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I'm being asked to prove that, for some coefficients $a_k$, $$ \cos^n(x) = \sum_{k=0}^{n} a_k \cos(kx) $$

I think I got pretty close, $$ \begin{align} \cos^n(x) & = { \left( \frac{e^{ix} + e^{-ix}}{2} \right) ^n} \\ & = \sum_{k=0}^{n} {n \choose k} {\left( \frac{e^{ix}}{2} \right)}^k {\left( \frac{e^{-ix}}{2} \right)}^{n-k} \\ & = \frac{1}{2^n} \sum_{k=0}^{n} {n \choose k} e^{i(2k-n)x} \\ & = \frac{1}{2^n} \sum_{k=0}^{n} {n \choose k}\cos((2k-n)x) \\ \end{align} $$

But I can't seem to get any further...

Mesmerized student
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galah92
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  • Why not use $\cos(a)\cos(b) = \frac{1}{2}(\cos(a+b) + \cos(a-b))$ ? Of course it would be better to finish your argument since as you said you seems pretty close but I don't really see how to finish ... –  May 21 '17 at 19:14
  • Did you try expanding the cosine term at the end using the difference angle formula? – Gregory May 21 '17 at 19:33
  • Yes, I got two sums and couldn't combine them at the end. – galah92 May 21 '17 at 19:35
  • Use induction. If you really want to understand how this works, see how $\cos nx$ is an n-th degree polynomial in $\cos x$. – P.K. May 21 '17 at 20:04
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    @galah92 I think you are almost done at that point. The sum contains terms $b_j = \cos(j,x)$ with $-n \le j \le n,$. But $\cos$ is an even function, and you can verify that $b_{-j}=b_j,$, so grouping the symmetric terms will give the form that you are after. – dxiv May 21 '17 at 20:08
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    @dxiv I didn't even think in that direction. Works perfectly, thanks! – galah92 May 21 '17 at 21:14

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