<s>[INST] What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right)   \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right]   \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x   \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$), ($2$) and ($3$):
\begin{align}
3\arcsin x &= \arcsin \left[x\sqrt{1-2x\sqrt{1-x^2}^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-4x^2(1-x^2)}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-2(2x^2)+(2x^2)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{(2x^2-1)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x|2x^2-1|+ 2x(1-x^2)\right]
\end{align}
If $2x^2-1$ is positive, then $|2x^2-1|$ is $2x^2 -1$.
If $2x^2-1$ is negative, then $|2x^2-1|$ is $-2x^2+1$. 
Range of $x$ is $-1\leq x \leq 1  \implies  0\leq x^2 \leq 1  \implies  0\leq 2x^2 \leq 2$.
For $x\in\left(\frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$, then $2x^2-1$ is negative. 
For $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}\ , 1\right)$, then $2x^2-1$ is positive. 
Thus for  $x\in\left( \frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$
\begin{align} 
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin \left[-2x^3 +x+ 2x(1-x^2)\right]\\
&= \arcsin [3x - 4x^3]
\end{align}
Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$
\begin{align} 
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin [2x^3- x+2x-2x^3]\\
&= \arcsin[x] 
\end{align}
But clearly, $3\arcsin x = \arcsin[3x-4x^3]$.
So what is wrong whith this proof?
 [/INST] For all positive $x$ (the case of negative $x$ is symmetric), $$\sin3x=3\sin x-4\sin^3x.$$
So with $x=\arcsin t$ we have
$$\sin(3\arcsin t)=3t-4t^3.$$
This allows us to write
$$3\arcsin t=\arcsin(3t-4t^3)\lor3\arcsin t=\pi-\arcsin(3t-4t^3).$$

As the range of the arc sine is $\left[0,\dfrac\pi2\right]$, the first identity holds up to $\arcsin t=\dfrac\pi6$, i.e. for $t\in\left[0,\dfrac12\right]$, then comes the second identity, for $t\in\left[\dfrac12,1\right]$.
$$3\arcsin t=\begin{cases}
t\le-\dfrac12&\to-\pi-\arcsin(3t-4t^3),
\\-\dfrac12\le t\le\dfrac12&\to\arcsin(3t-4t^3),
\\t\le\dfrac12&\to\pi-\arcsin(3t-4t^3).\end{cases}$$
</s>