Sine of Integer Multiple of Argument/Formulation 2

Theorem

For $n \in \Z_{>0}$:

$\ds \sin n \theta$

$=$

$\ds \cos^n \theta \paren {\dbinom n 1 \paren {\tan \theta} - \dbinom n 3 \paren {\tan \theta}^3 + \dbinom n 5 \paren {\tan \theta}^5 - \cdots}$

$\ds $

$=$

$\ds \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}$

$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$

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

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

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:

$\ds \sin n \theta$

$=$

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

$\ds $

$=$

$\ds \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}$

factor out $\cos^n \theta$

$\blacksquare$