Question

Evaluate sin $\left(\frac{1}{2}{\cos }^{-1}\frac{4}{5}\right)$.

Answer 1

We know that
$$\begin{gathered} {\cos }^{-1}x=2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}} \\ {\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}} \end{gathered}$$

$$\begin{gathered} \therefore \sin \left(\frac{1}{2}{\cos }^{-1}\frac{4}{5}\right)=\sin \left(\frac{1}{2}2{\tan }^{-1}\sqrt{\frac{1-{\displaystyle \frac{4}{5}}}{1+\frac{4}{5}}}\right) \\ =\sin \left({\tan }^{-1}\sqrt{\frac{{\displaystyle \frac{1}{5}}}{\frac{9}{5}}}\right) \\ =\sin \left({\tan }^{-1}\frac{1}{3}\right) \\ =\sin \left\{{\sin }^{-1}\left(\frac{\frac{1}{3}}{\sqrt{1+\frac{1}{9}}}\right)\right\} \\ =\sin \left({\sin }^{-1}\frac{1}{\sqrt{10}}\right) \\ =\frac{1}{\sqrt{10}}\left[\because \sin \left({\sin }^{-1}x\right)=x\right] \end{gathered}$$

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