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Verify the identity. sin? a – sin a = cos? a – cos“ a

Verify the identity. sin? a – sin a = cos? a – cos“ a

Calculus: Early Transcendentals
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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### Verifying the Trigonometric Identity The given identity to verify is: \[ \sin^2 \alpha - \sin^4 \alpha = \cos^2 \alpha - \cos^4 \alpha \] **Steps to Verify the Identity:** 1. **Rewrite the Expression on the Left Side:** The left side of the equation is \( \sin^2 \alpha - \sin^4 \alpha \). Factor out the common term \( \sin^2 \alpha \): \[ \sin^2 \alpha - \sin^4 \alpha = \sin^2 \alpha (1 - \sin^2 \alpha) \] 2. **Use the Pythagorean Identity:** Recall the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] Substitute \( \cos^2 \alpha \) for \( 1 - \sin^2 \alpha \): \[ \sin^2 \alpha (1 - \sin^2 \alpha) = \sin^2 \alpha \cos^2 \alpha \] 3. **Rewrite the Expression on the Right Side:** Likewise, the right side of the equation is \( \cos^2 \alpha - \cos^4 \alpha \). Factor out the common term \( \cos^2 \alpha \): \[ \cos^2 \alpha - \cos^4 \alpha = \cos^2 \alpha (1 - \cos^2 \alpha) \] 4. **Use the Pythagorean Identity Again:** Substitute \( \sin^2 \alpha \) for \( 1 - \cos^2 \alpha \): \[ \cos^2 \alpha (1 - \cos^2 \alpha) = \cos^2 \alpha \sin^2 \alpha \] 5. **Compare Both Sides:** Now, we have: \[ \sin^2 \alpha \cos^2 \alpha \quad \text{and} \quad \cos^2 \alpha \sin^2 \alpha \] Since multiplication is commutative, both expressions are equal: \[ \sin^2 \alpha \cos^2 \alpha = \cos^2 \alpha \sin^2 \alpha \] Thus, the identity \(
Transcribed Image Text:### Verifying the Trigonometric Identity The given identity to verify is: \[ \sin^2 \alpha - \sin^4 \alpha = \cos^2 \alpha - \cos^4 \alpha \] **Steps to Verify the Identity:** 1. **Rewrite the Expression on the Left Side:** The left side of the equation is \( \sin^2 \alpha - \sin^4 \alpha \). Factor out the common term \( \sin^2 \alpha \): \[ \sin^2 \alpha - \sin^4 \alpha = \sin^2 \alpha (1 - \sin^2 \alpha) \] 2. **Use the Pythagorean Identity:** Recall the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] Substitute \( \cos^2 \alpha \) for \( 1 - \sin^2 \alpha \): \[ \sin^2 \alpha (1 - \sin^2 \alpha) = \sin^2 \alpha \cos^2 \alpha \] 3. **Rewrite the Expression on the Right Side:** Likewise, the right side of the equation is \( \cos^2 \alpha - \cos^4 \alpha \). Factor out the common term \( \cos^2 \alpha \): \[ \cos^2 \alpha - \cos^4 \alpha = \cos^2 \alpha (1 - \cos^2 \alpha) \] 4. **Use the Pythagorean Identity Again:** Substitute \( \sin^2 \alpha \) for \( 1 - \cos^2 \alpha \): \[ \cos^2 \alpha (1 - \cos^2 \alpha) = \cos^2 \alpha \sin^2 \alpha \] 5. **Compare Both Sides:** Now, we have: \[ \sin^2 \alpha \cos^2 \alpha \quad \text{and} \quad \cos^2 \alpha \sin^2 \alpha \] Since multiplication is commutative, both expressions are equal: \[ \sin^2 \alpha \cos^2 \alpha = \cos^2 \alpha \sin^2 \alpha \] Thus, the identity \(
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