Question

In Problems 79-90, find the exact value of each expression.

${\sin }^2\left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$

Short answer

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

Step-by-step solution

Given information

Given expression${\sin }^2\left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)$

Rewriting the inverse function

Rewriting, we get

$$\begin{gathered} \alpha ={\cos }^{-1}\frac{3}{5} \\ \cos \alpha =\frac{3}{5} \end{gathered}$$

$$\begin{gathered} {\sin }^2\left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)={\sin }^2\frac{\alpha }{2} \\ {\sin }^2\left(\frac{1}{2}{\cos }^{-1}\frac{3}{5}\right)={\left(\sin \frac{\alpha }{2}\right)}^2 \end{gathered}$$

Using half angle formula and calculate

Using half angle, we get

$$\begin{gathered} \sin \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{2}} \\ \sin \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{2}} \\ \sin \frac{\alpha }{2}=\sqrt{\frac{1-\frac{3}{5}}{2}} \\ \sin \frac{\alpha }{2}=\sqrt{\frac{2}{10}} \\ \sin \frac{\alpha }{2}=\sqrt{\frac{1}{5}} \\ {\sin }^2\frac{\alpha }{2}=\frac{1}{5} \end{gathered}$$