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How to expand $\cos nx$ with $\cos x$?

Write $\cos(9x)$ in terms of powers of $\cos(x)$

I realize I could solve this by using De Moivre's and binomial expansion:

$\cos(9x) + i \sin(9x) = (\cos(x) + i\sin(x))^9$

then expanding using binomial and extracting the real part of the expansion and using a trig identity to transform any $\sin(x)$ terms.

However, this is going to take some time to do. I was wondering if this was the only way to handle this problem or if there was a more clever way of dealing with such problems, maybe using polar form?

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marked as duplicate by Did, Américo Tavares, t.b., Zev Chonoles Mar 31 '12 at 16:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Try using the binomial expansion forumla: $(x+y)^{n}=\sum_{k=0}^{n}{n \choose k}x^{k}y^{n-k}$. – Patch Mar 31 '12 at 8:17
Seems to be a standing problem, see also here and here. – draks ... Mar 31 '12 at 8:50
As some contributors on this page know well, the full answer was given there two days ago. In such cases, the close option might be preferable to the copy verbatim option. – Did Mar 31 '12 at 10:11
I just did the calculation you are suggesting, and it's not too bad. Just to get you started, the binomial coefficients for $n=9$ are $1,9,36,84,126,126,84,36,9,1$. As an alternative, @pedja's solution below is very good. – Donkey_2009 Mar 31 '12 at 11:21

Hint :

$\cos 9x =4\cos^33x-3\cos3x ~\text { and }~ \cos 3x =4\cos^3 x-3\cos x$

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You want the Chebyshev polynomials of the first kind. They satisfy $\cos(nx) = T_n(\cos x)$. You can compute them via the recurrence relation. It's not a short calculation, but is probably shorter than the method you're proposing.

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Try using the binomial expansion forumla: $(a+b)^{n}=\sum_{k=0}^{n}{n \choose k}a^{n-k}b^{k}$.

Letting $a=\cos(x)$ and $b=i\sin(x)$, along with the fact that $(i\sin(x))^{k}\in\mathbb{R}$ only when $k$ is even, we arrive at

$$\cos(9x)=\mathrm{Re}\left( \sum_{k=0}^{9}{9 \choose k}(\cos(x))^{9-k}(i\sin(x))^{k} \right)$$ $$=\sum_{j=0}^{4}{9 \choose 2j}(\cos(x))^{9-2j}(i\sin(x))^{2j}$$ $$=\left({9\choose0}\cos^{9}(x)+{9\choose4}\cos^{5}(x)\sin^{4}(x)+{9\choose8}\cos(x)\sin^{8}(x)\right)-\left({9\choose2}\cos^{7}(x)\sin^{2}(x)+{9\choose6}\cos^{2}(x)\sin^{6}(x)\right)$$

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