER
REVIEW FOR MATH PLACEMENT TEST
SOLUTIONS FOR TRIGONOMETRY PROBLEMS
1. Change 125° into radians
𝜋
Multiply by 180° to convert degrees into radians → 125° ×
2. What is
2𝜋
3
𝜋
180°
125
𝟐𝟓
= 180 𝜋 = 𝟑𝟔 𝝅
in degrees?
Multiply by
180°
to
𝜋
convert degrees into radians →
2𝜋
3
×
180
𝜋
=
2𝜋
3
×
180
𝜋
= 𝟏𝟐𝟎°
3. What is 192° in radians?
Again, multiply by
𝜋
to
180°
convert degrees into radians → 192° ×
𝜋
180
=
192
𝜋
180
=
𝟏𝟔
𝝅
𝟏𝟓
4. How many degrees is 4𝜋?
One of two ways:
180
𝜋

4𝜋 ×

4π is twice around the unit circle, thus 360° × 2 = 𝟕𝟐𝟎°
= 𝟕𝟐𝟎°
5. A beam of light from a spotlight sweeps through an angle of 125°. The spotlight covers an
area of 350 ft2. What is the range of the spotlight?
Finding the range is equivalent to finding the radius of the circle that is partially swept
out by the spotlight.
125⁰
r
We set up the proportion:
125°
360°
=
350𝑓𝑡 2
.
ð𝑟 2
Solving for r, we obtain r = 17.9 ft
Saginaw Valley State University – January 2011
6. A pulley belt is 2ft long. It takes 5 seconds to complete 1 revolution. The diameter of the
pulley is 6 inches. What is the angular velocity, in rpm, of the pulley?
Ə
We know the angular velocity is given by: ù = ∆𝑡 =
𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠
,
𝑡𝑖𝑚𝑒 (∆𝑡)
(the diameter of the
pulley and the length of the pulley belt are useless for this problem). First, we must
convert seconds to minutes: 5 seconds ×
Thus, ù =
7.
1 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1
𝑚𝑖𝑛
12
= 𝟏𝟐
1
60
=
1
min.
12
𝐫𝐞𝐯
𝐦𝐢𝐧
ð
What is the period, amplitude, and phase shift of the function 𝑓(𝑥) = 2 sin (3𝑥 − 2)?
The amplitude is the 2 since 2(1) = 2 is the largest the given expression can be. The
period is found by dividing 2ð by the coefficient of 𝑥, yielding
2ð
3
for the period. The
phase shift is found by taking the negative of the constant term of the argument
and dividing it by the coefficient of 𝑥, yielding
ð
𝟐
𝟑
ð
= .
𝟔
1
8. Let sin 𝑥 = 2 where x is in the second quadrant. What are the value of cos 𝑥 and tan 𝑥?
1
ð
Since sin 𝑥 = 2, 𝑥 is either 6 𝑜𝑟
5ð
6
(See “Quick Trigonometry Review” section on special
angles). Since the angle lies in the second quadrant, the angle measure, 𝑥, must be between
ð
2
and , so 𝑥 must
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5ð
be 6 .
Thus, cos 𝑥 =
√𝟑
− 𝟐
1
2
and tan 𝑥 =
√3
−
2
=−
1
√3
=−
√𝟑
𝟑
Trigonometry Review Solutions p. 2
3
ð
9. Let cos è = 5, where 0 ≤ è ≤ 2. What is the value of sin 2è?
Since we know the cosine, we know that the ratio of the adjacent leg to the hypotenuse
is 3/5. Consider the hypothetical triangle below. Using the Pythagorean Theorem, we
4
find the opposite leg to be 4. Thus we can conclude that sin è = 5. From the double
4
3
𝟐𝟒
angle formula, sin 2θ = 2 sin θ cosθ, we can conclude sin 2θ = 2 × 5 × 5 = 𝟐𝟓
10. Complete the following Pythagorean triples:
Use the Pythagorean Theorem or from memory
a. __, 4, 5 → 𝑥 2 +42 = 52 → 𝑥 = √25 − 16 → 3
b. 8, __ , 17→ 82 +𝑥 2 = 172 → 𝑥 = √225 → 15
c. 5, 12, __ → 52 +122 = 𝑥 2 → 𝑥 = √25 + 144 → 13
5
11. Given θ is in Quadrant II, and sinθ = 13, find cosθ.
Since we know the ratio of the opposite leg and hypotenuse is 5/13, we can consider
the hypothetical triangle below. The adjacent leg can be found to be 12 using the
Pythagorean Theorem. From the figure it is clear that the cosine is
second quadrant, it must be negative, therefore 𝐜𝐨𝐬 è =
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12
.
13
Since it is in the
−𝟏𝟐
𝟏𝟑
Trigonometry Review Solutions p. 3
12. Find exact values of x and y
10
x
a.)
b.)
30°
30°
y
a)
𝑦
10
= cos 30, so
4
b) 𝑥 = sin 30, so
1
√3
4
4
x
y
10
y
=
√3
2
solving for y give 𝟓√𝟑
1
= 2 solving for x gives 𝟖 For y,
4
y
= tan 30 so
4
y
=
, solving for y gives 𝟒√𝟑
13. Verify the following identities: (Other solutions may be correct.)
a.)
2
sin 𝑥 cot 𝑥
cos2 𝑥
𝐜𝐨𝐬𝟐 𝒙
=
sin
𝑥
cot
𝑥
=
sin
𝑥
=
tan 𝑥
𝐬𝐢𝐧 𝒙
sin2 𝑥
Or:
b.)
csc 𝑥
cot 𝑥
cos 𝑥
sin 𝑥
sin 𝑥
cos 𝑥
sin 𝑥
=
cos 𝑥
sin 𝑥
sin 𝑥
cos 𝑥
sin 𝑥
= csc 𝑥 tan 𝑥 =
1
= cos 𝑥
sin 𝑥
sin 𝑥 cos 𝑥
=
cos 𝑥
sin 𝑥
1
sec 𝑥 tan 𝑥
1 sin 𝑥 1
1
2
sin 𝑥 = cos 𝑥 cos 𝑥 sin 𝑥 = cos2 𝑥 = sec 𝑥
d.)
cot 𝑥 csc 𝑥 =
sin 𝑥
∙
1
sin 𝑥
=
cos 𝑥
sin2 𝑥
=
cos 𝑥
1−cos2 𝑥
𝒄𝒐𝒔𝟐 𝒙
𝐬𝐢𝐧 𝒙
= sec 𝑥
cos 𝑥
c.)
cos 𝑥
=
cos 𝑥(1+cos2 𝑥)
= (1−cos2
𝑥)(1+cos2 𝑥)
=
cos 𝑥+cos3 𝑥
1−cos4 𝑥
Saginaw Valley State University
Trigonometry Review Solutions p. 4
14. If
ð
2
1
< è < ð and sin è = 2, find cos è.
Since we know the ratio of the opposite leg and hypotenuse is ½, we can consider the triangle
below. The adjacent leg was found using the Pythagorean theorem. From the figure it is clear
that the cosine is
−
√3
.
2
Since it is in the second quadrant, it must be negative, therefore 𝐜𝐨𝐬 è =
√𝟑
𝟐
3
15. If è is acute and sin è = 5, find cos è.
Since we know the ratio of the opposite leg and hypotenuse is 3/5, we can consider the triangle
below. The adjacent leg was found using the Pythagorean theorem. It is clear that the cosine is
𝟒
𝟓
4/5. Since the angle is acute, cosine is positive, thus 𝐜𝐨𝐬 è = .
.
16. Find the quadrant containing è:
Recall that sine is positive in quadrants I and II, cosine is positive in quadrants I and IV, and
tangent is positive in quadrants I and III.
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Trigonometry Review Solutions p. 5
a.) cos è > 0 and sin è < 0
IV
b.) tan è > 0 and sin è > 0
I
c.) sin è < 0 and cot è > 0
III
17. Find the exact value of è if tan è = −√3 and sin è > 0.
ð
If |tan è| = √3, this relationship is satisfied by a reference angle of 3. Since tangent is
ð
negative and sine is positive, we are in quadrant II, thus è = ð − 3 =
𝟐ð
𝟑
18. Evaluate: (exact answers)
See the “Quick Trigonometry Review” . Be familiar with the special angles.
a.) sin 45° =
c.) tan
5ð
6
=
√𝟐
𝟐
5ð
6
5ð
cos
6
sin
19. If sin è =
√2
2
b.) cos
1⁄
2
√3⁄
2
=−
=−
and cos è = −
1
√3
=−
√2
,
2
√𝟑
d.) csc 30°
𝟑
3ð
4
=−
1
1
= sin 30° = 1⁄ = 𝟐
2
find cos 2è.
Recall that cos 2è = cos2 è − sin2 è, so cos 2è = (−
1
1−cos è
2
20. Evaluate sin 15° (Hint: Recall sin 2 è = ±√
1
1−cos 30°
2
sin (2 ∙ 30°) = √
√𝟐
𝟐
1−√3⁄2
=√
2
=√
2
√2
)
2
√2
2
2
−( ) =𝟎
)
𝟐−√𝟑
𝟒
=
√𝟐−√𝟑
𝟐
21. Given 30° < è < 60°, find the interval which contains x.
1
x
θ
sin è =
𝑥
, 𝑠𝑜
1
𝑥 = sin è where 30° < è < 60° → sin 30° < sin è < sin 60°
→ sin 30° < 𝑥 < sin 60°
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→
𝟏
√𝟑
<𝑥<
𝟐
𝟐
Trigonometry Review Solutions p. 6
3
1
22. Given cos 2 è = 4 and sin2 è = 4, find cos 2è
3
1
𝟏
cos 2è = cos 2 è − sin2 è = 4 − 4 = 𝟐
5
23. Given cos 2 è = 8, find cos 2è
5
5
𝟏
cos 2è = 2cos 2 è − 1 = 2 (8) − 1 = 4 − 1 = 𝟒
2
3
24. Given sin 2è = 3 and cos è = 4 , find sin è.
sin 2è = 2 sin è cos è
→
2
2
sin 2è
𝟒
3
sin è =
=
=3=
2 cos è 2 (3) 3 𝟗
4
2
25. A rocket is fired at sea level and climbs at a constant angle of 75 through a distance of 10,000.
Approximate the altitude of the rocket to the nearest foot.
Set up the right triangle below:
10,000 ft
x
75⁰
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
Using the identity 𝑠𝑖𝑛è = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒:
𝑋
𝑠𝑖𝑛75° = 10,000
→
𝑥 = 𝑠𝑖𝑛 75° × 10,000 ≈ 9659𝑓𝑡
26. A hiker is 30 feet from the base of a hill. At this distance from the base, the angle of
elevation from the ground to the peak of the hill is 60°. Calculate the height of the hill.
Since adjacent side is known and opposite side needed use
tangent
tan60 =
ℎ
30
Solve for ℎ
ℎ = 30tan60
Substitute value of tangent for special angle of 60°
ℎ = 30√3
Calculate
ℎ ≈ 51.96 𝑓𝑡
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Trigonometry Review Solutions p. 7
27. A ten meter lamp post casts a shadow on a woman that stands five meters away from the
lamp. If the woman is two meters tall, what is the length of her shadow?
The woman and her shadow and the rays of the lamp form a triangle that is similar to the
lamp and the ground and the rays of the lamp.
Create a proportion of corresponding parts for the two triangles
2
10
= 5+𝑥
𝑥
1
5
= 5+𝑥
Simplify and solve for 𝑥
𝑥
5 + 𝑥 = 5𝑥
5 = 4𝑥
𝑥=
5
4
= 1.25 𝑚
Saginaw Valley State University
Trigonometry Review Solutions p. 8