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 - Concept and Vocabulary Check - Page 679: 5


The required value of the ${{\cos }^{2}}\alpha $ is $\frac{1+\cos \,2\alpha }{2}$.

Work Step by Step

In order to find the value of ${{\cos }^{2}}\alpha $, the power reducing formula is used that is as shown below: $\cos 2\alpha =2\,{{\cos }^{2}}\alpha -1$ The aforementioned formula is proved in the earlier problems. Solve the aforementioned formula on the right side for the ${{\cos }^{2}}\alpha $: $\begin{align} & 2{{\cos }^{2}}\alpha =1+\cos 2\alpha \\ & {{\cos }^{2}}\alpha =\frac{1+\cos 2\alpha }{2} \end{align}$ Hence, the required value of the ${{\cos }^{2}}\alpha $ is $\frac{1+\cos \,2\alpha }{2}$.
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