Formula to calculate 2sin inverse x for different values of x

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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