Inverse trigonometric Problem For any $x \in [-1,0) \cup (0,1]$, how can I prove that:
$$\sin^{-1}(2x\sqrt{1-x^2})=2\cos^{-1}x$$ 
Also, can someone explain to me how to understand the graphs of $sin$ and $cos$ functions?
 A: Hint
Consider $x=\cos t$ and recall that $2\sin t\cos t=\sin 2t$.
A: Hint: If $\theta=2\cos^{-1}(x)$, then $x=\cos(\theta/2)$.
What is $\sqrt{1-x^2}$ ?
What is $2x\sqrt{1-x^2}$ ?
A: Sufficient care has to be taken while dealing with Inverse trigonometric functions
as the principal values of $\cos^{-1}x$ lies in $\left[0,\pi\right]$ whereas it lies in  $\left[-\frac\pi2,\frac\pi2\right]$ for $\sin^{-1}y$
$\displaystyle\iff0\le2\cos^{-1}x\le2\pi$
Case $\#1:\displaystyle2\cos^{-1}x$ will be $\displaystyle\sin^{-1}2x\sqrt{1-x^2}$ 
iff $\displaystyle-\frac\pi2\le2\cos^{-1}x\le\frac\pi2\iff -\frac\pi4\le\cos^{-1}x\le\frac\pi4$ 
$\displaystyle\implies0\le\cos^{-1}x\le\frac\pi4\iff1\ge x\ge\frac1{\sqrt2}$
Case $\#2:$
If $\displaystyle\frac\pi4<\cos^{-1}x\le\frac\pi2\iff\frac1{\sqrt2}> x\ge0$
$\displaystyle\iff\frac\pi2<2\cos^{-1}x\le\pi,\sin^{-1}2x\sqrt{1-x^2}=\pi-2\cos^{-1}x$
Similarly, Case $\#3:$
$\displaystyle\iff\frac\pi2<\cos^{-1}x\le\frac{3\pi}4,\sin^{-1}2x\sqrt{1-x^2}=\pi-2\cos^{-1}x$
and Case $\#4:$
$\displaystyle\iff\frac{3\pi}4<\cos^{-1}x\le\pi,\sin^{-1}2x\sqrt{1-x^2}=2\pi-2\cos^{-1}x$
For example, if $\displaystyle x=-1,\cos^{-1}x=\cos^{-1}(-1)=\pi$
But, $\displaystyle\sin^{-1}\{2x\sqrt{1-x^2}\}=\sin^{-1}0=0$
