Cosine to Power of Odd Integer/Proof 2

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

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

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

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 two terms of the sequence $0, 1, \ldots, n$ are $\dfrac {n - 1} 2$ and $\dfrac {n + 1} 2$.

Thus, when $k = \dfrac {n - 1} 2$:

Similarly, when $k = \dfrac {n + 1} 2$:

The binomial coefficient in each case is the same, because:

So:

Thus the two middle terms collapse 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 $n \in \Z$.