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

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

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.
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) .
Chapter 5 Solutions
Precalculus with Limits: A Graphing Approach
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