Sine of X over X as Infinite Product

Theorem
Let $z$ be a non-zero Complex Number.

Then:


 * $\displaystyle \frac {\sin z} z = \cos \frac z 2 \cos \frac z 4 \cos \frac z 8 \cdots = \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}$

where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.

Proof
Firstly, we will prove that $\displaystyle \frac {\sin z} z = \left({\frac {2^n} z}\right) \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$, where $n \in \N$.

Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \frac {\sin z} z = \left({\frac {2^n} z}\right) \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$

Basis for the Induction
$P(1)$ is true, as this says $\displaystyle \frac {\sin z} z = \frac {\sin z} z$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle \frac {\sin z} z = \left({\frac {2^k} z}\right) \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}$

Then we need to show:


 * $\displaystyle \frac {\sin z} z = \left({\frac {2^{k+1}} z}\right) \sin \frac z {2^{k+1}} \prod_{i \mathop = 1}^{k+1} \cos \frac z {2^i}$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \frac {\sin z} z = \left({\frac {2^n} z}\right) \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$

And then: