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: 7


The required value of the $\sin \frac{x}{2}$ is $\pm \sqrt{\frac{1-\cos x}{2}}$ .

Work Step by Step

In order to find the value of $\sin \frac{x}{2}$, the half angle formula is used that is as given below: ${{\sin }^{2}}\alpha =\frac{1-\cos 2\alpha }{2}$ Now, replace $\alpha $ with $\frac{x}{2}$ $\begin{align} & {{\sin }^{2}}\frac{x}{2}=\frac{1-\cos 2\left( \frac{x}{2} \right)}{2} \\ & =\frac{1-\cos x}{2} \\ & =\pm \sqrt{\frac{1-\cos x}{2}} \end{align}$
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