Sine of Integer Multiple of Argument/Formulation 3

Proof
The proof proceeds by induction.

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

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:
 * $\displaystyle \map \sin {n \theta} = \sin \theta \cos^{n - 1 } \theta \sum_{k \mathop = 0}^{n - 1 } \frac { \paren { \cos k \theta } } {\paren {\cos^{k } \theta } }$

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

Induction Step
This is our induction step:

For the first part:

The result follows by the Principle of Mathematical Induction.

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