Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises: 80

Answer

$\sec 2x=\frac{\sec^2 x+\sec^4 x}{2+\sec^2 x-\sec^4 x}$

Work Step by Step

Start with the right side: $\frac{\sec^2 x+\sec^4 x}{2+\sec^2 x-\sec^4 x}$ Multiply top and bottom by $\cos^4 x$: $=\frac{\sec^2 x+\sec^4 x}{2+\sec^2 x-\sec^4 x}*\frac{\cos^4 x}{\cos^4 x}$ $\frac{\sec^2 x\cos^4 x+\sec^4 x\cos^4 x}{2\cos^4 x+\sec^2 x\cos^4 x-\sec^4 x\cos^4 x}$ $=\frac{\cos^2 x+1}{2\cos^4 x+\cos^2 x-1}$ Factor the denominator and simplify: $=\frac{\cos^2 x+1}{(2\cos^2 x-1)(\cos^2 x+1)}$ $=\frac{1}{2\cos^2 x-1}$ Use the identity $2\cos^2 x-1=\cos 2x$: $=\frac{1}{\cos 2x}$ $=\sec 2x$ Since this equals the left side, the identity has been proven.
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