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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![**Write the product as a sum:**
\( 18 \sin(38z) \sin(23z) = \)
---
To solve this, we use the product-to-sum identities for sine functions. The identity for the product of sines is:
\[
\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]
\]
Applying this identity to \( \sin(38z) \sin(23z) \):
Let \( A = 38z \) and \( B = 23z \).
\[
\sin(38z) \sin(23z) = \frac{1}{2} [\cos((38z) - (23z)) - \cos((38z) + (23z))]
\]
\[
= \frac{1}{2} [\cos(15z) - \cos(61z)]
\]
Now multiply by 18:
\[
18 \sin(38z) \sin(23z) = 18 \times \frac{1}{2} [\cos(15z) - \cos(61z)]
\]
\[
= 9 [\cos(15z) - \cos(61z)]
\]
The solution is:
\[
9 \cos(15z) - 9 \cos(61z)
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fda8fc3ab-411b-46d9-806c-95f599ffb9ac%2Fbf6d33e0-0dbc-4e8f-ae6e-2797ef36f29c%2Fqoynnep_processed.png&w=3840&q=75)
Transcribed Image Text:**Write the product as a sum:**
\( 18 \sin(38z) \sin(23z) = \)
---
To solve this, we use the product-to-sum identities for sine functions. The identity for the product of sines is:
\[
\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]
\]
Applying this identity to \( \sin(38z) \sin(23z) \):
Let \( A = 38z \) and \( B = 23z \).
\[
\sin(38z) \sin(23z) = \frac{1}{2} [\cos((38z) - (23z)) - \cos((38z) + (23z))]
\]
\[
= \frac{1}{2} [\cos(15z) - \cos(61z)]
\]
Now multiply by 18:
\[
18 \sin(38z) \sin(23z) = 18 \times \frac{1}{2} [\cos(15z) - \cos(61z)]
\]
\[
= 9 [\cos(15z) - \cos(61z)]
\]
The solution is:
\[
9 \cos(15z) - 9 \cos(61z)
\]
Expert Solution
Step 1
The given expression is:
...(1)
Note that the formula used is:
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