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?
Best Answer
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.