# 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 $\displaystyle \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$, where $n \in \N$.

Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:

$\displaystyle \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$

### Basis for the Induction

$\map P 1$ is 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 $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

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

Then we need to show:

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

### Induction Step

This is our induction step:

 $\displaystyle \frac {\sin z} z$ $=$ $\displaystyle \paren {\frac {2^k} z} \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \paren {\frac {2^k} z} \paren {2 \sin \frac z {2^{k + 1} } \cos \frac z {2^{k + 1} } } \prod_{i \mathop = 1}^k \cos \frac z {2^i}$ Double Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \cos \frac z {2^{k + 1} } \prod_{i \mathop = 1}^k \cos \frac z {2^i}$ $\displaystyle$ $=$ $\displaystyle \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \prod_{i \mathop = 1}^{k + 1} \cos \frac z {2^i}$

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

Therefore:

$\displaystyle \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$

And then:

 $\displaystyle \frac {\sin z} z$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } \prod_{i \mathop = 1}^n \cos \frac z {2^i}$ $\displaystyle$ $=$ $\displaystyle \paren {\lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } } \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}$ $\displaystyle$ $=$ $\displaystyle \paren 1 \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}$ Limit of Sine of X over X $\displaystyle$ $=$ $\displaystyle \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}$

$\blacksquare$