Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 27


See the explanation below.

Work Step by Step

${{\sin }^{2}}x+\cos 2x={{\cos }^{2}}x$ Consider the left side of the given expression and apply the double angle formula. $\begin{align} & {{\sin }^{2}}x+\cos 2x={{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x \\ & ={{\cos }^{2}}x \end{align}$ Hence, it is proved that the given identity ${{\sin }^{2}}x+\cos 2x={{\cos }^{2}}x$ holds true.
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