# $\sin^2(2\theta)$ 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 :)

Then $$(\sin(2\theta))^2=(2\sin(\theta)\cos(\theta))^2=4\sin^2\theta\cos^2\theta.$$
• @Matt Yes. $(abc)^2 = a^2b^2c^2$, so $$(2\sin\theta\cos\theta)^2 = 2^2\sin^2\theta\cos^2\theta = 4\sin^2\theta\cos^2\theta$$ – N. F. Taussig Oct 19 '17 at 0:42
$\sin^2(2θ)$ is the same as $(\sin(2θ))^2$. Replace $\sin(2θ)$ with $2\sin(θ)\cos(θ)$ to get $(2\sin(θ)\cos(θ))^2=4\sin^2(θ)\cos^2(θ)$.