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
The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * $\displaystyle \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$

Basis for the Induction
$\map P 1$ is the case:

So $\map P 1$ is seen to hold.

$\map P 2$ is the case:

So $\map P 2$ is also seen to hold.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true.

So this is our induction hypothesis:
 * $\displaystyle \map \sin {n \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$

from which we are to show:
 * $\displaystyle \map \sin {\paren {n + 1} \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n + 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k + 1} } }$

Induction Step
This is our induction step:

To proceed, we will require the following lemma:

Lemma
The result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$