Statement-1: If $-1 ≤ x ≤ 1,$ then 

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

Statement-2: If $-1 ≤ x ≤ 1,$ then

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

Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1.

Clearly, statement-1 is true.

Putting x = cos $\theta $, we get

$2 sin^{-1} \sqrt{\frac{1-x}{2}}= 2cos^{-1} \sqrt{\frac{1+x}{2}}= \theta = cos^{-1}x$

So, statement-2 is also true.