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Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 28 Maths Textbbok Solution.

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Answer: \frac{1}{2 \sqrt{2}}

Hint: Try to solve \sin ^{-1} function, then move to sin function.

Given: \sin \left(\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8}\right)


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

\sin y=\left(\frac{\sqrt{63}}{8}\right)

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

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

                     \\ \\ \Rightarrow \hspace{1cm}\cos y=1-2\sin^{2}\frac{y}{2}\\ \\ \Rightarrow \hspace{1cm}\frac{1}{8}=1-2\sin^{2}\frac{y}{2}\Rightarrow \sin^{2}\frac{y}{2}=\frac{7}{16}\\ \\ \Rightarrow \hspace{1cm}\sin^{2}\frac{y}{2}+\cos^{2}\frac{y}{2}=1\Rightarrow \cos^{2}\frac{y}{2}=\frac{1}{2}\\ \\ \Rightarrow \hspace{1cm}\cos \frac{y}{2}=1-2\sin^{2}\frac{y}{4}\\ \\ \Rightarrow \hspace{1cm}\sin^{2}\frac{y}{4}=\frac{\sqrt{2}-1}{2\sqrt{2}}\\ \\ \Rightarrow \hspace{1cm}\sin \frac{y}{4}=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}                                   




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