Cosine to Power of Even Integer/Proof 2

Theorem
Let $n \in \Z$ be an even integer.

Then:
 * $\displaystyle \cos^n \theta = \frac 1 {2^{n - 1}} \sum_{k \mathop = 0}^{n / 2} \left({\binom n k \cos \left({n - 2 k}\right) \theta}\right)$

That is:
 * $\cos^n \theta = \dfrac 1 {2^{n-1}} \left({\cos n \theta + n \cos \left({n - 2}\right) \theta + \dfrac {n \left({n - 1}\right)} 2 \cos \left({n - 4}\right) \theta + \cdots + \dfrac {n!} {\left({\left({\frac n 2}\right)!}\right)^2} \cos \theta}\right)$

Proof
Matching up terms from the beginning of this expansion with those from the end:

Thus:


 * $\cos^n \theta = \dfrac 1 {2^{n-1}} \left({\cos n \theta + n \cos \left({n - 2}\right) \theta + \dfrac {n \left({n - 1}\right)} {2!} \cos \left({n - 4}\right) \theta + \cdots + R_n}\right)$

Now to determine $R_n$.

The middle term of the sequence $0, 1, \ldots, n$ is $\dfrac n 2$.

So when $k = \dfrac n 2$ we have $n - 2k = 0$ and $n - k = n - \dfrac n 2 = \dfrac n 2$.

Thus:

So the middle term collapses to:

Also defined as
This result is also reported in the form:
 * $\displaystyle \cos^{2n+1} \theta = \frac 1 {2^{2n}} \sum_{k \mathop = 0}^n \binom {2n+1} k \cos \left({2n - 2k + 1}\right) \theta$

for all $n \in \Z$.