Trigonometric Identities 2 PDF

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5-2 Verifying Trigonometric Identities Verify each identity.

1.(sec

2

– 1) cos

2

2

6.tan θ csc2

=sin

– tan

=cot

SOLUTION:

SOLUTION:

2.sec2

2

(1– cos

2

) = tan

7.

–

=cot

SOLUTION: SOLUTION:

3.sin

– sin

2

cos

3

=sin

SOLUTION:

 8. 4.csc

– cos

+

=2csc

cot =sin

SOLUTION:

SOLUTION:

5.

= cot

4

SOLUTION:

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

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5-2 Verifying Trigonometric Identities 9.

+ tan

12.

=sec

SOLUTION:

10.

+

+

=2sec

2

sin

SOLUTION:

=sin +cos

13.(csc

SOLUTION:

– cot

)(csc

+cot ) = 1

SOLUTION:

14.cos4

4

– sin

=cos

2

2

– sin

SOLUTION:

11.

+

=1

SOLUTION:

15.

+

=2sec

2

SOLUTION:

 

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5-2 Verifying Trigonometric Identities 16.

+

=2sec

19.FIREWORKS If a rocket is launched from ground level, the maximum height that it reaches is given by

SOLUTION:

h=

, where

istheanglebetweenthe

ground and the initial path of the rocket, v is the rocket’s initial speed, and g is the acceleration due to gravity, 9.8 meters per second squared.

17.csc4

4

– cot

=2cot

2

+1

SOLUTION:

a. Verify that

=

.

b. Suppose a second rocket is fired at an angle of 80 from the ground with an initial speed of 110 meters per second. Find the maximum height of the rocket.

SOLUTION: a.

18.

=

SOLUTION:   b. Evaluate the expression

forv = 110 m,

2

, and g = 9.8 m/s .  The maximum height of the rocket is about 598.7 meters.

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5-2 Verifying Trigonometric Identities Verify each identity. +cot )(1 – cos

20.(csc

) = sin

SOLUTION:

21.sin2

tan

2

=tan

22. SOLUTION:

2

2

– sin

SOLUTION:

 23.

= cos

+cot

SOLUTION:

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5-2 Verifying Trigonometric Identities 24.(csc

2

– cot

27.sec

) =

– cos

=tan sin

SOLUTION: SOLUTION:

28.1 – tan4 θ = 2 sec2 θ – sec4 θ SOLUTION: 25.

=

SOLUTION: 29.(csc

– cot

2

) =

SOLUTION:

30. 26.tan2

cos

2

2

=1– cos

= sec

SOLUTION:

SOLUTION:

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5-2 Verifying Trigonometric Identities 31.

= (sin

+cos )

2

GRAPHING CALCULATOR Test whether each equation is an identity by graphing. If it appears to be an identity, verify it. If not, find an x-value for which both sides are defined but not equal. 34. =

SOLUTION:

SOLUTION: Graph

andthengraph .



32.OPTICS If two prisms of the same power are placed next to each other, their total power can be determined using z = 2p cos , where z is the combined power of the prisms, p is the power of the individual prisms, and θ is the angle between the two 2

prisms. Verify that 2p cos θ = 2p (1 – sin

)sec

.

SOLUTION:

33.PHOTOGRAPHY The amount of light passing through a polarization filter can be modeled using I = 2

Im cos

, where I is the amount of light passing

through the filter, Im is the amount of light shined on the filter, and is the angle of rotation between the light source and the filter. Verify that .

SOLUTION:

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The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 = –1 and Y2 is undefined. Therefore, the equation is not an identity.

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5-2 Verifying Trigonometric Identities 36.sec2 x – 2 sec x tan x + tan2 x =

35.sec x + tan x = SOLUTION:

SOLUTION:

Graph

andthengraph .

The equation appears to be an identity because the graphs of the related functions coincide. Verify this algebraically.

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2

2

Graph Y1 = sec x – 2 sec x tan x + tan x and then graph Y2 =

.

 The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = 0, Y1 = 1 and Y2 = 0. Therefore, the equation is not an identity.

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5-2 Verifying Trigonometric Identities = 1 – 2 sin2 x

37.

SOLUTION: Graph Y1 =

38.

=

SOLUTION: andthengraphY2 = 1 – 2

Graph Y1 =

2

sin x .

andthengraphY2 = .

 

 The equation appears to be an identity because the graphs of the related functions coincide. Verify this algebraically.

The graphs of the related functions do not coincide for all values of x for which both functions are defined. Using the intersect feature from the CALC menu on the graphing calculator to find that when x = , Y1 –0.17 and Y2 = 0. Therefore, the equation is not an identity.

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5-2 Verifying Trigonometric Identities 39.cos2 x – sin2 x =

Verify each identity.

40.

SOLUTION: 2

2

Graph Y1 = cos x – sin x and then graph Y2 =

SOLUTION: 

.

The equation appears to be an identity because the graphs of the related functions coincide. Verify this algebraically.

41.

=

SOLUTION:

42.ln |csc x + cot x| + ln |csc x – cot x| = 0 SOLUTION:

43.ln |cot x| + ln |tan x cos x| = ln |cos x| SOLUTION:

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5-2 Verifying Trigonometric Identities Verify each identity. 44.sec2 +tan2 =sec4

48.sec4 x = 1 + 2 tan2 x + tan4 x

4

– tan

SOLUTION:

SOLUTION:

Start with the right side of the identity.

Start with the right side of the identity.

49.sec2 x csc2 x = sec2 x + csc2 x 45.–2 cos2

4

=sin

4

– cos

SOLUTION:

– 1

Start with the left side of the identity.

SOLUTION: Start with the right side of the identity.

46.sec2 θsin2 θ=sec4 θ – (tan4 θ+sec2 θ) SOLUTION: Start with the right side of the identity.

47.3 sec2

tan

2

+1=sec

6

6

– tan

SOLUTION:

50.ENVIRONMENT A biologist studying pollution situates a net across a river and positions instruments at two different stations on the river bank to collect samples. In the diagram shown, d is the distance between the stations and w is width of the river.

a. Determine an equation in terms of tangent α that can be used to find the distance between the stations. b. Verify that d = . c. Complete the table shown for d = 40 feet.

Start with the right side of the identity. d. If >60 or  120. If w = 35, then 45 < < 63.4 since 35 is between 20 and 40. The sites with widths of 5 and 140 feet could not be used because > 60 and  0

83.x 2 + 3x – 28 < 0 SOLUTION:

SOLUTION:

f(x ) has real zeros at x = –3 and x = 6. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.

f(x ) has real zeros at x = –7 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.





 









2

Thesolutionsofx – 3x – 18 > 0 are x -values such that f (x) is positive. From the sign chart, you can see that the solution set is .

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2

Thesolutionsofx + 3x – 28 < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the solution set is .

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5-2 Verifying Trigonometric Identities 84.x 2 – 4x ≤ 5

85.x 2 + 2x ≥ 24

SOLUTION:

SOLUTION:

2

2

2

2

First, write x – 4x ≤ 5 as x – 4x – 5 ≤ 0.

First, write x + 2x ≥ 24 as x + 2x – 24 ≥ 0.

f(x ) has real zeros at x = –1 and x = 5. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.

f(x ) has real zeros at x = –6 and x = 4. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.













2

The solutions of x – 4x –5 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .

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2

The solutions of x + 2x – 24 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is .

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5-2 Verifying Trigonometric Identities 86.− x 2 − x + 12 ≥ 0

87.− x 2 – 6x + 7 ≤ 0

SOLUTION:

SOLUTION:

2

2

2

2

First, write – x – x + 12 ≥ 0 as x + x – 12 ≤ 0.

First, write – x – 6x +7 ≤ 0 as x + 6x – 7 ≥ 0.

f(x ) has real zeros at x = –4 and x = 3. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.

f(x ) has real zeros at x = –7 and x = 1. Set up a sign chart. Substitute an x-value in each test interval into the polynomial to determine if f (x) is positive or negative at that point.





 

 

 

2

The solutions of x + x – 12 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that the solution set is .

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2

The solutions of x + 6x – 7 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that the solution set is .

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5-2 Verifying Trigonometric Identities 88.FOOD The manager of a bakery is randomly

89.SAT/ACT

SOLUTION:

a , b, a , b, b, a , b, b, b, a , b, b, b, b, a , …  If the sequence continues in this manner, how many bs are there between the 44th and 47th appearances of the letter a? A 91 B 135 C 138 D 182 E 230

The parent function of g(x) is f (x ) = |x |. The factor

SOLUTION:

checking slices of cake prepared by employees to ensure that the correct amount of flavor is in each slice. Each 12-ounce slice should contain half chocolate and half vanilla flavored cream. The amount of chocolate by which each slice varies can be represented by g(x) =

|x – 12|. Describe the

transformations in the function. Then graph the function.

of

willcausethegraphtobecompressedsince andthesubtractionof12willtranslatethe

graph 12 units to the right. Make a table of values for x and g(x).  x 4 8 12 16 g(x) 4 2 0 2  Plot the points and draw the graph of g(x).

The number of bs after each a is the same as the number of a in the list (i.e., after the 44th a there are 44 bs). Between the 44th and 47th appearances of a the number of bs will be 44 + 45 + 46 or 135. Therefore, the correct answer choice is B.

90.Which expression can be used to form an identity 20 4

with

,when

tan  −1? F sin  G cos H tan J csc

SOLUTION:

Therefore, the correct answer choice is J.

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5-2 Verifying Trigonometric Identities 91.REVIEW Which of the following is not equivalent to cos

, when 0 < θ...