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 5.1 Fundamental Identities - 5.1 Exercises - Page 201: 65

Answer

$$\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)}=\sin^2\theta\cos^2\theta$$

Work Step by Step

$$A=\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)}$$ $$A=\frac{1-[\cos(-\theta)]^2}{1+[\tan(-\theta)]^2}$$ - Negative-angle Identity: $$\cos(-\theta)=\cos\theta$$ $$\tan(-\theta)=-\tan\theta$$ Therefore, $[\cos(-\theta)]^2=[\cos\theta]^2=\cos^2\theta$ and $[\tan(-\theta)]^2=[-\tan\theta]^2=\tan^2\theta$ (since $[-A]^2=A^2$) Replace into $A$: $$A=\frac{1-\cos^2\theta}{1+\tan^2\theta}$$ - Pythagorean Identity: $$1-\cos^2\theta=\sin^2\theta$$ $$1+\tan^2\theta=\sec^2\theta$$ Replace into $A$: $$A=\frac{\sin^2\theta}{\sec^2\theta}$$ - Reciprocal Identity: $$\sec\theta=\frac{1}{\cos\theta}$$ Then, $$\sec^2\theta=\frac{1}{\cos^2\theta}$$ $$A=\frac{\sin^2\theta}{\frac{1}{\cos^2\theta}}$$ $$A=\sin^2\theta\cos^2\theta$$
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