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Does any body know a expansion of : $\cos^k(\theta)$ in function of $\cos$ and/or $\sin$ but without power? For example : $\cos^2(\theta)=\frac{1}{2}(\cos(2\theta)+1)$, but i would want a generalization to the power $k$. I searched in Abramowitz and Stegun but didn't find anything !

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Just write $\cos \theta = \frac12 (e^{i\theta} + e^{-i\theta})$ and expand using the binomial theorem... –  Hans Lundmark Mar 6 '12 at 13:41
Surely there is a better word for this than "development". –  Alex Becker Mar 6 '12 at 13:45
which word should I use? –  PanAkry Mar 6 '12 at 14:10
"Expansion" is better. –  Hans Lundmark Mar 7 '12 at 10:10
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3 Answers

$$ \cos^k \theta = \left( \frac{ e^{i\theta} + e^{-i\theta} }{2} \right)^k = \frac{1}{2^k} \sum_{n=0}^k \binom{k}{n} (e^{i\theta} )^n (e^{-i\theta})^{k-n} = \frac{1}{2^{k+1} } \sum_{n=0}^k 2\binom{k}{n} e^{i(2n-k)\theta} $$

Since $ \binom{k}{n} = \binom{k}{k-n} $ we split up the sum and reverse the index of summation in the second term as follows:

$$ = \frac{1}{2^{k+1} } \left( \sum_{n=0}^k \binom{k}{n} e^{i(2n-k)\theta} + \sum_{n=0}^k \binom{k}{k-n} e^{i(2n-k)\theta}\right) = \frac{1}{2^{k+1} } \left( \sum_{n=0}^k \binom{k}{n} e^{i(2n-k)\theta} + \sum_{n=0}^k \binom{k}{n} e^{i(k-2n)\theta}\right) $$

$$ = \frac{1}{ 2^{k} } \sum_{n=0}^k \binom{k}{n} \left( \frac{ e^{i(2n-k)\theta} + e^{-i(2n-k)\theta} }{2} \right) =\frac{1}{ 2^{k} } \sum_{n=0}^k \binom{k}{n} \cos ( (2n-k)\theta). $$

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of course I didn't try this way. Thanks –  PanAkry Mar 6 '12 at 13:58
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You can repeatedly replace powers of the cosine by a combination of smaller powers and $\cos(n\theta)$ using Chebyshev polynomials of the first kind. I'm not sure the resulting coefficients have any conventional name, though.

Alternatively, express $\cos x$ as a Fourier series and use the second convolution theorem to raise it to the $k$th power.

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thank you! maybe it will be useful later for me. –  PanAkry Mar 6 '12 at 13:59
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The example you give can be generalized using trigonometric identities. For odd $k$ we get $$\cos^k(\theta)=2^{1-k}\sum\limits_{n=0}^{\frac{k-1}{2}}\binom{k}{n}\cos((k-2n)\theta)$$ while for even $k$ the formula becomes $$\cos^k(\theta)=2^{-k}\binom{k}{\frac{k}{2}}+2^{1-k}\sum\limits_{n=0}^{\frac{k}{2}-1}\binom{k}{n}\cos((k-2n)\theta).$$

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