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?

 A: I think that we can establish the identity without too much work; it certainly isn't short though. Your statement is equivalent to $$\cos^2 4x-\cos^2 2x=-\sin 6x\sin2x$$We'll start with left hand side; my strategy will be to make everything in terms of $\cos 2x$. I will make use of the following identities:
$$\begin{align}
\sin^2\theta&\equiv1-\cos^2\theta\\
\cos2\theta&\equiv2\cos^2\theta-1\\
\sin3\theta&\equiv3\sin\theta-4\sin^3\theta
\end{align}$$
Here we go:
$$\begin{align}
\cos^2 4x-\cos^2 2x&\equiv(2\cos^2 2x-1)^2-\cos^22x\\
&\equiv4\cos^42x-5\cos^22x+1
\end{align}$$
Now for the right hand side:
$$\begin{align}
-\sin6x\sin2x&\equiv-\sin2x(3\sin2x-4\sin^3 2x)\\
&\equiv4\sin^42x-3\sin^2x\\
&\equiv4(1-\cos^22x)^2-3(1-\cos^22x)\\
&\equiv4(\cos^4 2x-2\cos^2x+1)-3+3\cos^22x\\
&\equiv4\cos^42x-5\cos^22x+1
\end{align}$$
Hence we have $$\cos^2 4x-\cos^2 2x\equiv-\sin 6x\sin2x$$ as required.

I personally would also be very interested in any shortcut or trick that could be used here instead; if you find one please let me know.
A: OK, I think I have something a little more straightforward. If I am using some of the forbidden fruit, I apologize. First of all, as hamam already noted, I'm going to let $2\alpha = x$ to save my sanity. So I want to show that
$$\cos^2 2x = \cos^2 x - \sin 3x\sin x.$$
Aside from the double-angle formula for $\cos 2x$, I will need the triple angle formula
\begin{align*}
\cos 3x &= \cos 2x\cos x -\sin 2x\sin x \\ &= (2\cos^2 x-1)\cos x - 2\sin^2x\cos x \\ &= 2\cos^3 x-\cos x - 2(1-\cos^2 x)\cos x \\ &= 4\cos^3 x-3\cos x.
\end{align*}
Note that $\cos 2x=\cos (3x-x) = \cos 3x\cos x + \sin 3x\sin x$, so the original formula is equivalent to
\begin{equation*}
\cos^2x -\cos^2 2x + \cos3x\cos x = \cos 2x. \tag{$\star$}
\end{equation*}
Well, by the triple angle formula above,
\begin{align*}
\cos 2x+ \cos^2 2x - \cos^2 x &= (2\cos^2 x-1)+(2\cos^2 x -1)^2 - \cos^2  x \\ & = 4\cos^4x - 3\cos^2 x = \cos3x\cos x,
\end{align*}
and this establishes ($\star$).
Having looked carefully at A-level student's, he's using the triple angle formula for sin, so these solutions are almost surely equivalent. :)
A: hint
Put $$a=2\alpha$$
and prove simply that
$$\cos^2(2a)=\cos^2(a)-\sin(3a)\sin(a)$$
using
$$2\sin(3a)\sin(a)=$$
$$\cos(3a-a)-\cos(3a+a)$$
A: $$\sin^22\theta=4\sin^2\theta\ (1-\sin^2\theta)$$
$$=\sin\theta\ (\sin\theta+(3\sin\theta-4\sin^3\theta))$$
$$=\sin\theta\ (\sin\theta+\sin 3\theta)=\sin^2\theta+\sin 3\theta\sin \theta.$$
Therefore, $\cos^22\theta=\cos^2\theta-\sin 3\theta\sin\theta.$ Finally,
replace $\theta$ by $2\alpha.$
