Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.2 Verifying Trigonometric Identities - 5.2 Exercises - Page 209: 68


$$\frac{\sin^4\alpha-\cos^4\alpha}{\sin^2\alpha-\cos^2\alpha}=1$$ The expression is an identity.

Work Step by Step

$$\frac{\sin^4\alpha-\cos^4\alpha}{\sin^2\alpha-\cos^2\alpha}=1$$ The left side is more complicated. We would simplify it. $$A=\frac{\sin^4\alpha-\cos^4\alpha}{\sin^2\alpha-\cos^2\alpha}$$ We have $a^4-b^4=(a^2-b^2)(a^2+b^2)$. So, $$A=\frac{(\sin^2\alpha-\cos^2\alpha)(\sin^2\alpha+\cos^2\alpha)}{\sin^2\alpha-\cos^2\alpha}$$ $$A=\sin^2\alpha+\cos^2\alpha$$ From Pythagorean Identity: $$A=\sin^2\alpha+\cos^2\alpha=1$$ They are thus equal. The expression is an identity.
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