Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Expressing Trigonometric Products as Sums or Differences
In this exercise, we are tasked with expressing the given trigonometric product as a sum or difference that contains only sines or cosines.
The given expression is:
\[ \sin(9x) \sin(3x) \]
To find the solution, we use product-to-sum identities for trigonometric functions. Specifically, for sine functions, the identity is:
\[ \sin(A)\sin(B) = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
By applying this identity, we can simplify our expression:
\[ \sin(9x) \sin(3x) = \frac{1}{2} [\cos(9x - 3x) - \cos(9x + 3x)] \]
Simplify inside the cosine functions:
\[ \cos(9x - 3x) = \cos(6x) \]
\[ \cos(9x + 3x) = \cos(12x) \]
Therefore, the expression simplifies to:
\[ \sin(9x) \sin(3x) = \frac{1}{2} [\cos(6x) - \cos(12x)] \]
#### Final Answer:
\[ \sin(9x) \sin(3x) = \frac{1}{2} [\cos(6x) - \cos(12x)] \]
Now you can use this simplified expression in further calculations or analyses.
(Simplify your answer.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc87a9968-3e16-4a34-a66a-7e5dcd8bf48b%2F7356b084-561c-4bf2-bca7-1233d35fd26e%2Fx4ohrlv.jpeg&w=3840&q=75)
Transcribed Image Text:### Expressing Trigonometric Products as Sums or Differences
In this exercise, we are tasked with expressing the given trigonometric product as a sum or difference that contains only sines or cosines.
The given expression is:
\[ \sin(9x) \sin(3x) \]
To find the solution, we use product-to-sum identities for trigonometric functions. Specifically, for sine functions, the identity is:
\[ \sin(A)\sin(B) = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
By applying this identity, we can simplify our expression:
\[ \sin(9x) \sin(3x) = \frac{1}{2} [\cos(9x - 3x) - \cos(9x + 3x)] \]
Simplify inside the cosine functions:
\[ \cos(9x - 3x) = \cos(6x) \]
\[ \cos(9x + 3x) = \cos(12x) \]
Therefore, the expression simplifies to:
\[ \sin(9x) \sin(3x) = \frac{1}{2} [\cos(6x) - \cos(12x)] \]
#### Final Answer:
\[ \sin(9x) \sin(3x) = \frac{1}{2} [\cos(6x) - \cos(12x)] \]
Now you can use this simplified expression in further calculations or analyses.
(Simplify your answer.)
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