## Precalculus (6th Edition) Blitzer

The value is $\frac{9}{10}$.
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, ${{\cos }^{2}}\left[ \frac{1}{2}{{\sin }^{-1}}\left( \frac{3}{5} \right) \right]=\frac{9}{10}$.