Question

Show that ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2{\sin }^{-1}x$ for $\frac{-1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}$.

Answer 1

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

Let's take derivative of LHS w.r.t $x$ i.e.

$\frac{d}{dx}{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

$=\frac{1}{\sqrt{1-(2x\sqrt{1-x^2}{)}^2}}\times d\left(2x\sqrt{1-x^2}\right)$

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

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

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

$=\frac{2(1-2x^2)}{(1-2x^2)\sqrt{1-x^2}}$

$=\frac{2}{\sqrt{1-x^2}}$

$⟹\frac{d}{dx}{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=\frac{2}{\sqrt{1-x^2}}$

$⟹\int \frac{d}{dx}{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=\int \frac{2}{\sqrt{1-x^2}}$

$⟹\int d\left({\sin }^{-1}\right(2x\sqrt{1-x^2}\left)\right)=\int \frac{2}{\sqrt{1-x^2}}dx$

$⟹{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2{\sin }^{-1}x$ for $\frac{-1}{\sqrt{2}}\le x\le \frac{1}{\sqrt{2}}$