The value of sin
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) $\frac{1}{3 \sqrt{3}}$

Solution:

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

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

Then,

$\sin y=\frac{\sqrt{63}}{8}$

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

Now, we have

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

$=\sqrt{\frac{1-\cos \frac{y}{2}}{2}}\left[\because \cos 2 x=1-2 \sin ^{2} x\right]$

$=\sqrt{\frac{1-\sqrt{\frac{1+\cos y}{2}}}{2}}\left[\because \cos 2 x=2 \cos ^{2} x-1\right]$

$=\sqrt{\frac{1-\sqrt{\frac{1+\frac{1}{8}}{2}}}{2}}$

$=\sqrt{\frac{1-\sqrt{\frac{9}{16}}}{2}}$

$=\sqrt{\frac{1-\frac{3}{4}}{2}}$

$=\sqrt{\frac{1}{8}}$

$=\frac{1}{2 \sqrt{2}}$

 

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