Write the product as a sum: 20 sin(23u)sin(9v)
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![### Converting a Product to a Sum
The following task involves rewriting a trigonometric product as a sum using trigonometric identities.
**Problem Statement:**
Write the product as a sum:
\[ 20 \sin(32x) \sin(9x) = \]
**Explanation:**
To express the product of sine functions as a sum, we can use the product-to-sum identities in trigonometry:
\[ \sin(A)\sin(B) = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \]
For our problem:
- \( A = 32x \)
- \( B = 9x \)
Applying the identity:
\[ \sin(32x)\sin(9x) = \frac{1}{2} [\cos(32x - 9x) - \cos(32x + 9x)] \]
\[ = \frac{1}{2} [\cos(23x) - \cos(41x)] \]
Now, multiplying by 20:
\[ 20 \sin(32x) \sin(9x) = 20 \cdot \frac{1}{2} [\cos(23x) - \cos(41x)] \]
\[ = 10 [\cos(23x) - \cos(41x)] \]
Therefore:
\[ 20 \sin(32x) \sin(9x) = 10 \cos(23x) - 10 \cos(41x) \]
**Final Answer:**
\[ 20 \sin(32x) \sin(9x) = 10 \cos(23x) - 10 \cos(41x) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cdbc022-1ecb-430f-8c37-0abecd7b3ab0%2F18e64715-289e-4db4-bb0b-e28c4534030b%2F0tay6o_processed.png&w=3840&q=75)
Transcribed Image Text:### Converting a Product to a Sum
The following task involves rewriting a trigonometric product as a sum using trigonometric identities.
**Problem Statement:**
Write the product as a sum:
\[ 20 \sin(32x) \sin(9x) = \]
**Explanation:**
To express the product of sine functions as a sum, we can use the product-to-sum identities in trigonometry:
\[ \sin(A)\sin(B) = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \]
For our problem:
- \( A = 32x \)
- \( B = 9x \)
Applying the identity:
\[ \sin(32x)\sin(9x) = \frac{1}{2} [\cos(32x - 9x) - \cos(32x + 9x)] \]
\[ = \frac{1}{2} [\cos(23x) - \cos(41x)] \]
Now, multiplying by 20:
\[ 20 \sin(32x) \sin(9x) = 20 \cdot \frac{1}{2} [\cos(23x) - \cos(41x)] \]
\[ = 10 [\cos(23x) - \cos(41x)] \]
Therefore:
\[ 20 \sin(32x) \sin(9x) = 10 \cos(23x) - 10 \cos(41x) \]
**Final Answer:**
\[ 20 \sin(32x) \sin(9x) = 10 \cos(23x) - 10 \cos(41x) \]
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