<s>[INST] Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$ I'm trying to prove that
$$
\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}.
$$
Let $\alpha = 2 \arcsin x$ and $\beta = \arccos x$; meaning $\sin\alpha = \frac{x}{2}, \cos\beta = x$. We know that:
$$
\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.
$$
Finding $\cos\alpha$ and $\sin\beta$:
$$\begin{align}
\cos\alpha & = \frac{\sqrt{4-x^2}}{2}, \\[0.1in]
\sin\beta & = \sqrt{1-x^2}.
\end{align}$$
So plugging everything in:
$$\begin{align}
\sin(2\arcsin x + \arccos x) & = \frac{x}{2} \cdot x + \frac{\sqrt{4-x^2}}{2} \cdot \sqrt{1-x^2}\\[0.1in]
& = \frac{x^2 + \sqrt{4-x^2}\sqrt{1-x^2}}{2}.
\end{align}$$
But this doesn't seem to lead to the right side. Is my method incorrect?
 [/INST] $\alpha=2\arcsin x$  means that $x=\sin(\frac\alpha2)$, not that $\frac x2=\sin\alpha$
$\frac\alpha2=\arcsin x$
$x=\sin(\frac\alpha2)$
an alternate and faster way is to use 
$\arcsin x+\arccos x=\frac\pi2$
</s>