#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 28 Maths Textbbok Solution.

Answer: $\frac{1}{2 \sqrt{2}}$

Hint: Try to solve $\sin ^{-1}$ function, then move to sin function.

Given: $\sin \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)$

Solution:

$\text { Let } \sin ^{-1} \frac{\sqrt{63}}{8}=\mathrm{y}$

$\sin y=\left(\frac{\sqrt{63}}{8}\right)$

$\cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\frac{63}{64}}=\frac{1}{8}$

We have, $\sin \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)=\sin \left(\frac{1}{4} y\right)$

$\\ \\ \Rightarrow \hspace{1cm}\cos y=1-2\sin^{2}\frac{y}{2}\\ \\ \Rightarrow \hspace{1cm}\frac{1}{8}=1-2\sin^{2}\frac{y}{2}\Rightarrow \sin^{2}\frac{y}{2}=\frac{7}{16}\\ \\ \Rightarrow \hspace{1cm}\sin^{2}\frac{y}{2}+\cos^{2}\frac{y}{2}=1\Rightarrow \cos^{2}\frac{y}{2}=\frac{1}{2}\\ \\ \Rightarrow \hspace{1cm}\cos \frac{y}{2}=1-2\sin^{2}\frac{y}{4}\\ \\ \Rightarrow \hspace{1cm}\sin^{2}\frac{y}{4}=\frac{\sqrt{2}-1}{2\sqrt{2}}\\ \\ \Rightarrow \hspace{1cm}\sin \frac{y}{4}=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$