Sine of Integer Multiple of Argument/Formulation 5

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

Proof
The proof proceeds by induction.

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

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.

$\map P 3$ is the case:

So $\map P 3$ 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 $n > 3$ and $n$ even, then it logically follows that $\map P {n + 2}$ is true.

We also need to show that, if $\map P n$ is true, where $n > 3$ and $n$ odd, then it logically follows that $\map P {n + 2}$ is true.

So this is our induction hypothesis:
 * $\ds \map \sin {n \theta} = \paren {\sin \frac {n \pi } 2 } \paren {\sin \theta } + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n } \paren {\sin \frac {k \pi }  2 } \map \sin {\paren {n - k } \theta} }$

from which we are to show:
 * $\ds \map \sin {\paren {n + 2} \theta} = \paren {\sin \frac {\paren {n + 2} \pi } 2 } \paren {\sin \theta } + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n + 2 } \paren {\sin \frac {k \pi }  2 } \map \sin {\paren {n + 2 - k } \theta} }$

Induction Step
This is our induction step:

For $n$ even or for $n$ odd: (Identical argument)

The result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall n \in \Z_{>0}: \sin n \theta = \paren {\sin \frac {n \pi } 2 } \paren {\sin \theta } + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n } \paren {\sin \frac {k \pi }  2 } \map \sin {\paren {n - k } \theta} }$