Download Trigonometry (Chapters 4-5) – Sample Test #1

Trigonometry(Chapters4‐5)–SampleTest#1
First, a couple of things to help out: Page 1 of 24
Trigonometry (Chapters 4‐5) – Sample Test #1 Use periodic properties of the trigonometric functions to find the exact value of the expression. 1. cos
cos
2
cos
2. sin
sin
4
sin
3. cot
cot 2
√
Use the unit circle or the chart at the front of this packet. cot
√
√
√
Sin t and cos t are given. Use identities to find the indicated value. Where necessary, rationalize denominators. √
4. sin
, cos
. Find sec . Nothing fancy here. We don’t even need a drawing. sec
and cos t is given. Use the Pythagorean identity √
5. cos
sin
to find sin t. cos
1 ⇒ sin
sin
sin
√
1 1 in Q1 ⇒ √
Find a cofunction with the same value as the given expression. 6. sin
7. csc 52°
cos
sec 90°
52°
° sin
tan
sec
cos 90°
cot 90°
csc 90°
cos
cot
csc
sin 90°
tan 90°
sec 90°
Find all six trig functions for the angle . √
8. √4
25
√29 √
√
√
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√
√
Trigonometry (Chapters 4‐5) – Sample Test #1 A point on the terminal side of angle is given. Find the exact value of the six trigonometric functions of . 2, 3 9.
√
√
√
√
√
√
Solve the problem. 10. A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 500 feet to a mountain lake at an elevation of 8,300 feet. What is the length of the trail (to the nearest foot)? The height of the triangle is found by measuring the distance between the lake and the lodge. 8,300 500 7,800. sin 16°
7,800
7,800
sin 16°
,
ft. 11. A building 200 feet tall casts a 90 foot long shadow. If a person looks down from the top of the building, what is the measure of the angle between the end of the shadow and the vertical side of the building (to the nearest degree)? (Assume the person's eyes are level with the top of the building.) tan x°
tan
90
200
0.45 0.45
° Find the exact value of the indicated trigonometric function of . 12. The key on this type of problem is to draw the correct triangle. Notice that sin
0 , tan
0. Therefore is in 3. Notice that the horizontal leg must have length: Then, sec
√
√
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3
2
√5. Trigonometry (Chapters 4‐5) – Sample Test #1 13. Notice that cot
The hypotenuse has length: 4
Then, csc
0 , cos
√
0. Therefore is in 2. 9
√97. Use reference angles to find the exact value of the expression. Do not use a calculator. 14. sin
I like to draw the given angle so I can visualize the reference angle and the quadrant it is in. terminates in Q3. The reference angle is The sine function is negative in Q3. So, sin
15. sec
. 4
3
sin
√
3
terminates in Q2. The reference angle is . The secant (and cosine) functions are negative in Q3. So, sec
5
4
sec
1
4
cos
√ 4
16. csc 660° The given angle terminates in Q4. The reference angle is 720° 660° 60°. Also, the cosecant (and sine) functions are negative in Q4. So, csc 660°
csc 60°
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1
sin 60°
2
√3
√
Trigonometry (Chapters 4‐5) – Sample Test #1 Graphs of Trigonometric Functions Six Functions – Reference Guide The sine and cosecant functions are inverses. So: sin
1
csc
and
csc
1
sin
The cosine and secant functions are inverses. So: cos
1
sec
and
sec
1
cos
The tangent and cotangent functions are inverses. So: tan
1
cot
and
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cot
1
tan
Trigonometry (Chapters 4‐5) – Sample Test #1 Graph the function. 17. 3 sin 3 Standard Form: Characteristic Period | | Amplitude 3 2 Vertical Shift 1 Phase Shift 2
2
3
3
0 0 0 0 Note: the problem does not require us to show the parent function. I show it for comparison purposes only. 18. 3 sin
Standard Form: Characteristic Period | | Amplitude Vertical Shift 1 2 Phase Shift 3 2
0
/4
0 0 1
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2 Note: the problem does not require us to show the parent function. I show it for comparison purposes only. Trigonometry (Chapters 4‐5) – Sample Test #1 19. sin
Standard Form: Characteristic | | Amplitude Period 1 1
0 Vertical Shift 2
2 0 1/3 2 Phase Shift Note: the problem does not require us to show the parent function. I show it for comparison purposes only. 0 20. 3 cos
Standard Form: Characteristic Period | | Amplitude Vertical Shift 1 2 Phase Shift 3 2
1
2
0 0 0 0 Page 7 of 24
4 Note: the problem does not require us to show the parent function. I show it for comparison purposes only. Trigonometry (Chapters 4‐5) – Sample Test #1 21. 3 cos 3
Standard Form: Characteristic Period 1 3 (negative) 2 Phase Shift | | Amplitude Vertical Shift 2
2
3
3
0 /3 0 0 Note: the problem does not require us to show the parent function. I show it for comparison purposes only. 22. tan
Standard Form: Characteristic Stretch | | Period Phase Shift Vertical Shift 1 1 (negative) 1
0 0 0 Page 8 of 24
Note: the problem does not require us to show the parent function. I show it for comparison purposes only. Trigonometry (Chapters 4‐5) – Sample Test #1 23. 4 cot 3 Standard Form: | | Period 1 Vertical Shift 3 sec
0 0 0 | | 1 Period 2 Vertical Shift 3 2
1
0 0 0 0 Page 9 of 24
Note: the problem does not require us to show the parent function. I show it for comparison purposes only. 1
3
Standard Form: Stretch Phase Shift 3
0 Characteristic 4 Phase Shift 24. Characteristic Stretch 2 Note: the problem does not require us to show the parent function. I show it for comparison purposes only. Trigonometry (Chapters 4‐5) – Sample Test #1 4 csc
25.
Standard Form: Characteristic Stretch | | 1 Period 2 Phase Shift 4 (negative) 2
1
0 Vertical Shift /4 0 2 0 Find the exact value of the expression. Use the unit circle or the chart at the front of this packet. 26. sin
√
√
27. cos
28. cos
1
29. tan
√
Know where the primary values for the inverse trig functions are defined. sin
θ is defined in Q1 and Q4. cos
θ is defined in Q1 and Q2. tan
θ is defined in Q1 and Q4. Page 10 of 24
Note: the problem does not require us to show the parent function. I show it for comparison purposes only. Trigonometry (Chapters 4‐5) – Sample Test #1 Find the exact value of the expression, if possible. Do not use a calculator. 30. tan
tan
The angle is in Q2, but tangent is defined only in Q1 and Q4. Further, tan
0 in Q2. 0, with the same tangent value as . So we seek the angle in Q4, where tangent is also Recall that the tangent function has a period of radians. Then, tan
tan
Use a sketch to find the exact value of the expression. √
31. cot sin
5√61
61
First, calculate the horizontal leg of the triangle: 6√61. Then draw. Based on the diagram, then, cot sin
√
√
cot θ
√
32. cot sin
√
2
First, calculate the horizontal leg of the triangle: √2
√2. Then draw. Based on the diagram, then, cot sin
√
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cot θ
√
√
1 Trigonometry (Chapters 4‐5) – Sample Test #1 Use a right triangle to write the expression as an algebraic expression. Assume that x is positive and in the domain of the given inverse trigonometric function. 33. cos tan
Since the tangent value is , let’s set up a triangle with the side opposite θ equal to , and the 1. side adjacent to θ equal to 1. The hypotenuse, then is √
Then, cos tan
√
1
cos θ
1
√
34. sin sec
√
The cosine of the angle is √
and the hypotenuse equal to √
right triangle. , so let’s set up a triangle with the side adjacent to θ equal to , 9. The side opposite θ, then, would be 3 in order to have a Then, sin sec
√
sin θ
√
35. cos sin
The sine of the angle is , so let’s set up a triangle with the side opposite θ equal to 3, and the hypotenuse equal to 5. The side adjacent to θ, then, would be 4 in order to have a right triangle. Then, cos sin
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cos θ
Trigonometry (Chapters 4‐5) – Sample Test #1 Find the exact value of the expression, if possible. Do not use a calculator. 36. sin
sin
The angle is in Q2, but sine is defined only in Q1 and Q4. Further, sin
0 with the same tangent value as . So we seek the angle in Q1, where sine is also sin
sin
0 in Q2. Solve the right triangle shown in the figure. Round lengths to one decimal place and express angles to the nearest tenth of a degree. 37. a = 3.8 cm, b = 2.4 cm √3.8
∠
90°
2.4
. 57.7°
∠
.
tan
.
. ° . ° 38. a = 3.3 in, A = 55.1° sin 55.1°
tan 55.1°
∠
90°
3.3
3.3
sin 55.1°
3.3
tan 55.1°
⇒
3.3
⇒
55.1°
. . . ° Using a calculator, solve the following problems. Round your answers to the nearest tenth. 39. A ship is 50 miles west and 31 miles south of a harbor. What bearing should the Captain set to sail directly to harbor? θ tan
31.8° φ 90° 31.8° 58.2° . °
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Trigonometry (Chapters 4‐5) – Sample Test #1 40. A boat leaves the entrance of a harbor and travels 16 miles on a bearing of N 22° E. How many miles north and how many miles east from the harbor has the boat traveled? θ
90°
22°
68° 16 ∙ cos 68°
.
16 ∙ sin 68°
.
CHAPTER 5 Complete the identity. 41. ? A) 1 + cot x sin
cos
B) sin x tan x cos cos
∙
sin cos
sin
cos
sin cos
1
sin cos
42. tan cot
A) ‐ sec2 x tan
cot
csc sec
cos
B) 1 ‐ sin x C) 0 AnswerB cos sin cos
sin
∙
∙ cos cos sin
cos
?
sin
cos
∙
cos
cos
sin
D) ‐2 tan2 x cos
sin
sin sin
∙
sin cos
C) sec x csc x Page 14 of 24
D) 1 Trigonometry (Chapters 4‐5) – Sample Test #1 43.
? A) 1 ‐ sec x csc x B) 2 ‐ sec x csc x C) 2 + sec x csc x D) sec x csc x sin cos
cos sin
sin
cos
sin cos sin
sin cos cos
∙
∙
sin
cos
sin
cos
sin cos
sin
sin cos
sin cos
2sin cos
sin
sin cos
1
sin cos
2
cos
2
cos
2sin cos
sin cos
csc sec 1
44. cos (α + β) + cos (α ‐ β) = ? A) sin β cos α cos
B) 2cos α cos β cos
cos cos
sin sin
C) 2sin α cos β D) cos α cos β cos cos
sin sin 2 cos cos AnswerB
45.
? A) cot α + cot β sin
cos cos
sin
cos
B) tan β + tan α sin cos sin cos
cos cos
sin
cos
C) ‐tan α + cot β sin cos cos cos
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D) tan α + tan β sin cos
cos cos
Trigonometry (Chapters 4‐5) – Sample Test #1 ? 46.
A) cos2 x 1
cos 2
sin 2
B) sin2 x 1
1
2 cos
2 sin cos
C) tan x 2 cos
2 sin cos
D) cot x cos
sin
47. sin
? A) ‐cos x B) ‐sin x C) sin x D) cos x We might be tempted to use the angle addition formula for sin
to solve this, but the form of the formula indicates that we would likely be barking up the wrong tree. So, the question boils down to which of the answers provided is correct. A look at the graphs of the sine and cosine functions reveals that the cosine function is, in fact, equal to the sine function with a phase shift of (i.e., to the left). Therefore the correct AnswerD. solution is: Find the exact value by using a sum or difference identity. sin sin cos
sin cos sin sin cos sin cos cos cos cos
sin sin 48. sin 215°
95°
sin 210°
sin 210° cos 90°
49. sin 165°
sin 120°
sin 120° cos 45°
50. cos 285°
cos 240°
cos 240° cos 45°
90° 1
2
sin 90° cos 210°
0
1
√3
2
√2
2
1
2
√
45° sin 45° cos 120°
√3
2
√2
2
1
2
√2
2
√ √
45° sin 240° sin 45°
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√3
2
√2
2
√ √
Trigonometry (Chapters 4‐5) – Sample Test #1 Find the exact value of the expression. 51. sin265°cos25°
52. cos
sin
cos
53. sin185°cos65°
cos265°sin25°
sin
sin 265°
sin
25° sin 240° sin
cos185°sin65°
sin 185°
√
65° sin 120° √
Use the figure to find the exact value of sin 2 , cos 2 , and tan 2 . sin 2 2sin cos cos 2 cos
sin
tan 2 54. sin 2
2 sin cos
2∙
cos 2 cos
tan 2
sin 2
cos 2
sin 2
2 sin cos
2∙
cos 2 cos
tan 2
sin 2
cos 2
sin
7 24
∙
25 25
24
25
7
25
55. sin
12 5
∙
13 13
5
13
12
13
Use the given information to find the sin 2 , cos 2 , and tan 2 .
56. sin
, and lies in quadrant I
sin 2
2 sin cos
4 3
2∙ ∙
5 5
cos 2 cos
tan 2
sin 2
cos 2
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sin
3
5
4
5
Trigonometry (Chapters 4‐5) – Sample Test #1 57. tan
, and lies in quadrant III
sin 2
2 sin cos
2∙
15
∙
17
cos 2 cos
8
17
tan 2
sin 2
cos 2
sin
8
17
15
17
Write the expression as the sine, cosine, or tangent of a double angle. Then find the exact value of the expression.
58. 2sin120°cos120°
sin 2 ∙ 120°
°
√
59. tan 2 ∙
5
8
Find all solutions of the equation.
60. 2sin 2 sin
sin
√3 0
The drawing at left illustrates the two angles in 0, 2
√3 √
. To get all solutions, we need to add all integer multiples of 2 to these solutions. So, √3
2
∈
for which sin
Page 18 of 24
∪
Trigonometry (Chapters 4‐5) – Sample Test #1 61. tan sec
2 tan tan sec
2 tan
tan sec
tan
2
0 0or sec
0
0 sec
2 2
0 cos
Collecting the various solutions, ∈ sec
2
2
∪
0 2
or ∪
Note: the solution involving the tangent function has two answers in the interval 0, 2 . However, they are radians apart, as most solutions involving the tangent function are. Therefore, we can simplify the answers by showing only one base answer and adding , instead of showing two base answers that are apart, and adding 2 to each. 0 are telescoped into the single solution For example, the following two solutions for tan
given above: 0
2
… , 4 , 2 , 0, 2 , 4 , … 2
…, 3 ,
0
, ,3 ,5 … …, 2 ,
, 0, , 2 , … Solve the equation on the interval [0, 2 ). 62. sin 4
√
When working with a problem in the interval 0, 2 that involves a function of expand the interval to 0, 2
for the first steps of the solution. So, we want all solutions to sin
√
, it is useful to 4 is an angle in the interval 0, 8
where . Note that, beyond the two solutions suggested by the diagram, additional solutions are obtained by adding multiples of 2 to those two solutions. Using the diagram at left, we get the following solutions:
,
Then, dividing by 4, we get: 2 7 8 13 14 19 20
,
,
,
,
,
,
12 12 12 12 12 12 12 12
2 7 8 13 14 19 20
,
,
,
,
,
,
3 3 3 3 3
3
3
3
4
,
Note that there are 8 solutions because the usual number of solutions (i.e., 2) is increased by a factor of 4. And simplifying, we get: ,
Page 19 of 24
,
,
,
,
,
,
Trigonometry (Chapters 4‐5) – Sample Test #1 63. cos 2
√
√
So, we want all solutions to cos
2 is an angle in the interval 0, 4
where . Note that, beyond the two solutions suggested by the diagram, additional solutions are obtained by adding 2 to those two solutions. Using the diagram at left, we get the following solutions:
,
Then, dividing by 2, we get: ,
11 13 23
,
,
6 6
6
6
2
2cos
1
,
We cannot simplify these solutions any further. Note that there are 4 solutions because the usual number of solutions (i.e., 2) is increased by a factor of 2. 64. cos
,
0 The trick on this problem is to replace the trigonometric function, in this case, cos , with a variable, like , that will make it easier to see how to factor the expression. If you can see how to factor the expression without the trick, by all means proceed without it. Let 2
cos , and our equation becomes: This equation factors to get: And finally: cos
cos
The only solution for this on the interval 0, 2
0. 0 1
Substituting cos back in for gives: 1
1
0 0 ⇒ cos
1
is: 1 65. cos
sin This problem is most easily solved by inspection. Where are the cosine and sine functions equal? At the angles with a reference angle of in Q1 and Q3. Therefore, ,
Another method that can be used to solve this kind of problem is shown in the solution to Problem 66, below. Page 20 of 24
Trigonometry (Chapters 4‐5) – Sample Test #1 66. sin
cos
sin
0 cos
sin
cos
0 sin
cos
0 or sin
cos
sin
cos sin
cos tan
1 tan
1 ,
,
,
,
,
0 In this problem, we take a different approach to solving sin
cos , which could, as in Problem 65, above, be solved by inspection. Since sin and cos are never both zero, we can divide both sides by cos to get the resulting tan equations. 67. sin
sin
0 sin sin
sin
1
0 0 or 0, π sin
sin
1
0 1 , ,
68. tan
sin
tan
tan
sin
tan
tan
sin
1
tan
0 or Be extra careful when dealing with functions other than sine and cosine, because there are values at which these functions are undefined. 0 0 0, π sin
sin
1
1 0 While sin
1, tan is undefined at so is not a solution to this equation. , is a solution to the equation Page 21 of 24
, Trigonometry (Chapters 4‐5) – Sample Test #1 69. cos
2 cos sin
cos 1
0 2 sin
0 cos
0 or ,
1
sin
2 sin
,
,
,
,
0 70. cos 2
√2
2cos 2
cos 2
cos 2 √2 √2
2
Recall that working with a problem in the interval 0, 2 that involves a function of useful to expand the interval to 0, 2
for the first steps of the solution. So, we want all solutions to cos
√
2 is an angle in the interval 0, 4
where , it is . Note that, beyond the two solutions suggested by the diagram, additional solutions are obtained by adding 2 to those two solutions. Using the diagram at left, we get the following solutions:
7 9 15
,
,
4 4 4 4
2
,
Then, dividing by 2, we get: Note that there are 4 solutions because the usual number of solutions (i.e., 2) is increased by a factor of 2. ,
,
,
We cannot simplify these answers any further. Page 22 of 24
Trigonometry (Chapters 4‐5) – Sample Test #1 71. 2 cos
sin
2
sin
2 cos
2 1
2
0 2
sin
sin
2 sin
2
sin
2 sin
sin
sin 0 2
When an equation contains more than one function, try to convert it to one that contains only one function. 0 0 0 2 sin
1
0 sin
0 or 0, π 2 sin
sin
,
,
,
1
0 , 72. cos
cos
1 The following formulas will help us solve this problem. cos cos cos
cos
cos
3
cos cos 3
2cos cos
2cos ∙
cos
sin sin 3
cos 1 sin sin cos cos 3
3
sin sin 3
1 3
1 1 cos cos
Page 23 of 24
1 sin sin Trigonometry (Chapters 4‐5) – Sample Test #1 Use a calculator to solve the equation on the interval [0, 2 ). Round the answer to two decimal places. 73. cos
.74 0.738 radians (by calculator) 2
.738
6.283
.738
5.545 radians .
Rounding to 2 decimal places gives: , .
Use a calculator to solve the equation on the interval [0, 2 ). Round to the nearest hundredth of a radian. sin
74. sin 2
sin 2
0 sin
2 sin cos
sin
0 sin
2 cos
1
0 0 sin
0 or 0, π 2 cos
1
cos
,
0 5π
0, , , 3
3
Rounding to the nearest hundredth of a radian gives: Page 24 of 24
, .
, .
, .