Cosine to Power of Even Integer/Proof 2

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

Then:
 * $\ds \cos^n \theta = \frac 1 {2^{n - 1} } \sum_{k \mathop = 0}^{n / 2} \paren {\binom n k \cos \paren {n - 2 k} \theta}$

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

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

Thus:


 * $\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } {2!} \cos \paren {n - 4} \theta + \cdots + R_n}$

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:
 * $\ds \cos^{2 n + 1} \theta = \frac 1 {2^{2 n} } \sum_{k \mathop = 0}^n \binom {2 n + 1} k \cos \paren {2 n - 2 k + 1} \theta$

for all $n \in \Z$.