Statement-1: If $\frac{1}{2} ≤ x ≤ 1.$ then 

$cos^{-1} x - sin^{-1} \begin{Bmatrix}\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}\end{Bmatrix}$ is equal to $-\frac{\pi}{6}$

Statement-2: $sin^{-1}(2x\sqrt{1-x^2})= 2sin^{-1} x , $ if $x ∈ [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

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

Let $x = cos \theta .$ Then,

$\frac{1}{2}≤ x ≤ \frac{1}{2}⇒ \frac{1}{2}≤cos \theta ≤ 1  ⇒  0 ≤ \theta ≤ \frac{\pi}{3}$

$∴ cos^{-1} x - sin^{-1} \begin{Bmatrix} \frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}\end{Bmatrix}$

$= \theta  - sin^{-1} \begin{Bmatrix} \frac{1}{2}cos \theta +\frac{\sqrt{3}}{2}sin \theta \end{Bmatrix}$

$= \theta  - sin^{-1} \begin{Bmatrix}sin\frac{\pi}{6} cos \theta + cos \frac{\pi}{6} sin \theta \end{Bmatrix}$

$= \theta  - sin^{-1} \begin{Bmatrix} sin \left(\theta +\frac{\pi}{6}\right) \end{Bmatrix}$

$= \theta - \left(\theta +\frac{\pi}{6}\right)=-\frac{\pi}{6} $   $\left[ ∵0 ≤ \theta ≤ \frac{\pi}{3}⇒ \frac{\pi}{6} ≤ \theta  + \frac{\pi}{6} ≤ \frac{\pi}{2}\right]$

So, statement -1 is true.

Let x = $sin \theta $. Then,

$-\frac{1}{\sqrt{2}}≤ x ≤ \frac{1}{\sqrt{2}}⇒ -\frac{1}{\sqrt{2}}≤ sin \theta ≤\frac{1}{\sqrt{2}}⇒ - \frac{\pi}{4} ≤ \theta ≤ \frac{\pi}{4}$

$∴ sin^{-1}(2x \sqrt{1-x^2})= sin^{-1} sin(2\theta) = 2\theta = 2 sin^{-1} x $

So, statement-2 is true.