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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![### Section 4.4: Problem 17
**Problem Statement:**
Express the given sum as a product of sines and/or cosines.
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right)
\]
Rewrite the expression using trigonometric identities:
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right) = \boxed{\phantom{answer}}
\]
(Simplify your answer.)
---
**Solution Strategy:**
To simplify the given trigonometric sum, you can use the sum-to-product identities for cosines:
\[
\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)
\]
Where, in our case:
- \( A = \frac{x}{2} \)
- \( B = \frac{7x}{2} \)
Applying the identity, we have:
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right) = 2 \cos\left(\frac{\frac{x}{2} + \frac{7x}{2}}{2}\right) \cos\left(\frac{\frac{x}{2} - \frac{7x}{2}}{2}\right)
\]
Simplify the angles inside the cosines:
\[
A + B = \frac{x}{2} + \frac{7x}{2} = 4x \implies \frac{A + B}{2} = 2x
\]
\[
A - B = \frac{x}{2} - \frac{7x}{2} = -3x \implies \frac{A - B}{2} = -\frac{3x}{2}
\]
Thus, the expression simplifies to:
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right) = 2 \cos(2x) \cos\left(-\frac{3x}{2}\right)
\]
By using the even property of cosine, \(\cos(-\theta) = \cos(\theta)\), we](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc87a9968-3e16-4a34-a66a-7e5dcd8bf48b%2F104f79aa-927a-49bf-9b29-0b28bdedd134%2Frr8z10l.jpeg&w=3840&q=75)
Transcribed Image Text:### Section 4.4: Problem 17
**Problem Statement:**
Express the given sum as a product of sines and/or cosines.
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right)
\]
Rewrite the expression using trigonometric identities:
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right) = \boxed{\phantom{answer}}
\]
(Simplify your answer.)
---
**Solution Strategy:**
To simplify the given trigonometric sum, you can use the sum-to-product identities for cosines:
\[
\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)
\]
Where, in our case:
- \( A = \frac{x}{2} \)
- \( B = \frac{7x}{2} \)
Applying the identity, we have:
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right) = 2 \cos\left(\frac{\frac{x}{2} + \frac{7x}{2}}{2}\right) \cos\left(\frac{\frac{x}{2} - \frac{7x}{2}}{2}\right)
\]
Simplify the angles inside the cosines:
\[
A + B = \frac{x}{2} + \frac{7x}{2} = 4x \implies \frac{A + B}{2} = 2x
\]
\[
A - B = \frac{x}{2} - \frac{7x}{2} = -3x \implies \frac{A - B}{2} = -\frac{3x}{2}
\]
Thus, the expression simplifies to:
\[
\cos\left(\frac{x}{2}\right) + \cos\left(\frac{7x}{2}\right) = 2 \cos(2x) \cos\left(-\frac{3x}{2}\right)
\]
By using the even property of cosine, \(\cos(-\theta) = \cos(\theta)\), we
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