Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Question:**

Write the product as a sum:

\[ 18 \cos(35u) \cos(12u) = \]

---

**Solution:**

To express the product as a sum, use the trigonometric identity for the product-to-sum formula for cosine:

\[
\cos A \cos B = \frac{1}{2} [\cos (A + B) + \cos (A - B)]
\]

Given:

\[
A = 35u, \quad B = 12u
\]

Substitute into the formula:

\[
\cos(35u) \cos(12u) = \frac{1}{2} [\cos(35u + 12u) + \cos(35u - 12u)]
\]

\[
= \frac{1}{2} [\cos(47u) + \cos(23u)]
\]

Therefore:

\[
18 \cos(35u) \cos(12u) = 18 \times \frac{1}{2} [\cos(47u) + \cos(23u)]
\]

\[
= 9[\cos(47u) + \cos(23u)]
\]

Final result:

\[
18 \cos(35u) \cos(12u) = 9 \cos(47u) + 9 \cos(23u)
\]
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Transcribed Image Text:**Question:** Write the product as a sum: \[ 18 \cos(35u) \cos(12u) = \] --- **Solution:** To express the product as a sum, use the trigonometric identity for the product-to-sum formula for cosine: \[ \cos A \cos B = \frac{1}{2} [\cos (A + B) + \cos (A - B)] \] Given: \[ A = 35u, \quad B = 12u \] Substitute into the formula: \[ \cos(35u) \cos(12u) = \frac{1}{2} [\cos(35u + 12u) + \cos(35u - 12u)] \] \[ = \frac{1}{2} [\cos(47u) + \cos(23u)] \] Therefore: \[ 18 \cos(35u) \cos(12u) = 18 \times \frac{1}{2} [\cos(47u) + \cos(23u)] \] \[ = 9[\cos(47u) + \cos(23u)] \] Final result: \[ 18 \cos(35u) \cos(12u) = 9 \cos(47u) + 9 \cos(23u) \]
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