 I have this identity: \cos^2 4\alpha=\cos^2 2\alpha−\ compagnia04 2022-01-03 Answered
I have this identity:
$$\displaystyle{{\cos}^{{2}}{4}}\alpha={{\cos}^{{2}}{2}}\alpha−{\sin{{6}}}\alpha{\sin{{2}}}\alpha$$
If I write this like as:
$$\displaystyle{\left({\cos{{4}}}\alpha+{\cos{{2}}}\alpha\right)}\cdot{\left({\cos{{4}}}\alpha−{\cos{{2}}}\alpha\right)}=−{\sin{{6}}}\alpha{\sin{{2}}}\alpha$$
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:
$$\displaystyle{{\cos}^{{2}}{4}}\alpha={{\cos}^{{2}}{2}}\alpha-{\sin{{6}}}\alpha{\sin{{2}}}\alpha$$ (1)
$$\displaystyle{\left[{\cos{{\left({2}{\left({2}\alpha\right)}\right)}}}\right]}^{{2}}={\left({2}{\cos{{2}}}\alpha−{1}\right)}^{{2}}−{\sin{{6}}}\alpha{\sin{{2}}}\alpha$$ (2)
$$\displaystyle{{\cos}^{{4}}{2}}\alpha−{2}{{\sin}^{{2}}{2}}\alpha{{\cos}^{{2}}{2}}\alpha+{{\sin}^{{4}}{2}}\alpha={\left({2}{{\cos}^{{2}}\alpha}−{1}\right)}^{{2}}−{\sin{{6}}}\alpha{\sin{{2}}}\alpha$$ (3)
Then I have abandoned because I should do long computations and I think that is not the right way.

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I think that we can establish the identity without too much work; it certainly isn't short though. Your statement is equivalent to
$$\displaystyle{{\cos}^{{2}}{4}}{x}−{{\cos}^{{2}}{2}}{x}=−{\sin{{6}}}{x}{\sin{{2}}}{x}$$
We'll start with left hand side; my strategy will be to make everything in terms of $$\displaystyle{\cos{{2}}}{x}$$. I will make use of the following identities:
$$\displaystyle{\sin{{2}}}\theta\equiv{1}-{\cos{{2}}}\theta$$
$$\displaystyle{\cos{{2}}}\theta\equiv{2}{\cos{{2}}}\theta−{1}$$
$$\displaystyle{\sin{{3}}}\theta\equiv{3}{\sin{\theta}}−{4}{\sin{{3}}}\theta$$
Here we go:
$$\displaystyle{{\cos}^{{24}}{x}}−{{\cos}^{{22}}{x}}\equiv{\left({2}{{\cos}^{{22}}{x}}−{1}\right)}^{{2}}−{{\cos}^{{22}}{x}}$$
$$\displaystyle\equiv{4}{{\cos}^{{42}}{x}}−{5}{{\cos}^{{22}}{x}}+{1}$$
Now for the right hand side:
$$\displaystyle−{\sin{{6}}}{x}{\sin{{2}}}{x}\equiv−{\sin{{2}}}{x}{\left({3}{\sin{{2}}}{x}−{4}{{\sin}^{{3}}{2}}{x}\right)}$$
$$\displaystyle\equiv{4}{{\sin}^{{42}}{x}}−{3}{{\sin}^{{2}}{x}}$$
$$\displaystyle\equiv{4}{\left({1}−{{\cos}^{{22}}{x}}\right)}^{{2}}−{3}{\left({1}−{{\cos}^{{22}}{x}}\right)}$$
$$\displaystyle\equiv{4}{\left({{\cos}^{{42}}{x}}−{2}{{\cos}^{{2}}{x}}+{1}\right)}−{3}+{3}{{\cos}^{{22}}{x}}$$
$$\displaystyle\equiv{4}{{\cos}^{{42}}{x}}−{5}{{\cos}^{{22}}{x}}+{1}$$
Hence we have
$$\displaystyle{{\cos}^{{2}}{4}}{x}−{{\cos}^{{2}}{2}}{x}\equiv−{\sin{{6}}}{x}{\sin{{2}}}{x}$$
as required.
Not exactly what you’re looking for? lalilulelo2k3eq
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 $$\displaystyle{2}\alpha={x}$$ to save my sanity. So I want to show that
$$\displaystyle{{\cos}^{{2}}{2}}{x}={{\cos}^{{2}}{x}}-{\sin{{3}}}{x}{\sin{{x}}}$$
Aside from the double-angle formula for $$\displaystyle{\cos{{2}}}{x}$$, I will need the triple angle formula
$$\displaystyle{\cos{{3}}}{x}={\cos{{2}}}{x}{\cos{{x}}}-{\sin{{2}}}{x}{\sin{{x}}}$$
$$\displaystyle={\left({2}{{\cos}^{{2}}{x}}-{1}\right)}{\cos{{x}}}-{2}{{\sin}^{{2}}{x}}{\cos{{x}}}$$
$$\displaystyle={2}{{\cos}^{{3}}{x}}-{\cos{{x}}}-{2}{\left({1}-{{\cos}^{{2}}{x}}\right)}{\cos{{x}}}$$
$$\displaystyle={4}{{\cos}^{{3}}{x}}-{3}{\cos{{x}}}$$
Note that $$\displaystyle{\cos{{2}}}{x}={\cos{{\left({3}{x}−{x}\right)}}}={\cos{{3}}}{x}{\cos{{x}}}+{\sin{{3}}}{x}{\sin{{x}}}$$, so the original formula is equivalent to
$$\displaystyle{{\cos}^{{2}}{x}}−{{\cos}^{{2}}{2}}{x}+{\cos{{3}}}{x}{\cos{{x}}}={\cos{{2}}}{x}$$ (*)
Well, by the triple angle formula above,
$$\displaystyle{\cos{{2}}}{x}+{\cos{{22}}}{x}−{\cos{{2}}}{x}={\left({2}{{\cos}^{{2}}{x}}−{1}\right)}+{\left({2}{{\cos}^{{2}}{x}}−{1}\right)}^{{2}}−{{\cos}^{{2}}{x}}$$
$$\displaystyle={4}{{\cos}^{{4}}{x}}−{3}{{\cos}^{{2}}{x}}={\cos{{3}}}{x}{\cos{{x}}}$$
and this establishes (*) Vasquez

hint
Put
$$a=2\alpha$$
and prove simply that
$$\cos^2(2a)=\cos^2(a)-\sin(3a)\sin(a)$$
using
$$2\sin(3a)\sin(a)=$$
$$\cos(3a-a)-\cos(3a+a)$$