Question

The value of sin$\left(\frac{1}{4}{\sin }^{-1}\frac{\sqrt{63}}{8}\right)$ is
(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{1}{\sqrt{3}}$

(c) $\frac{1}{2\sqrt{2}}$

(d) $\frac{1}{3\sqrt{3}}$

Answer 1

(c) $\frac{1}{2\sqrt{2}}$

Let ${\sin }^{-1}\frac{\sqrt{63}}{8}=y$
Then,
$$\begin{gathered} \sin y=\frac{\sqrt{63}}{8} \\ \cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-\frac{63}{64}}=\frac{1}{8} \end{gathered}$$
Now, we have

$$\begin{gathered} \sin \left(\frac{1}{4}{\sin }^{-1}\frac{\sqrt{63}}{8}\right)=\sin \left(\frac{1}{4}y\right) \\ =\sqrt{\frac{1-\cos {\displaystyle \frac{y}{2}}}{2}}\left[\because \cos 2x=1-2{\sin }^{2}x\right] \\ =\sqrt{\frac{1-\sqrt{\frac{1+\cos y}{2}}}{2}}\left[\because \cos 2x=2{\cos }^{2}x-1\right] \\ =\sqrt{\frac{1-\sqrt{\frac{1+{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{2}}}{2}} \\ =\sqrt{\frac{1-\sqrt{\frac{9}{16}}}{2}} \\ =\sqrt{\frac{1-{\displaystyle \frac{3}{4}}}{2}} \\ =\sqrt{\frac{1}{8}} \\ =\frac{1}{2\sqrt{2}} \end{gathered}$$