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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Expert Answer

Donald Cheek
Answered 2022-01-04 Author has 363 answers
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.
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lalilulelo2k3eq
Answered 2022-01-05 Author has 2680 answers
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 (*)
0
Vasquez
Answered 2022-01-08 Author has 9499 answers

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

0

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