The value of (1 4 ^ -1 63 8 ) is: - Vidyadip

MCQ

The value of $\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$ is:

A
$\frac{1}{\sqrt2}$
B
$\frac{1}{\sqrt3}$
✓
$\frac{1}{2\sqrt2}$
D
$\frac{1}{3\sqrt3}$

✓

Answer

Correct option: C.

$\frac{1}{2\sqrt2}$

$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
Let, $\sin^{-1}\frac{\sqrt{63}}{8}=\text{x}$
$\sin\text{x}=\frac{\sqrt{63}}{8}$
$\cos\text{x}\sqrt{1-\sin^2\text{x}}$
$\cos\text{x}=\sqrt{1-\frac{63}{64}}$
$\cos\text{x}=\frac{1}{8}$
Consider,
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
$=\sin\Big(\frac{1}{4}\text{x}\Big)$
$=\sqrt{\frac{1-\cos\frac{\text{x}}{2}}{2}}$ $\Big(\because \sin\text{x}=\frac{1-\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{\frac{1+\cos\text{x}}{2}}}{2}}$ $\Big(\because \cos\text{x}=\frac{1+\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{1-\frac{1}{8}}}{2}}$
$=\sqrt{\frac{1-\frac{3}{4}}{2}}$
$=\sqrt{\frac{1}{8}}$
$=\frac{1}{2\sqrt2}$

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