TITLE: Power-reduction formula
QUESTION [13 upvotes]: According to the Power-reduction formula, one can interchange between $\cos(x)^n$ and $\cos(nx)$ like the following:
$$
\cos^n\theta = \frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}} \binom{n}{k} \cos{((n-2k)\theta)} \tag{odd}\\
$$
$$
\cos^n\theta = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} \binom{n}{k} \cos{((n-2k)\theta)} \tag{even}
$$
To me this looks like an Binomial transform. Is this true?
May I think of it as a change of basis of a vector space?

REPLY [2 votes]: You can generalize your formulae for odd and even values of $n$. 
$$ \cos^n\theta = 2^{-n} \sum_{k=0}^n \bigl( \begin{smallmatrix} n \\ k \end{smallmatrix} \bigr) \cos (2k-n)\theta.$$
That looks nice and compact, but if you want to identify the individual cosines you'll need to add $\bigl( \begin{smallmatrix} n \\ k \end{smallmatrix} \bigr) \cos (-\alpha) + \bigl( \begin{smallmatrix} n \\ n-k \end{smallmatrix} \bigr) \cos \alpha = 2\bigl( \begin{smallmatrix} n \\ k \end{smallmatrix} \bigr) \cos \alpha $.