Question

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

 

Options

• `1/sqrt2`

• `1/sqrt3`

• `1/(2sqrt2)`

• `1/(3sqrt3)`

Answer 1

(c) `1/(2sqrt2)`

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
\[\sin\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right) = \sin\left( \frac{1}{4}y \right)\]
\[ = \sqrt{\frac{1 - \cos\frac{y}{2}}{2}} \left[ \because \cos2x = 1 - 2 \sin^2 x \right]\]
\[ = \sqrt{\frac{1 - \sqrt{\frac{1 + \cos{y}}{2}}}{2}} \left[ \because \cos2x = 2 \cos^2 x - 1 \right]\]
\[ = \sqrt{\frac{1 - \sqrt{\frac{1 + \frac{1}{8}}{2}}}{2}}\]
\[ = \sqrt{\frac{1 - \sqrt{\frac{9}{16}}}{2}}\]
\[ = \sqrt{\frac{1 - \frac{3}{4}}{2}}\]
\[ = \sqrt{\frac{1}{8}}\]
\[ = \frac{1}{2\sqrt{2}}\]