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Possible Duplicate:
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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    Try using the binomial expansion forumla: $(x+y)^{n}=\sum_{k=0}^{n}{n \choose k}x^{k}y^{n-k}$.2012-03-31
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    Seems to be a standing problem, see also [here](http://math.stackexchange.com/q/117061/19341) and [here](http://math.stackexchange.com/a/125814/19341).2012-03-31
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    As some contributors on this page know well, the full answer was given [there](http://math.stackexchange.com/a/125826/6179) two days ago. In such cases, the *close* option might be preferable to the *copy verbatim* option.2012-03-31
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    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.2012-03-31

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