Sine of X over X as Infinite Product

Theorem
Let $z \in \C$ be a non-zero complex number.

Then:


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

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

Proof
First we prove that:
 * $\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$

for $n \in \N$.

The proof proceeds by induction.

For all $n \in \N$, let $\map P n$ be the proposition:
 * $\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$

Basis for the Induction
$\map P 1$ is the case:
 * $\ds \frac {\sin z} z = \paren {\frac {2^1} z} \sin \frac z {2^1} \prod_{i \mathop = 1}^1 \cos \frac z {2^i}$

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

This is our basis for the induction.

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

So this is our induction hypothesis:


 * $\ds \frac {\sin z} z = \paren {\frac {2^k} z} \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}$

Then we need to show:


 * $\ds \frac {\sin z} z = \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \prod_{i \mathop = 1}^{k + 1} \cos \frac z {2^i}$

Induction Step
This is our induction step:

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

Therefore:
 * $\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$

And then: