Sine to Power of Odd Integer

Theorem

 * $\displaystyle \sin^{2 n - 1} \theta = \frac {\left({-1}\right)^{n - 1}} {2^{2 n - 2}} \left({\sin \left({2 n - 1}\right) \theta - \binom {2 n - 1} 1 \sin \left({2 n - 3}\right) \theta + \cdots + \left({-1}\right)^{n - 1} \binom {2 n - 1} {n - 1} \sin \theta}\right)$

That is:


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