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

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

The Value of Sin ( 1 4 Sin − 1 √ 63 8 ) is (A) 1 √ 2 (B) 1 √ 3 (C) 1 2 √ 2 (D) 1 3 √ 3 - Mathematics

Question

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

Options

$\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{3}}$
$\frac{1}{2 \sqrt{2}}$
$\frac{1}{3 \sqrt{3}}$

MCQ

Solution

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

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Inverse Trigonometric Functions (Simplification and Examples)

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Chapter 4: Inverse Trigonometric Functions - Exercise 4.16 [Page 121]

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