Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$. 
Prove the following identity:
  $$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$

How can I express $\cos(4\theta) $ in other terms?

$$\cos (4\theta)=2\cos^2 (2\theta)-1=2(1-2\sin^2 \theta)^2-1=8\sin^4 \theta-8\sin^2 \theta+1$$
$$4\cos 2\theta = 4(1-2\sin^2 \theta)=4-8\sin^2 \theta$$

$$\cos 4\theta-4\cos 2\theta= (8\sin^4 \theta-8\sin^2 \theta+1)-(4-8\sin^2 \theta)=8\sin^4 \theta-3$$

