Evaluate

We know that

$\cos ^{-1} x=2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}$

$\tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}$

$\therefore \sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \left(\frac{1}{2} 2 \tan ^{-1} \sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}\right)$

$=\sin \left(\tan ^{-1} \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}}\right)$

$=\sin \left(\tan ^{-1} \frac{1}{3}\right)$

$=\sin \left\{\sin ^{-1}\left(\frac{\frac{1}{3}}{\sqrt{1+\frac{1}{9}}}\right)\right\}$

$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$

$=\frac{1}{\sqrt{10}} \quad\left[\because \sin \left(\sin ^{-1} x\right)=x\right]$

$\therefore \sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\frac{1}{\sqrt{10}}$