Sine of X over X as Infinite Product

From ProofWiki
Jump to navigation Jump to search


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


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

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


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.


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

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
\(\displaystyle \) \(=\) \(\displaystyle \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}\)