Sine to Power of Even Integer

Theorem

 * $\displaystyle \sin^{2 n} \theta = \frac 1 {2^{2 n}} \binom {2 n} n + \frac {\left({-1}\right)^n} {2^{2 n - 1}} \left({\cos 2 n \theta - \binom {2 n} 1 \cos \left({2 n - 2}\right) \theta + \cdots + \left({-1}\right)^{n - 1} \binom {2 n} {n - 1} \cos 2 \theta}\right)$

That is:


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