Cosine of Integer Multiple of Argument/Formulation 1

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:
 * $\ds \cos n \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac n k \dbinom {n - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n - 2 k} }$

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:
 * $\ds \map \cos {n \theta} = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac n k \dbinom {n - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n - 2 k} }$

from which we are to show:
 * $\ds \map \cos {\paren {n + 1} \theta} = \dfrac 1 2 \paren {\paren {2 \cos \theta}^{n + 1} + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac {n + 1} k \dbinom {n + 1 - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n + 1 - 2 k} }$

Induction Step
This is our induction step:

To proceed, we will require the following Lemma:

Lemma
Dividing through by $\sin \theta$, we obtain:

We now have:

The result follows by the Principle of Mathematical Induction.

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