Sine to Power of Even Integer

Theorem

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

That is:


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

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

Therefore by the product formula for powers:
 * $\sin^{2n} \theta = \dfrac 1 {\left({ 2i } \right)^{2n}}\left({ e^{i\theta} - e^{-i\theta} }\right)^{2n}$

Now by the product formula for powers and the power formula for powers:
 * $\dfrac 1 {\left({ 2i } \right)^{2n}} = \dfrac 1 {2^{2n} \left({ -1 }\right)^n} = \dfrac{\left({ -1 }\right)^n} {2^{2n}}$

Thus:
 * $\sin^{2n} \theta = \dfrac{\left({ -1 }\right)^n} {2^{2n}} \left({ e^{i\theta} - e^{-i\theta} }\right)^{2n}$

Now:

So we have:
 * $\displaystyle \sin^{2n} \theta = \dfrac{\left({ -1 }\right)^n} {2^{2n}} \left({ \sum_{k \mathop = 0}^{2n} {2n \choose k} \left({ -1 }\right)^k \cos\left({ 2n - 2k }\right)\theta + i \sum_{k \mathop = 0}^{2n} {2n \choose k} \left({ -1 }\right)^k \sin\left({ 2n - 2k }\right)\theta}\right)$

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

We find, for the remaining term:

Thus we have:

as required.