### Formula to calculate 2cos inverse x for different x

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$cos^{-1}(x)=\theta$ then $x=cos\theta$................................................(1)

Case 1 $0\leqslant x\leq 1$

$0\leqslant x\leq 1$

$0 \leq cos\theta \leq 1$

$0 \leq \theta \leq \frac{\Pi }{2}$

$0 \leq 2\theta \leq \Pi$

Also $0 \leq x \leq 1$

$-1\leq 2x^2-1 \leq 1$

$cos2\theta =2x^2-1$

$2\theta = cos^{-1}(2x^2-1 )$

$2cos^{-1}(x) = cos^{-1}(2x^2-1 )$

Case 2  $-1\leq x\leq 0$

That implies $-1\leq cos\theta \leq 0$              from equation 1

$\frac{\Pi }{2}\leq \theta \le\Pi$

$\Pi \leq 2\theta \le2\Pi$

$-2\Pi \leq -2\theta \le-\Pi$

$0 \leq2\pi -2\theta \le\pi$

Also $-1\leq x\leq 0$

hence $-1\leq 2x^2-1\leq1$

$cos2\theta= 2x^2-1$

$cos(2\pi-2\theta)= 2x^2-1$

$(2\pi-2\theta)= cos^{-1}(2x^2-1)$

$2\pi-2cos^{-1}(x)=cos^{-1}(2x^2-1)$

Therefore $2cos^{-1}(x)=2\pi-cos^{-1}(2x^2-1)$

hence proved