Use suitable sum-to-product formulas to rewrite the following expression in terms of tan3x. sin 2x + sin 4x cos 2x + cos 4x Use the paperclip button below to attach files.

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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)
   \
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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