Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas I have this identity:
$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$
If I write this like as:
$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$
I can use Werner and prostaferesis formulas and I find the identity $(1)$.
But if we suppose of not to use these formulas I have done a try writing:
$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$
$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$
$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$
Then I have abandoned because I should do long computations and I think that is not the right way.

Is there any trick without to use the prostapheresis or Werner's formulas?


$$\sin^22\theta=4\sin^2\theta\ (1-\sin^2\theta)$$
$$=\sin\theta\ (\sin\theta+(3\sin\theta-4\sin^3\theta))$$
$$=\sin\theta\ (\sin\theta+\sin 3\theta)=\sin^2\theta+\sin 3\theta\sin \theta.$$
Therefore, $\cos^22\theta=\cos^2\theta-\sin 3\theta\sin\theta.$ Finally,
replace $\theta$ by $2\alpha.$
