Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![### Trigonometric Transformations and Identitites
#### Problem Statement
**Use suitable sum-to-product formulas to rewrite the following expression in terms of \(\tan 3x\):**
\[
\frac{\sin 2x + \sin 4x}{\cos 2x + \cos 4x}
\]
#### Solution
To transform the given expression using sum-to-product identities, follow these steps:
1. **Sum-to-Product Identity for Sine:**
\[
\sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)
\]
Apply the identity to \(\sin 2x + \sin 4x\):
\[
\sin 2x + \sin 4x = 2 \sin \left( \frac{2x + 4x}{2} \right) \cos \left( \frac{2x - 4x}{2} \right)
= 2 \sin(3x) \cos(-x)
\]
Since \(\cos(-x) = \cos x\), we have:
\[
\sin 2x + \sin 4x = 2 \sin(3x) \cos(x)
\]
2. **Sum-to-Product Identity for Cosine:**
\[
\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)
\]
Apply the identity to \(\cos 2x + \cos 4x\):
\[
\cos 2x + \cos 4x = 2 \cos \left( \frac{2x + 4x}{2} \right) \cos \left( \frac{2x - 4x}{2} \right)
= 2 \cos(3x) \cos(-x)
\]
Since \(\cos(-x) = \cos x\), we have:
\[
\cos 2x + \cos 4x = 2 \cos(3x) \cos(x)
\](https://content.bartleby.com/qna-images/question/fd95c06f-a3ea-4120-9338-f910c4caee1b/295b9536-7d31-4efd-812c-8c75ce3fb9b8/c31vm3l_processed.jpeg)
Transcribed Image Text:### Trigonometric Transformations and Identitites
#### Problem Statement
**Use suitable sum-to-product formulas to rewrite the following expression in terms of \(\tan 3x\):**
\[
\frac{\sin 2x + \sin 4x}{\cos 2x + \cos 4x}
\]
#### Solution
To transform the given expression using sum-to-product identities, follow these steps:
1. **Sum-to-Product Identity for Sine:**
\[
\sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)
\]
Apply the identity to \(\sin 2x + \sin 4x\):
\[
\sin 2x + \sin 4x = 2 \sin \left( \frac{2x + 4x}{2} \right) \cos \left( \frac{2x - 4x}{2} \right)
= 2 \sin(3x) \cos(-x)
\]
Since \(\cos(-x) = \cos x\), we have:
\[
\sin 2x + \sin 4x = 2 \sin(3x) \cos(x)
\]
2. **Sum-to-Product Identity for Cosine:**
\[
\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)
\]
Apply the identity to \(\cos 2x + \cos 4x\):
\[
\cos 2x + \cos 4x = 2 \cos \left( \frac{2x + 4x}{2} \right) \cos \left( \frac{2x - 4x}{2} \right)
= 2 \cos(3x) \cos(-x)
\]
Since \(\cos(-x) = \cos x\), we have:
\[
\cos 2x + \cos 4x = 2 \cos(3x) \cos(x)
\
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