Sine to Power of Even Integer

Proof
First, by Sine Exponential Formulation we have:
 * $\sin \theta = \dfrac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }$

Therefore by Power of Product:
 * $\sin^{2 n} \theta = \dfrac 1 {\paren {2 i}^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$

Now by Power of Product and the Power of Power:
 * $\dfrac 1 {\paren {2 i}^{2 n} } = \dfrac 1 {2^{2 n} \paren {-1}^n} = \dfrac {\paren {-1}^n} {2^{2 n} }$

Thus:
 * $\sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$

Now:

So we have:


 * $\displaystyle \sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta}$

Now we look at each of the terms in the parentheses:

We find, for the remaining term:

Thus we have:

as required.