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?
Best Answer
A geometric point of view might be illuminating.
Suppose $0\leq x \leq \frac{1}{\sqrt 2}$. Consider the figure below, where we start from a right-angled triangle $\triangle ABC$ with sides $\overline{AB} = \sqrt{1-x^2}$ and $\overline{BC} = x$, and hypotenuse $\overline{AC}=1$. The choice of $x$ we have made guarantees that $$ \alpha = \angle BAC$$ is in the range $\left[0, \frac{\pi}{4}\right].$
By definition is
$$ \alpha = \arcsin x.$$
Extend first $CB$ to a segment $BD \cong BC$, then draw from $D$ the line perpendicular to $AD$ that intersects the extension of $AC$ in $E$. Finally draw from $C$ the perpendicular to $BC$ that meets $ED$ in $F$.
Define \begin{equation}\beta = \angle CAD = 2\alpha.\tag{1}\label{eq:1}\end{equation} We have, by definition, \begin{equation}\beta = \arcsin \left(\frac{\overline{ED}}{\overline{AE}}\right).\tag{2}\label{eq:2}\end{equation}
From $\triangle DCF \sim \triangle ABC$ find $$\overline{CF} = \frac{2x^2}{\sqrt{1-x^2}}$$ and $$\overline{DF} = \frac{2x}{\sqrt{1-x^2}}.$$
By Pythagorean Theorem on $\triangle ADE$, and by $\triangle CEF \sim \triangle CED$ \begin{equation} \begin{cases} 1+\overline{ED}^2 = (1+ \overline{EC})^2\\ \overline{EC} = \frac{x}{\sqrt{1-x^2}}\overline{ED}. \end{cases} \end{equation} Solving the system yields $$\overline{ED} = \frac{2x\sqrt{1-x^2}}{1-2x^2}$$ and $$\overline{AC} = 1 + \overline{EC} = \frac{1}{1-2x^2}.$$ Using these results in \eqref{eq:2} and then cosidering the identity \eqref{eq:1} leads to $$2\arcsin x = \arcsin \left(2x\sqrt{1-x^2}\right).$$
For $\frac{1}{\sqrt 2} \leq x \leq 1$, I would consider the triangle below, where again $\overline{BC} = x$, $\overline{AC} = 1$ and $D$ is the point symmetrical to $C$ with respect to line $AB$. Draw then from $C$ the perpendicular to $AD$ that meets its extension in $E$. Define then $\alpha$ as before and $$\beta = \angle CAE = \pi - \angle CAD = \pi -2\alpha.$$ Use then the fact that $\sin \beta = \sin 2\alpha$.
Finally, for negative $x$ just define $\overline{BC} = -x$ and proceed as above, taking advantage of the sine odd symmetry.