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

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

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