What is wrong with this proof of $3\arcsin x$?

Question

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?

Answer 1

The problem is a conflict between these two lines:

$2\arcsin x= \arcsin(2x\sqrt{1-x^2}) \tag1$

and

Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$

The problem is that if $\frac1{\sqrt2} < \lvert x \rvert < 1,$ then $\frac\pi4 < \lvert\arcsin x\rvert < \frac\pi2,$ and therefore the left-hand side of equation $(1)$ obeys $$ \frac\pi2 < \lvert2\arcsin x\rvert < \pi.$$ But the right-hand side of $(1)$ is just an application of the arc sine to a number, which must obey the conditions $$ \lvert \arcsin (2x\sqrt{1-x^2}) \rvert \leq \frac\pi2.$$ Hence it is impossible for $(1)$ to be true when $x\in\left(-1, -\,\frac{1}{\sqrt2}\right) \cup \left(\frac{1}{\sqrt2}, 1\right).$ That means that everything you concluded by using $(1)$ -- which is to say, everything after the first few lines -- is inapplicable to the case $x\in\left(-1, -\,\frac{1}{\sqrt2}\right) \cup \left(\frac{1}{\sqrt2}, 1\right).$ By using $(1),$ you restrict the validity of your argument to only the case where $-\frac\pi4 \leq \arcsin x \leq \frac\pi4,$ that is, $x \in \left(-\,\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right).$

The line

$ 3\arcsin x= \arcsin\left[x\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2}+ 2x(1-x^2)\right] $

restricts the applicability of your argument even further, because again the right-hand side has magnitude at most $\frac\pi2,$ which means $\arcsin x$ can only have magnitude at most $\frac\pi6.$ So even the line

Thus for $x\in\left(\frac{-1}{\sqrt2},\frac{+1}{\sqrt2}\right)$

is not right; everything you wrote after the first few lines is applicable only for $-\frac12 \leq x \leq \frac12,$ so you really should have written, "Thus for $x \in \left[-\frac12,\frac12\right]$". For other $x$ you can follow Yves Daoust's answer.

Answer 2

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

$$3\arcsin t=\arcsin(3t-4t^3)\lor3\arcsin t=\pi-\arcsin(3t-4t^3).$$

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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 $t\in\left[\dfrac12,1\right]$.

$$3\arcsin t=\begin{cases} t\le-\dfrac12&\to-\pi-\arcsin(3t-4t^3), \\-\dfrac12\le t\le\dfrac12&\to\arcsin(3t-4t^3), \\t\le\dfrac12&\to\pi-\arcsin(3t-4t^3).\end{cases}$$

Answer 3

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 $$2\arcsin x = \arcsin \left(2x\sqrt{1-x^2}\right).$$

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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.