Sine of Integer Multiple of Argument/Formulation 1

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

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

Proof
By De Moivre's Formula:
 * $\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$

As $n \in \Z_{>0}$, we use the Binomial Theorem on the, resulting in:


 * $\displaystyle \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$

When $k$ is odd, the expression being summed is imaginary.

Equating the imaginary parts of both sides of the equation, replacing $k$ with $2 k + 1$ to make $k$ odd, gives:


 * $\displaystyle \sin n \theta = \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\cos^{n - \paren {2 k + 1} } \theta} \paren {\sin^{2 k + 1} \theta}$