Sine of Integer Multiple of Argument

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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:

$\cos n \theta + i \sin n \theta = \left({\cos \theta + i \sin \theta}\right)^n$

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

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

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:

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



Sources