Inverse trignometric functions $\arcsin\left(2x\sqrt{1 - x^2}\right)$ I recently learnt about inverse trigonometric functions. While studying, I came across a formula
$$\arcsin\left(2x\sqrt{1-x^2}\right)=2\arcsin x$$
The proof of this was given by substituting $x$ by $\sin\phi$.
However if we substitute $x$ by $\cos\phi$, we can prove that the above expression equals $2\arccos x$. And indeed while going through another book I found that the above expression equals $2\arcsin x$ for $x$ between $-1/\sqrt{2}$ and $1/\sqrt{2}$ and $2\arccos x$ for $x$ between $1/\sqrt{2}$ and $1$.  I graphed the function in GeoGebra and wanted to post the images but I can't.
Links:
arcsin(2x√1-x²)
2arcsin(x)
2arccos(x)
But it was clear that $\arcsin\left(2x\sqrt{1-x^2}\right)$ is not equal to $\arcsin x$ throughout its domain.
So the question is:why isn't $\arcsin\left(2x\sqrt{1-x^2}\right)=2\arcsin x$ throughout its domain? Why is there another definition for it?
Also as a bonus question, what can be a definition for the expression for $x$ between $-1/\sqrt{2}$ and $-1$?
Please explain in a simple way.
I'm new to high level mathematics that is common here in mathstackexchange :)

Substituting $x = \sin \phi$ gives on the LHS, by trig addition theorem:
$\arcsin(2x\sqrt{1-x^2}) = \arcsin(2\sin \phi \cos \phi) = \arcsin(\sin ( 2\phi))$
Note that $\sin ( 2\phi)$ is increasing with $\phi$ only for $0 \le \phi \le \pi/4$ so  
$\arcsin(\sin ( 2\phi))  = 2 \phi = 2 \arcsin(x)$ is only valid for the corresponding $0\le x \le \sin \pi/4 = \frac{1}{\sqrt 2}$ . 
However, for $\pi/4 \le \phi \le \pi/2$,  $\sin ( 2\phi)$ is decreasing, so 
substituting, $x = \sin \phi$  gives in this range $\frac{1}{\sqrt 2} \le x \le 1$ and 
$\arcsin(2x\sqrt{1-x^2}) = \arcsin(2\sin \phi \cos \phi) = \arcsin(\sin ( 2\phi))  = \pi - 2 \phi = \pi - 2 \arcsin x = 2 \arccos x$ 
So the answer in your "other book" is accurate.
