Precalculus with Limits: A Graphing Approach
Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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Fundamentals of Trigonometric Identities
Trigonometric Identities are the equalities that involve trigonometry functions and hold for all the values of variables given in the equation.
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Chapter 5.5, Problem 100E
To determine

To find: The value of the given function in two ways, using the given figure and trigonometric identities.

Expert Solution & Answer
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Answer to Problem 100E

Value of the given function is 3665

Explanation of Solution

Given information:

The following figure:

  Precalculus with Limits: A Graphing Approach, Chapter 5.5, Problem 100E

The following function:

  cosαsinβ

Formula used:

The following Pythagorean identity:

  sin2θ+cos2θ=1

The following Pythagorean relations:

  Hypotenuse2=Adjacent2+Opposite2cosθ=AdjacentHypotenusesinθ=OppositeHypotenuse

Calculation:

From the given figure,

  Adjacent=12                                             [For α]Opposite=5        [For α]Hypotenuse2=Adjacent2+Opposite2       [Pythagorean relation]Hypotenuse2=(12)2+(5)2        [Substitute values]Hypotenuse2=144+25        [Find square]Hypotenuse2=169        [Add]Hypotenuse=13        [Find square root of both sides]cosα=AdjacentHypotenuse=1213        [Pythagorean relation]sinα=OppositeHypotenuse=513        [Pythagorean relation]Adjacent=4                                               [For β]Opposite=3        [For β]Hypotenuse2=Adjacent2+Opposite2       [Pythagorean relation]Hypotenuse2=(4)2+(3)2        [Substitute values]Hypotenuse2=16+9                    [Find square]Hypotenuse2=25        [Add]Hypotenuse=5        [Find square root of both sides]cosβ=AdjacentHypotenuse=45        [Pythagorean relation]sinβ=OppositeHypotenuse=35        [Pythagorean relation]

Value of cosαsinβ from the figure is calculated as:

  cosαsinβ=1213⋅35 [Substitute]    =3665           [Simplify]

Value of cosαsinβ from trigonometric identities is calculated as:

  cosαsinβ=1−sin2α⋅1−cos2β                            [Pythagorean identity]       =1−(513)2⋅1−(45)2    [Substitute]       =1−25169.1−1625                 [Square]        =144169.925    [Subtract]      =1213⋅35    [Find square root]      =3665    [Simplify]

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Chapter 5.5, Problem 99E
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Chapter 5.5, Problem 101E

Chapter 5 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 5.1 - Prob. 1ECh. 5.1 - Prob. 2ECh. 5.1 - Prob. 3ECh. 5.1 - Prob. 4ECh. 5.1 - Prob. 5ECh. 5.1 - Prob. 6ECh. 5.1 - Prob. 7ECh. 5.1 - Prob. 8ECh. 5.1 - Prob. 9ECh. 5.1 - Prob. 10E
Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - Prob. 27ECh. 5.1 - Prob. 28ECh. 5.1 - Prob. 29ECh. 5.1 - Prob. 30ECh. 5.1 - Prob. 31ECh. 5.1 - Prob. 32ECh. 5.1 - Prob. 33ECh. 5.1 - Prob. 34ECh. 5.1 - Prob. 35ECh. 5.1 - Prob. 36ECh. 5.1 - Prob. 37ECh. 5.1 - Prob. 38ECh. 5.1 - Prob. 39ECh. 5.1 - Prob. 40ECh. 5.1 - Prob. 41ECh. 5.1 - Prob. 42ECh. 5.1 - Prob. 43ECh. 5.1 - Prob. 44ECh. 5.1 - Prob. 45ECh. 5.1 - Prob. 46ECh. 5.1 - Prob. 47ECh. 5.1 - Prob. 48ECh. 5.1 - Prob. 49ECh. 5.1 - Prob. 50ECh. 5.1 - Prob. 51ECh. 5.1 - Prob. 52ECh. 5.1 - Prob. 53ECh. 5.1 - Prob. 54ECh. 5.1 - Prob. 55ECh. 5.1 - Prob. 56ECh. 5.1 - Prob. 57ECh. 5.1 - Prob. 58ECh. 5.1 - Prob. 59ECh. 5.1 - Prob. 60ECh. 5.1 - Prob. 61ECh. 5.1 - Prob. 62ECh. 5.1 - Prob. 63ECh. 5.1 - Prob. 64ECh. 5.1 - Prob. 65ECh. 5.1 - Prob. 66ECh. 5.1 - Prob. 67ECh. 5.1 - Prob. 68ECh. 5.1 - Prob. 69ECh. 5.1 - Prob. 70ECh. 5.1 - Prob. 71ECh. 5.1 - Prob. 72ECh. 5.1 - Prob. 73ECh. 5.1 - Prob. 74ECh. 5.1 - Prob. 75ECh. 5.1 - Prob. 76ECh. 5.1 - Prob. 77ECh. 5.1 - Prob. 78ECh. 5.1 - Prob. 79ECh. 5.1 - Prob. 80ECh. 5.1 - Prob. 81ECh. 5.1 - Prob. 82ECh. 5.1 - Prob. 83ECh. 5.1 - Prob. 84ECh. 5.1 - Prob. 85ECh. 5.1 - Prob. 86ECh. 5.1 - Prob. 87ECh. 5.1 - Prob. 88ECh. 5.1 - Prob. 89ECh. 5.1 - Prob. 90ECh. 5.1 - Prob. 91ECh. 5.1 - Prob. 92ECh. 5.1 - Prob. 93ECh. 5.1 - Prob. 94ECh. 5.1 - Prob. 95ECh. 5.1 - Prob. 96ECh. 5.1 - Prob. 97ECh. 5.1 - Prob. 98ECh. 5.1 - Prob. 99ECh. 5.1 - Prob. 100ECh. 5.1 - Prob. 101ECh. 5.1 - Prob. 102ECh. 5.1 - Prob. 103ECh. 5.1 - Prob. 104ECh. 5.1 - Prob. 105ECh. 5.1 - Prob. 106ECh. 5.1 - Prob. 107ECh. 5.1 - Prob. 108ECh. 5.1 - Prob. 109ECh. 5.1 - Prob. 110ECh. 5.1 - Prob. 111ECh. 5.1 - Prob. 112ECh. 5.1 - Prob. 113ECh. 5.1 - Prob. 114ECh. 5.1 - Prob. 115ECh. 5.1 - 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Prob. 105ECh. 5.3 - Prob. 106ECh. 5.3 - Prob. 107ECh. 5.3 - Prob. 108ECh. 5.3 - Prob. 109ECh. 5.3 - Prob. 110ECh. 5.3 - Prob. 111ECh. 5.3 - Prob. 112ECh. 5.3 - Prob. 113ECh. 5.3 - Prob. 114ECh. 5.3 - Prob. 115ECh. 5.3 - Prob. 116ECh. 5.3 - Prob. 117ECh. 5.3 - Prob. 118ECh. 5.3 - Prob. 119ECh. 5.4 - Prob. 1ECh. 5.4 - Prob. 2ECh. 5.4 - Prob. 3ECh. 5.4 - Prob. 4ECh. 5.4 - Prob. 5ECh. 5.4 - Prob. 6ECh. 5.4 - Prob. 7ECh. 5.4 - Prob. 8ECh. 5.4 - Prob. 9ECh. 5.4 - Prob. 10ECh. 5.4 - Prob. 11ECh. 5.4 - Prob. 12ECh. 5.4 - Prob. 13ECh. 5.4 - Prob. 14ECh. 5.4 - Prob. 15ECh. 5.4 - Prob. 16ECh. 5.4 - Prob. 17ECh. 5.4 - Prob. 18ECh. 5.4 - Prob. 19ECh. 5.4 - Prob. 20ECh. 5.4 - Prob. 21ECh. 5.4 - Prob. 22ECh. 5.4 - Prob. 23ECh. 5.4 - Prob. 24ECh. 5.4 - Prob. 25ECh. 5.4 - Prob. 26ECh. 5.4 - Prob. 27ECh. 5.4 - Prob. 28ECh. 5.4 - Prob. 29ECh. 5.4 - Prob. 30ECh. 5.4 - Prob. 31ECh. 5.4 - Prob. 32ECh. 5.4 - Prob. 33ECh. 5.4 - Prob. 34ECh. 5.4 - Prob. 35ECh. 5.4 - Prob. 36ECh. 5.4 - Prob. 37ECh. 5.4 - Prob. 38ECh. 5.4 - 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