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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Question
![**Problem Statement:**
Write the sum as a product:
\[
\cos(10.7q) - \cos(5.7q) =
\]
**Explanation:**
This expression uses a trigonometric identity that allows the difference of two cosines to be written as a product. The formula used is:
\[
\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
\]
In this case:
- \( A = 10.7q \)
- \( B = 5.7q \)
To apply the identity, calculate:
1. The average: \(\frac{A+B}{2} = \frac{10.7q + 5.7q}{2} = 8.2q\)
2. The difference: \(\frac{A-B}{2} = \frac{10.7q - 5.7q}{2} = 2.5q\)
Thus, the expression can be rewritten as:
\[
-2 \sin(8.2q) \sin(2.5q)
\]
Students can use this identity to simplify expressions and solve trigonometric problems more efficiently.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c16c5d2-e997-4e04-a749-57ee722e7387%2Fe2f1efab-4641-49ce-bfc8-ee77ae133e4f%2Ftcnhtae_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Write the sum as a product:
\[
\cos(10.7q) - \cos(5.7q) =
\]
**Explanation:**
This expression uses a trigonometric identity that allows the difference of two cosines to be written as a product. The formula used is:
\[
\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
\]
In this case:
- \( A = 10.7q \)
- \( B = 5.7q \)
To apply the identity, calculate:
1. The average: \(\frac{A+B}{2} = \frac{10.7q + 5.7q}{2} = 8.2q\)
2. The difference: \(\frac{A-B}{2} = \frac{10.7q - 5.7q}{2} = 2.5q\)
Thus, the expression can be rewritten as:
\[
-2 \sin(8.2q) \sin(2.5q)
\]
Students can use this identity to simplify expressions and solve trigonometric problems more efficiently.
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