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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![**Verify the Identity**
\( 6 \cos^2(x) - 3 = 3 - 6 \sin^2(x) \)
**Step-by-step Verification:**
1. Start with the left side of the equation:
\[
6 \cos^2(x) - 3
\]
2. Factor out the 6:
\[
= 6 \left(1 - (\text{blank space})\right) - 3
\]
3. Replace the blank space using the Pythagorean identity \(\sin^2(x) = 1 - \cos^2(x)\):
\[
= 6 - (\text{blank space}) - 3
\]
4. Substitute \(\sin^2(x)\) into the equation:
\[
= 3 - 6 \sin^2(x)
\]
This demonstrates the original identity is correct:
\(6 \cos^2(x) - 3 = 3 - 6 \sin^2(x)\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcce4fb07-e12e-416f-8942-a1a17d96b17d%2F0dda78c4-5bfb-44ba-b7d0-e5c04219ec0f%2F4qyvra_processed.png&w=3840&q=75)
Transcribed Image Text:**Verify the Identity**
\( 6 \cos^2(x) - 3 = 3 - 6 \sin^2(x) \)
**Step-by-step Verification:**
1. Start with the left side of the equation:
\[
6 \cos^2(x) - 3
\]
2. Factor out the 6:
\[
= 6 \left(1 - (\text{blank space})\right) - 3
\]
3. Replace the blank space using the Pythagorean identity \(\sin^2(x) = 1 - \cos^2(x)\):
\[
= 6 - (\text{blank space}) - 3
\]
4. Substitute \(\sin^2(x)\) into the equation:
\[
= 3 - 6 \sin^2(x)
\]
This demonstrates the original identity is correct:
\(6 \cos^2(x) - 3 = 3 - 6 \sin^2(x)\)
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