trigonometry (chapters 4-5) – sample test #1

Trigonometry(Chapters4‐5)–SampleTest#1

First, a couple of things to help out:

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Trigonometry (Chapters 4‐5) – Sample Test #1

√4 25 √29

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 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 to find sin t.

5. cos √

sin cos 1 ⇒ sin √ 1

sin 1

sin in Q1 ⇒

√

Find a cofunction with the same value as the given expression.

6. sin cos

7. csc 52° sec 90° 52° °

Find all six trig functions for the angle .

8.

√

√

√

√

√

√

Use the unit circle or the chart

at the front of this packet.

sin cos 90° cos sin 90°

tan cot 90° cot tan 90°

sec csc 90° csc sec 90°

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A point on the terminal side of angle is given. Find the exact value of the six trigonometric

functions of .

9. 2, 3

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)?

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.)

Find the exact value of the indicated trigonometric function of .

12.

√

√

√

√

√

√

sin 16°7,800

7,800sin 16°

, ft.

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: 3 2 √5.

Then, sec √

√

tan x°90200

0.45

tan 0.45 °

The height of the triangle

is found by measuring the

distance between the

lake and the lodge.

8,300 500 7,800.

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13.

Use reference angles to find the exact value of the expression. Do not use a calculator.

14. sin

15. sec

16. csc 660°

Notice that cot 0 , cos 0. Therefore is in 2.

The hypotenuse has length: 4 9 √97.

Then, csc √

sin43

sin3

√

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,

sec54

sec4

1

cos 4

√

terminates in Q2. The reference angle is .

The secant (and cosine) functions are negative in Q3. So,

csc 660° csc 60°1

sin 60°2

√3

√

The given angle terminates in Q4. The reference angle is

720° 660° 60°. Also, the cosecant (and sine) functions are negative in Q4. So,

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The sine and cosecant functions are inverses. So:

sin1

cscand csc

1sin

The cosine and secant functions are inverses. So:

cos1

secand sec

1cos

The tangent and cotangent functions are inverses. So:

tan1

cotand cot

1tan

Graphs of Trigonometric Functions

Six Functions – Reference Guide

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Graph the function.

17. 3 sin 3 Standard Form:

18. 3 sin Standard Form:

Characteristic

Amplitude | | 1 3

Period 2 2 323

Phase Shift 0 0

Vertical Shift 0 0

Characteristic

Amplitude | | 1 3

Period 2 2 1 2

Phase Shift 0 /4

Vertical Shift 0 0

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

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19. sin Standard Form:

20. 3 cos Standard Form:

Characteristic

Amplitude | | 1 1/3

Period 2 2 1 2

Phase Shift 0

Vertical Shift 0 0

Characteristic

Amplitude | | 1 3

Period 2 212

4

Phase Shift 0 0

Vertical Shift 0 0

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

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21. 3 cos 3 Standard Form:

22. tan Standard Form:

Characteristic

Amplitude | | 1 3 (negative)

Period 2 2 323

Phase Shift 0 /3

Vertical Shift 0 0

Characteristic

Stretch | | 1 1 (negative)

Period 1

Phase Shift 0

Vertical Shift 0 0

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

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23. 4 cot 3 Standard Form:

24. 3 sec Standard Form:

Characteristic

Stretch | | 1 4

Period 313

Phase Shift 0 0

Vertical Shift 0 0

Characteristic

Stretch | | 1 3

Period 2 2 1 2

Phase Shift 0 0

Vertical Shift 0 0

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

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25. 4 csc Standard Form:

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 √

Characteristic

Stretch | | 1 4 (negative)

Period 2 2 1 2

Phase Shift 0 /4

Vertical Shift 0 0

Note: the problem

does not require us

to show the parent

function. I show it

for comparison

purposes only.

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.

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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.

So we seek the angle in Q4, where tangent is also 0, with the same tangent value as .

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 √

First, calculate the horizontal leg of the triangle: 61 5√61 6√61. Then draw.

32. cot sin √

First, calculate the horizontal leg of the triangle: 2 √2 √2. Then draw.

Based on the diagram, then,

cot sin √ cot θ √

√

Based on the diagram, then,

cot sin √ cot θ √

√1

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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

side adjacent to θ equal to 1. The hypotenuse, then is √ 1.

34. sin sec√

The cosine of the angle is √

, so let’s set up a triangle with the side adjacent to θ equal to ,

and the hypotenuse equal to √ 9. The side opposite θ, then, would be 3 in order to have a right triangle.

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.

cos tan cos θ1

√ 1

√

Then,

Then,

sin sec√

sin θ√

Then,

cos sin cos θ

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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 in Q2.

So we seek the angle in Q1, where sine is also 0 with the same tangent value as .

sin sin

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 2.4 . ∠ tan .

.. °

∠ 90° 57.7° . °

38. a = 3.3 in, A = 55.1°

sin 55.1°3.3

⇒

3.3sin 55.1°

.

tan 55.1°3.3

⇒

3.3tan 55.1°

.

∠ 90° 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?

φ 90° 31.8° 58.2°

. °

θ tan 31.8°

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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?

CHAPTER 5

Complete the identity.

41. ?

A) 1 + cot x B) sin x tan x C) sec x csc x D) ‐2 tan2 x

sincos

cossin

sinsin

∙sincos

cossin

∙coscos

sin cossin cos

1sin cos

csc sec

42. tan cot cos ?

A) ‐ sec2 x B) 1 ‐ sin x C) 0 D) 1

tan cot cos

sincos

∙cossin

cos

sincos

∙cossin

sincos

∙ cos

AnswerB

16 ∙ cos 68° .

16 ∙ sin 68° .

θ 90° 22° 68°

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43. ?

A) 1 ‐ sec x csc x B) 2 ‐ sec x csc x C) 2 + sec x csc x D) sec x csc x

cos sincos

sin cos

sin

sinsin

∙cos sin

cos

sin cossin

∙coscos

sin cos sin sin cos cossin cos

2sin cos sin cossin cos

2sin cos 1sin cos

21

sin cos 2 csc sec

44. cos (α + β) + cos (α ‐ β) = ?

A) sin β cos α B) 2cos α cos β C) 2sin α cos β D) cos α cos β

cos cos

cos cos sin sin cos cos sin sin

2 cos cos AnswerB

45.

?

A) cot α + cot β B) tan β + tan α C) ‐tan α + cot β D) tan α + tan β

sincos cos

sin cos sin cos

cos cos

sin cos cos cos

sin coscos cos

sincos

sincos

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46. ?

A) cos2 x B) sin2 x C) tan x D) cot x

1 cos 2sin 2

1 2 cos 12 sin cos

2 cos2 sin cos

cossin

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

solution is: AnswerD.

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° 90°

sin 210° cos 90° sin 90° cos 210°12

0 1√32

√

49. sin 165° sin 120° 45°

sin 120° cos 45° sin 45° cos 120°√32

√22

√22

12

√ √

50. cos 285° cos 240° 45°

cos 240° cos 45° sin 240° sin 45°12

√22

√32

√22

√ √

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Find the exact value of the expression.

51. sin265°cos25° cos265°sin25° sin 265° 25° sin 240° √

52. cos sin cos sin sin sin

53. sin185°cos65° 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.

55.

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 2 ∙725

∙2425

cos 2 cos sin 2425

725

tan 2sin 2cos 2

sin 2 2 sin cos 2 ∙1213

∙513

cos 2 cos sin 513

1213

tan 2sin 2cos 2

sin 2 2 sin cos 2 ∙45∙35

cos 2 cos sin 35

45

tan 2sin 2cos 2

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57. tan , and lies in quadrant III

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 ∙ 58

Find all solutions of the equation.

60. 2sin √3 0

2 sin √3

sin√32

sin 2 2 sin cos 2 ∙1517

∙817

cos 2 cos sin 817

1517

tan 2sin 2cos 2

The drawing at left illustrates the two

angles in 0, 2 for which sin √ . To

get all solutions, we need to add all

integer multiples of 2 to these solutions.

So,

∈ ∪

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61. tan sec 2 tan

tan sec 2 tan 0 sec 2 0

tan sec 2 0 sec 2

tan 0or sec 2 0 cos

0 2 or 2

Collecting the various solutions, ∈ ∪ ∪

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.

For example, the following two solutions for tan 0 are telescoped into the single solution given above:

0 2 … , 4 , 2 , 0, 2 , 4 , …

2 … , 3 , , , 3 , 5 …

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 , it is useful to

expand the interval to 0, 2 for the first steps of the solution.

So, we want all solutions to sin √ where 4 is an angle in the interval 0, 8 . Note

that, beyond the two solutions suggested by the diagram, additional solutions are obtained by

adding multiples of 2 to those two solutions.

0 … , 2 , , 0, , 2 , …

43,23,73,83,133

,143

,193

,203

12,212

,712

,812

,1312

,1412

,1912

,2012

, , , , , , ,

Using the diagram at left, we get the following solutions:

Then, dividing by 4, we get:

And simplifying, we get: Note that there are 8 solutions because the usual number of

solutions (i.e., 2) is increased by a factor of 4.

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63. cos 2 √

So, we want all solutions to cos √ where 2 is an angle in the interval 0, 4 . Note


adding 2 to those two solutions.

64. cos 2cos 1 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 cos , and our equation becomes: 2 1 0.

This equation factors to get: 1 0

Substituting cos back in for gives: cos 1 0

And finally: cos 1 0⇒ cos 1

The only solution for this on the interval 0, 2 is:

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.

26,116

,136

,236

, , ,



We cannot simplify these solutions any further. Note that there are 4 solutions because the usual number of


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66. sin cos 0

sin cos sin cos 0

sin cos 0 or sin cos 0

sin cos sin cos

tan 1 tan 1

, ,

, , ,

67. sin sin 0

sin sin 1 0

sin 0 or sin 1 0

0, π sin 1

, ,

68. tan sin tan

tan sin tan 0

tan sin 1 0

tan 0 or sin 1 0

0, π sin 1

,

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.

While is a solution to the equation

sin 1, tan is undefined at ,

so is not a solution to this equation.

Be extra careful when dealing with functions

other than sine and cosine, because there are

values at which these functions are undefined.

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69. cos 2 cos sin 0

cos 1 2 sin 0

cos 0 or 1 2 sin 0

, sin

,

, , ,

70. cos 2 √2 cos 2

2cos 2 √2

cos 2√22

Recall that working with a problem in the interval 0, 2 that involves a function of , it is

useful to expand the interval to 0, 2 for the first steps of the solution.

So, we want all solutions to cos √ where 2 is an angle in the interval 0, 4 . Note


adding 2 to those two solutions.

24,74,94,154

, , ,



We cannot simplify these answers any further.

Note that there are 4 solutions because the usual number of


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71. 2 cos sin 2 0

2 cos sin 2 0

2 1 sin sin 2 0

2 2 sin sin 2 0

2 sin sin 0

sin 2 sin 1 0

sin 0 or 2 sin 1 0

0, π sin

,

, , ,

72. cos cos 1

The following formulas will help us solve this problem.

cos cos cos sin sin cos cos cos sin sin

cos3

cos3

1

cos cos3 sin sin

3 cos cos

3 sin sin

3 1

2cos cos3

1

2cos ∙ 1

cos 1

When an equation contains more

than one function, try to convert it to

one that contains only one function.

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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.

74. sin 2 sin 0

sin 2 sin 0

2 sin cos sin 0

sin 2 cos 1 0

sin 0 or 2 cos 1 0

0, π cos

,

0,3, ,

5π3

Rounding to the nearest hundredth of a radian gives: , . , . , .

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trigonometry (chapters 4-5) – sample test #1

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