Formulae Boundary conditions
\[2sin^{-1}(x)= sin^{-1}(2x\sqrt{1-x^2}) \] for \[ -\frac{1}{\sqrt{2}}< x<\frac{1}{\sqrt{2}}\]
\[2sin^{-1}(x)= \Pi -sin^{-1}(2x\sqrt{1-x^2}) \] for \[ \frac{1}{\sqrt{2}}\leq x\leq 1\]
\[2sin^{-1}(x)=- \Pi -sin^{-1}(2x\sqrt{1-x^2}) \] for \[ -1\leq x\leq \frac{-1}{\sqrt{2}}\]
Proof
let \[sin^{-1}(x)=\theta \] then x =sin\[\theta \]
\[cos\theta =\sqrt{1-x^2}\]
\[ sin2\theta =2sin\theta cos\theta \]\[ sin2\theta =2x\sqrt{1-x^2}\]
Case 1 when \[ -\frac{1}{\sqrt{2}}< x<\frac{1}{\sqrt{2}}\]
\[ -\frac{1}{\sqrt{2}}< x<\frac{1}{\sqrt{2}}\] implies \[ \frac{-\pi }{4}\leq x\leq \frac{\pi }{4}\]
or \[ \frac{-\pi }{2}\leq 2x\leq \frac{\pi }{2}\]
Also \[ \frac{1 }{\sqrt{2}}\leq x\leq \frac{1 }{\sqrt{2}}\] That implies \[-1\leq 2x\sqrt{1-x^2}\leq 1\]
\[ sin2\theta =2x\sqrt{1-x^2}\]
\[2\theta =2sin^{-1}(2x\sqrt{1-x^2})\]
\[2sin^{-1}(x) =2sin^{-1}(2x\sqrt{1-x^2})\] hence proved
