Sine of Integer Multiple of Argument/Formulation 2

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: