Show that sin(2 arc sin x) = 2x / (7-4x²)

Let $\theta = \arcsin x$.
This implies that $\sin \theta = x$.
The range of the principal value of $\arcsin x$ is $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

The LHS is $\sin(2 \arcsin x)$, which can be written as $\sin(2\theta)$.
The double angle identity for sine is:

We use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.
Substitute $\sin \theta = x$:

Taking the square root of both sides:

Since $\theta$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the cosine function is non-negative ($\cos \theta \ge 0$).
Therefore, we take the positive square root:

Substitute $x$ for $\sin \theta$ and $\sqrt{1 - x^2}$ for $\cos \theta$ into the identity from Step 2:

The problem asks to show that $\sin(2 \arcsin x) = \frac{2x}{\sqrt{1 - x^2}}$.
However, our derivation shows that $\sin(2 \arcsin x) = 2x\sqrt{1 - x^2}$.
These two expressions are generally not equal. For them to be equal:

If $x \ne 0$, we can divide both sides by $2x$:

Squaring both sides:

This means the given identity $\sin(2 \arcsin x) = \frac{2x}{\sqrt{1 - x^2}}$ is only true for $x=0$. For any other value of $x$ within the domain of $\arcsin x$ (i.e., $-1 \le x \le 1$), the statement is false.

Therefore, the statement $\sin(2 \arcsin x) = \frac{2x}{\sqrt{1 - x^2}}$ is not generally true. The correct identity is $\sin(2 \arcsin x) = 2x\sqrt{1 - x^2}$.