Sine of X over X as Infinite Product

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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:

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

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


Therefore:

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


And then:

\(\ds \frac {\sin z} z\) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } \prod_{i \mathop = 1}^n \cos \frac z {2^i}\)
\(\ds \) \(=\) \(\ds \paren {\lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } } \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}\)
\(\ds \) \(=\) \(\ds \paren 1 \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}\) Limit of $\dfrac {\sin x} x$ at Zero
\(\ds \) \(=\) \(\ds \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}\)

$\blacksquare$


Sources

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