## Precalculus (6th Edition) Blitzer

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