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 683: 110


The value is $\frac{1}{5}$.

Work Step by Step

We have to compute the value in the following manner: Now, let us consider, $\begin{align} & \theta ={{\cos }^{-1}}\frac{3}{5} \\ & \cos \theta =\frac{3}{5} \end{align}$ Then, solve the equation: $\begin{align} & {{\sin }^{2}}\frac{1}{2}{{\cos }^{-1}}\frac{3}{5}={{\sin }^{2}}\frac{1}{2}\theta \\ & =\frac{1-\cos 2\times \frac{1}{2}\theta }{2} \\ & =\frac{1-\cos \theta }{2} \end{align}$ By further simplify the equation: $\begin{align} & {{\sin }^{2}}\frac{1}{2}{{\cos }^{-1}}\frac{3}{5}=\frac{1-\frac{3}{5}}{2} \\ & =\frac{\frac{2}{5}}{2} \\ & =\frac{1}{5} \end{align}$ The value is $\frac{1}{5}$.
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