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: 109


The value is $\frac{9}{10}$.

Work Step by Step

Let us assume, $\begin{align} & {{\sin }^{-1}}\left( \frac{3}{5} \right)=x \\ & \sin x=\left( \frac{3}{5} \right) \end{align}$ And the sine is computed by dividing the perpendicular and hypotenuse. So, the perpendicular (y) is 3 and the hypotenuse (r) is 5. Compute the base (x) as follows: $\begin{align} & {{x}^{2}}={{r}^{2}}-{{y}^{2}} \\ & ={{5}^{2}}-{{3}^{2}} \\ & =25-9 \end{align}$ Therefore, the value is $\begin{align} & x=\sqrt{16} \\ & =4 \end{align}$ So, the base is 4. And compute the value as follows: $\begin{align} & {{\cos }^{2}}\left[ \frac{1}{2}{{\sin }^{-1}}\left( \frac{3}{5} \right) \right]={{\cos }^{2}}\frac{1}{2}\theta \\ & =\frac{1+\cos 2\times \frac{1}{2}\theta }{2} \\ & =\frac{1+\cos \theta }{2} \\ & =\frac{1+\frac{4}{5}}{2} \end{align}$ By simplifying the equation, the result will be $\begin{align} & {{\cos }^{2}}\left[ \frac{1}{2}{{\sin }^{-1}}\left( \frac{3}{5} \right) \right]=\frac{\frac{9}{5}}{2} \\ & =\frac{9}{10} \end{align}$ So, ${{\cos }^{2}}\left[ \frac{1}{2}{{\sin }^{-1}}\left( \frac{3}{5} \right) \right]=\frac{9}{10}$.
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