Showing $\frac14 (1 -\cos 2\theta)^2 = \frac18(3 -5\cos 2\theta)$

Question

I have an example which I have:

$$D = \frac{2}{\sin^4\theta} \tag{1}$$

Which in the notes goes to

$$D = \frac{2}{\frac14 (1 -\cos 2\theta)^2} \tag{2}$$

I understand that part.. but the next part in the notes... it goes to

$$D = \frac{16}{3 -5\cos 2\theta} \tag{3}$$

I have having some serous trouble getting from $(2)$ to $(3)$, as I keep getting stuck in a loop of trig identities where $\cos 2\theta$ end up with a $\cos^2\theta$ identity, and visa versa...

Can anyone give an idea how to get from $(2)$ to $(3)$?

Answer 1

Use that $$\sin^2(x)=1-\cos^2(x)$$ and $$\cos(2x)=2\cos^2(x)-1$$