Question

Solve this problem

${ \orange{ \boxed{ \boxed{ \large{ \purple{sin \left( \frac{1}{4} \sin^{ - 1} \frac{ \sqrt{63} }{8} \right)}}}}}}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: sin \left( \dfrac{1}{4} \sin^{ - 1} \dfrac{ \sqrt{63} }{8} \right)$

Let assume that

$\rm \: {sin}^{ - 1}\dfrac{ \sqrt{63} }{8} = x$

$\rm\implies \:x \: \in \: \bigg(0, \: \dfrac{\pi}{2} \bigg)$

$\rm\implies \:sinx, \: cosx \: > 0$

So, it means, we have to find the value of

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\sf{ \: \: \:sin \frac{x}{4} \: \: \: }}$

Now,

$\rm \: {sin}^{ - 1}\dfrac{ \sqrt{63} }{8} = x$

$\rm \: sinx = \dfrac{ \sqrt{63} }{8}$

$\rm\implies \:\rm \: {sin}^{2} x = \dfrac{ 63 }{64}$

$\rm \: 1 - {cos}^{2}x = \dfrac{63}{64}$

$\rm \: - {cos}^{2}x = \dfrac{63}{64} - 1$

$\rm \: - {cos}^{2}x = \dfrac{63 - 64}{64}$

$\rm \: - {cos}^{2}x \: = \: - \: \dfrac{1}{64}$

$\rm \: {cos}^{2}x \: = \: \: \dfrac{1}{64}$

$\rm\implies \:\rm \: cosx \: = \: \: \dfrac{1}{8}$

We know,

$\boxed{\sf{ \:cos2x = {2cos}^{2}x - 1 \: }}$

So, using this identity, we get

$\rm \: {2cos}^{2}\dfrac{x}{2} - 1 = \dfrac{1}{8}$

$\rm \: {2cos}^{2}\dfrac{x}{2} = \dfrac{1}{8} + 1$

$\rm \: {2cos}^{2}\dfrac{x}{2} = \dfrac{1 + 8}{8}$

$\rm \: {2cos}^{2}\dfrac{x}{2} = \dfrac{9}{8}$

$\rm \: {cos}^{2}\dfrac{x}{2} = \dfrac{9}{16}$

$\rm\implies \:\rm \: {cos}\dfrac{x}{2} = \dfrac{3}{4}$

We know,

$\boxed{\sf{ \:cos2x = 1 - {2sin}^{2}x \: }}$

So, using this identity, we get

$\rm \: 1 - {2sin}^{2}\dfrac{x}{4} = \dfrac{3}{4}$

$\rm \: - {2sin}^{2}\dfrac{x}{4} = \dfrac{3}{4} - 1$

$\rm \: - {2sin}^{2}\dfrac{x}{4} = \dfrac{3 - 4}{4}$

$\rm \: - {2sin}^{2}\dfrac{x}{4} = \: - \: \dfrac{1}{4}$

$\rm \: {2sin}^{2}\dfrac{x}{4} = \: \: \dfrac{1}{4}$

$\rm \: {sin}^{2}\dfrac{x}{4} = \: \: \dfrac{1}{8}$

$\rm \: sin\dfrac{x}{4} = \: \: \dfrac{1}{ \sqrt{8} }$

$\rm\implies \:\rm \: sin\dfrac{x}{4} = \: \: \dfrac{1}{ 2\sqrt{2} }$

Hence,

$\rm\implies \: \:\boxed{\sf{ \: \: \rm \: sin \left( \dfrac{1}{4} \sin^{ - 1} \dfrac{ \sqrt{63} }{8} \right) \: = \: \frac{1}{2 \sqrt{2} } \: }}$

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y =  {sin}^{ - 1}(sinx) & \sf  x \:  \: if -\dfrac{\pi  }{2} \leqslant x \leqslant \dfrac{\pi  }{2}\\ \\ \sf y =  {cos}^{ - 1}(cosx) & \sf x \:  \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y =  {tan}^{ - 1}(tanx) & \sf x \:  \: if \:  - \dfrac{\pi  }{2} < x < \dfrac{\pi  }{2}\\ \\ \sf y =  {cosec}^{ - 1}(cosecx) & \sf x \:  \: if \: x \:  \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi  }{2}\bigg] -  \{0 \}\\ \\ \sf y =  {sec}^{ - 1}(secx) & \sf x \:  \: if \: x \:  \in \: [0, \: \pi] \:   -  \: \bigg\{\dfrac{\pi  }{2}\bigg\}\\ \\ \sf y =  {cot}^{ - 1}(cotx) & \sf x \:  \: if \:  \:  \in \: \bigg( -  \dfrac{\pi  }{2} , \dfrac{\pi  }{2}\bigg) -  \{0 \} \end{array}} \\ \end{gathered}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf   - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf  - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \:  \in \: ( -  \infty , \infty )\\ \\ \sf y = cosecx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \:  \in \: ( -  \infty , \infty ) \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

$\begin{aligned}\\& \\ \\& \tt \color{darkred} Let \sin ^{-1} \frac{\sqrt{63}}{8}=x \qquad\Rightarrow \sin x=\frac{\sqrt{63}}{8} \\ \\& \tt \color{red}\left[\because \sin ^{2} x+\cos ^{2} x=1 \Rightarrow \cos ^{2} x=1-\sin ^{2} x\right. \\ \\&\\ \\& \tt \color{maroon} \cos ^{2} x=1-\frac{63}{64}=\frac{1}{64} \\ \\& \\ \\& \tt \color{orange}\Rightarrow \cos x=\frac{1}{8} \\ \\&\\ \\& \tt \color{navy}\left[\because 1+\cos x=2 \cos ^{2} \frac{x}{2} \Rightarrow \cos ^{2} \frac{x}{2}=\frac{1+\frac{1}{8}}{2}=\frac{9}{16} \Rightarrow \cos \frac{x}{2}=\frac{3}{4}\right] \\ \\&\\ \\& \tt \color{blue}\left[\because 1+\cos \frac{x}{2}=2 \cos ^{2} \frac{x}{4} \Rightarrow \cos ^{2} \frac{x}{4}=\frac{1+\frac{3}{4}}{2}=\frac{7}{8}\right] \\ \\\\\ & & \\ \\& \tt \color{green}Now, \Rightarrow \sin ^{2} \frac{x}{4}+\cos ^{2} \frac{x}{4}=1 \\ \\& \\ \\& \tt \color{violet}\Rightarrow \sin ^{2} \frac{x}{4}=1-\frac{7}{8}=\frac{1}{8} \\ \\&\\ \\& \tt \color{blueviolet} \Rightarrow \sin \frac{x}{4}=\frac{1}{\sqrt{8}} \\ \\& \\ \\& \tt \color{purple}Now, \Rightarrow \sin \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)=\sin \frac{x}{4} \\ \\&\\ \\& \tt \color{darkcyan}=\frac{1}{\sqrt{8}} \\ \\&\\ \\& \tt \color{olive} \Rightarrow \sin \frac{x}{4}=\frac{1}{2 \sqrt{2}} \end{aligned}$