Simplifying $\cos(2\arcsin(x))$ using only pythagorean trigonometric identity

$\cos(2\arcsin x)$ will be $\ge0,$ if

$-\dfrac\pi2\le2\arcsin x\le\dfrac\pi2$

$\iff -\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$

In that case $$|2x^2-1|=-(2x^2-1)$$

Check if $x>\dfrac1{\sqrt2}$

or $x<-\dfrac1{\sqrt2}$

At this point, we know $$ \cos(2\sin^{-1}x) = \pm(2x^2-1). $$ Since both these functions are continuous, we can just check one value in each region where $2x^2-1 > 0$ and $2x^2-1 < 0$ to verify the sign to use in that region. For $|x|<1/\sqrt{2}$, check $x = 0$, and for $|x|>1/\sqrt{2}$, check $x = \pm 1$. You'll find they all work out to $\cos(2\sin^{-1}x) = 1-2x^2$.