Rewriting $\sin(2\cos^{-1}{(x/3)})$ as a fraction? I'm looking to simplify
$$\sin(2\cos^{-1}{(x/3)})$$
I know it simplifies to $\frac{2x}{3}\sqrt{1-\frac{x^2}{9}}$, but I am unsure of the required steps.
Thanks for any help
 A: Use the double-angle formula,
$$ \sin{2\theta} = 2\sin{\theta} \cos{\theta}. $$
What happens to the $\cos$ is clear. To deal with the sine, recall that
$$ \cos^{2}{\theta}+\sin^2{\theta} = 1, $$
with which you can rewrite the sine, after you verify which sign of the square root. The answer you should find is
$$ \frac{2x}{3}\sqrt{1-\frac{x^2}{9}}. $$
A: $$\sin (2\cos^{-1}y)=2y\sin(2\cos^{-1}y)=2y\sqrt{1-y^2}$$where in the first step I have used $\sin 2\theta =2\sin \theta \cos \theta$ and in the second step I have used $\sin^2\theta+\cos^2\theta=1$
A: (If you don't remember that $\sin({2\theta})=2\sin\theta\cos\theta$, you can look here for helpful proofs.)
An alternative after converting the equation to $2\sin\theta\cos\theta$, where $\theta=\cos^{-1}(x/3)$, is to draw a triangle to solve for the missing pieces.
Assume the reference angle is $\theta$.  Then draw $\cos^{-1}(x/3)$ where $x$ is the adjacent side, and $3$ is the hypotenuse.  Using Pythagorean's Theorem, we see the opposite side is $\sqrt{3^2-x^2}$.  Thus, $\sin\theta$, being opposite over hypotenuse, is $(\sqrt{3^2-x^2})/3$.  The $\cos$ will cancel out with each other to get $x/3$.
From there, multiply the terms together: $2 * ((\sqrt{3^2-x^2})/3) * (x/3)$.

