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 5.4, Problem 89E

(a)

To determine

To represent:Thetrigonometric expression 12sin3θ+5cos3θ in the form of a2+b2sin(Bθ+C) and verify using graph.

(a)

Expert Solution
Check Mark

Answer to Problem 89E

The trigonometric expression 12sin3θ+5cos3θ can be represented as 13sin(3θ+0.3948) .

Explanation of Solution

Given information:

The given equationis 12sin3θ+5cos3θ .

Formula Used:

  a2+b2=r2

Proof:

Compare the expression 12sin3θ+5cos3θ with a2+b2sin(Bθ+C) .

  a=12,b=5,B=3

Calculate the value of a2+b2 .

  a2+b2=122+52=169=13

Calculate the value of C .

  C=tan−1(ba)=tan−1(512)=0.39479≈0.3948

Consider the left hand side of the equation.

  12sin3θ+5cos3θ=a2+b2sin(Bθ+C)=13sin(3θ+0.3948)

Draw the graph for 13sin(3θ+0.3948) and 12sin3θ+5cos3θ using graphical utility.

  Precalculus with Limits: A Graphing Approach, Chapter 5.4, Problem 89E , additional homework tip 1

Figure (1)

From the above, it is clear that both the graph coincides. Therefore, the trigonometric expression 12sin3θ+5cos3θ can be represented as 13sin(3θ+0.3948) .

(b)

To determine

To represent:The trigonometric expression 12sin3θ+5cos3θ in the form of a2+b2cos(Bθ−C) and verify using graph.

(b)

Expert Solution
Check Mark

Answer to Problem 89E

The trigonometric expression 12sin3θ+5cos3θ can be represented as 13cos(3θ−1.1760) .

Explanation of Solution

Given information:

The given equationis 12sin3θ+5cos3θ .

Formula Used:

  a2+b2=r2

Proof:

Compare the expression 12sin3θ+5cos3θ with a2+b2cos(Bθ−C) .

  a=1,b=1,B=1

Calculate the value of a2+b2 .

  a2+b2=12+12=2

Calculate the value of C .

  C=tan−1(ab)=tan−1(125)=1.17601≈1.1760

Consider the left hand side of the equation.

  12sin3θ+5cos3θ=a2+b2cos(Bθ−C)=13cos(3θ−1.1760)

Draw the graph for 13cos(3θ−1.1760) and 12sin3θ+5cos3θ using graphical utility.

  Precalculus with Limits: A Graphing Approach, Chapter 5.4, Problem 89E , additional homework tip 2

Figure (2)

From the above, it is clear that both the graph coincides. Therefore, the trigonometric expression 12sin3θ+5cos3θ can be represented as 13cos(3θ−1.1760) .

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Chapter 5.4, Problem 88E
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Chapter 5.4, Problem 90E

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
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Prob. 67RECh. 5 - Prob. 68RECh. 5 - Prob. 69RECh. 5 - Prob. 70RECh. 5 - Prob. 71RECh. 5 - Prob. 72RECh. 5 - Prob. 73RECh. 5 - Prob. 74RECh. 5 - Prob. 75RECh. 5 - Prob. 76RECh. 5 - Prob. 77RECh. 5 - Prob. 78RECh. 5 - Prob. 79RECh. 5 - Prob. 80RECh. 5 - Prob. 81RECh. 5 - Prob. 82RECh. 5 - Prob. 83RECh. 5 - Prob. 84RECh. 5 - Prob. 85RECh. 5 - Prob. 86RECh. 5 - Prob. 87RECh. 5 - Prob. 88RECh. 5 - Prob. 89RECh. 5 - Prob. 90RECh. 5 - Prob. 91RECh. 5 - Prob. 92RECh. 5 - Prob. 93RECh. 5 - Prob. 94RECh. 5 - Prob. 95RECh. 5 - Prob. 96RECh. 5 - Prob. 97RECh. 5 - Prob. 98RECh. 5 - Prob. 99RECh. 5 - Prob. 100RECh. 5 - Prob. 101RECh. 5 - Prob. 102RECh. 5 - Prob. 103RECh. 5 - Prob. 104RECh. 5 - Prob. 105RECh. 5 - Prob. 106RECh. 5 - Prob. 107RECh. 5 - Prob. 108RECh. 5 - Prob. 109RECh. 5 - Prob. 110RECh. 5 - Prob. 111RECh. 5 - Prob. 112RECh. 5 - Prob. 113RECh. 5 - Prob. 114RECh. 5 - Prob. 115RECh. 5 - Prob. 116RECh. 5 - Prob. 117RECh. 5 - Prob. 118RECh. 5 - Prob. 119RECh. 5 - Prob. 120RECh. 5 - Prob. 121RECh. 5 - Prob. 122RECh. 5 - Prob. 123RECh. 5 - Prob. 124RECh. 5 - Prob. 125RECh. 5 - Prob. 126RECh. 5 - Prob. 127RECh. 5 - Prob. 128RECh. 5 - Prob. 129RECh. 5 - Prob. 130RECh. 5 - Prob. 131RECh. 5 - Prob. 132RECh. 5 - Prob. 133RECh. 5 - Prob. 134RECh. 5 - Prob. 135RECh. 5 - Prob. 136RECh. 5 - Prob. 137RECh. 5 - Prob. 138RECh. 5 - Prob. 139RECh. 5 - Prob. 140RECh. 5 - Prob. 1CTCh. 5 - Prob. 2CTCh. 5 - Prob. 3CTCh. 5 - Prob. 4CTCh. 5 - Prob. 5CTCh. 5 - Prob. 6CTCh. 5 - Prob. 7CTCh. 5 - Prob. 8CTCh. 5 - Prob. 9CTCh. 5 - Prob. 10CTCh. 5 - Prob. 11CTCh. 5 - Prob. 12CTCh. 5 - Prob. 13CTCh. 5 - Prob. 14CTCh. 5 - Prob. 15CTCh. 5 - Prob. 16CTCh. 5 - Prob. 17CTCh. 5 - Prob. 18CTCh. 5 - Prob. 19CTCh. 5 - Prob. 20CTCh. 5 - Prob. 21CTCh. 5 - Prob. 22CTCh. 5 - Prob. 23CTCh. 5 - Prob. 24CT
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Fundamental Trigonometric Identities: Reciprocal, Quotient, and Pythagorean Identities; Author: Mathispower4u;https://www.youtube.com/watch?v=OmJ5fxyXrfg;License: Standard YouTube License, CC-BY