$\sin^2(2\theta)$ Trig Identitiy? I'm sure there is some very basic Algebra I'm missing out on but. . . 
how does $\sin^2(2\theta)$ end up equaling $4\sin^2(\theta)\cos^2(\theta)$?
I assume this is derived from the $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
This is in a Khan academy Calculus example I'm working through. I'm about 15 years out of my most recent algebra class, so please be kind :)

Your assumption is correct.
Then 
$$
(\sin(2\theta))^2=(2\sin(\theta)\cos(\theta))^2=4\sin^2\theta\cos^2\theta.
$$
