Sine of Integer Multiple of Argument/Formulation 1

Theorem
For $n \in \Z_{>0}$:
 * $\sin n \theta = \sin \theta \left({\left({2 \cos \theta}\right)^{n - 1} - \dbinom {n - 2} 1 \left({2 \cos \theta}\right)^{n - 3} + \dbinom {n - 3} 2 \left({2 \cos \theta}\right)^{n - 5} - \cdots}\right)$

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

Proof
By De Moivre's Formula, we have $\cos n\theta+i\sin n\theta=(\cos\theta+i\sin\theta)^n$. As $n\in\Z_{>0}$, we use the Binomial Theorem on the right side of the equation, resulting in:

$\displaystyle\cos n\theta+i\sin n\theta=\sum_{k\ge 0}\binom{n}{k}\cos^{n-k}(\theta)(i\sin(\theta))^k$

Note that when $k$ is odd, the expression being summed is imaginary. Equating the imaginary parts of both sides of the equation, replacing $k$ with $2k+1$ to make $k$ odd, gives:

$\displaystyle\sin n\theta=\sum_{k\ge 0}(-1)^k\binom{n}{2k+1}\cos^{n-(2k+1)}(\theta)\sin^{2k+1}(\theta)$