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...

  • $\begingroup$ 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 ... $\endgroup$ – user171326 May 21 '17 at 19:14
  • $\begingroup$ Did you try expanding the cosine term at the end using the difference angle formula? $\endgroup$ – Gregory May 21 '17 at 19:33
  • $\begingroup$ Yes, I got two sums and couldn't combine them at the end. $\endgroup$ – galah92 May 21 '17 at 19:35
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    $\begingroup$ @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. $\endgroup$ – dxiv May 21 '17 at 20:08
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    $\begingroup$ @dxiv I didn't even think in that direction. Works perfectly, thanks! $\endgroup$ – galah92 May 21 '17 at 21:14

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