Question

Find the value of: $\sin \left(\frac{1}{4}{\sin }^{-1}\frac{\sqrt{63}}{8}\right)$

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{3\sqrt{3}}$

Answer 1

The correct option is C $\frac{1}{2\sqrt{2}}$

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

Let ${\sin }^{-1}\frac{\sqrt{63}}{8}=x$

$\Rightarrow \sin x=\frac{\sqrt{63}}{8}$

$\Rightarrow {\cos }^{2}x=1-{\sin }^{2}x=1-{\left(\frac{\sqrt{63}}{8}\right)}^2=\frac{64-63}{64}=\frac{1}{64}$

$\Rightarrow \cos x=\frac{1}{8}$

$\sin \left(\frac{x}{4}\right)$ where $x={\sin }^{-1}\frac{\sqrt{63}}{8}$

We know that

$1+\cos 2\theta =2{\cos }^2\theta$

$\Rightarrow 1+\cos x=2{\cos }^2\frac{x}{2}$

$\Rightarrow 1+\frac{1}{8}=2{\cos }^2\frac{x}{2}$ $\because \cos x=\frac{1}{8}$

$\Rightarrow \frac{9}{8}=2{\cos }^2\frac{x}{2}$

$\Rightarrow {\cos }^2\frac{x}{2}=\frac{9}{16}$

$\Rightarrow \cos \frac{x}{2}=\frac{3}{4}$

$\Rightarrow 2{\cos }^2\frac{x}{4}=1+\cos \frac{x}{2}=1+\frac{3}{4}=\frac{4+3}{4}=\frac{7}{4}$

$\Rightarrow {\cos }^2\frac{x}{4}=\frac{7}{8}$

$\Rightarrow {\sin }^2\frac{x}{4}=1-{\cos }^2\frac{x}{4}=1-\frac{7}{8}=\frac{8-7}{8}$

$\Rightarrow \sin \frac{x}{4}=\sqrt{\frac{1}{8}}=\frac{1}{2\sqrt{2}}$

$\therefore \sin \left(\frac{1}{4}{\sin }^{-1}\frac{\sqrt{63}}{8}\right)=\sin \left(\frac{x}{4}\right)=\frac{1}{2\sqrt{2}}$