Question

$\sin \left[\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{4}{5}\right)\right]=$

• A. $\frac{1}{\sqrt{10}}$
• B. $-\frac{1}{\sqrt{10}}$
• C. $\frac{1}{10}$
• D. $-\frac{1}{10}$

Answer 1

The correct option is A

$\frac{1}{\sqrt{10}}$

Explanation for the correct answer:

Simplifying the given expression:

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

Let ${\cos }^{-1}\left(\frac{4}{5}\right)=x...\left(i\right)$

$\therefore \cos x=\frac{4}{5}$

Substituting the value,

$$\begin{gathered} \Rightarrow \sin \left[\left(\frac{1}{2}\right){\cos }^{-1}\left(\frac{4}{5}\right)\right] \\ \Rightarrow \sin \left[\left(\frac{1}{2}\right)x\right]\left[\mathrm{From}\left(i\right)\right] \\ \Rightarrow \sin \left(\frac{x}{2}\right)...\left(ii\right) \end{gathered}$$

From $\left(i\right)$, $\cos x=\frac{4}{5}$

$$\begin{gathered} \Rightarrow 1-2{\sin }^2\left(\frac{x}{2}\right)=\frac{4}{5}\left[\because \cos x=1-2{\sin }^2\left(\frac{x}{2}\right)\right] \\ \Rightarrow 2{\sin }^2\left(\frac{x}{2}\right)=1-\frac{4}{5} \\ \Rightarrow 2{\sin }^2\left(\frac{x}{2}\right)=\frac{1}{5} \\ \Rightarrow {\sin }^2\left(\frac{x}{2}\right)=\frac{1}{10} \\ \text{Applying square root on both the sides} \\ \Rightarrow \sin \left(\frac{x}{2}\right)=\sqrt{\frac{1}{10}}=\frac{1}{\sqrt{10}} \end{gathered}$$

Hence, option (A) is correct.