Question

If $0 < x <\frac{1}{\sqrt{2}}$ and $\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta}$, then a value of $\sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)$ is

• A. $4 \sqrt{\left(1-x^2\right)}\left(1-2 x^2\right)$
• B. $4 x \sqrt{\left(1-x^2\right)}\left(1-2 x^2\right)$
• C. $2 x \sqrt{\left(1-x^2\right)}\left(1-4 x^2\right)$
• D. $4 \sqrt{\left(1-x^2\right)}\left(1-4 x^2\right)$

Answer 1

Answer: (B)

$ \frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta}= k$
$\sin ^{-1} x=k \alpha$
$ \cos ^{-1} x= k \beta $
$ k =\frac{\pi}{2(\alpha+\beta)} .....$(i)
$ \sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)=\sin \left(4 \sin ^{-1} x\right) $
$ =2 \sin \left(2 \sin ^{-1} x\right) \cos \left(2 \sin ^{-1} x\right) $
$ =4 x \sqrt{1-x^2}\left(1-2 x^2\right)$