Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Let $\frac{{\sin }^{-1}x}{\alpha }=\frac{{\cos }^{-1}x}{\beta }=k$

i.e. ${\sin }^{-1}x=k\alpha$ and ${\cos }^{-1}x=k\beta$

$\Rightarrow {\sin }^{-1}x+{\cos }^{-1}x=k\left(\alpha +\beta \right)$

i.e. $k\left(\alpha +\beta \right)=\frac{\mathrm{\pi }}{2}$

$\Rightarrow k=\frac{\pi }{2\left(\alpha +\beta \right)}$

Now $\frac{2\pi \alpha }{\alpha +\beta }=4k\alpha =4{\sin }^{-1}x$

So $\sin \left(\frac{2\pi \alpha }{\alpha +\beta }\right)=\sin \left(4{\sin }^{-1}x\right)$

$=2\sin \left(2{\sin }^{-1}x\right)\cos \left(2{\sin }^{-1}x\right)$

$=2\sin \left({\sin }^{-1}2x\sqrt{1-x^2}\right)\cos \left({\sin }^{-1}2x\sqrt{1-x^2}\right)$

$=2\sin \left({\sin }^{-1}2x\sqrt{1-x^2}\right)\cos \left({\cos }^{-1}\left(1-2x^2\right)\right)$

$=4x\sqrt{1-x^2}\left(1-2x^2\right)$