Continuing from this post

We start with $cos( \dfrac{\xi}{2})cos(\dfrac{\xi}{4}) ... cos(\dfrac{\xi}{2^n})$. Recall the double angle identity for sin: $sin2 \theta = 2sin \theta cos \theta$. We will use this a lot.

Multiply our expression by $sin(\dfrac{\xi}{2^n})$. Then we have

$cos( \dfrac{\xi}{2})cos(\dfrac{\xi}{4}) ... cos(\dfrac{\xi}{2^n})sin(\dfrac{\xi}{2^n})$

Using the double angle identity, we can reduce this:

$= \dfrac{1}{2} cos( \dfrac{\xi}{2})cos(\dfrac{\xi}{4}) ... cos(\dfrac{\xi}{2^{n-1}})sin(\dfrac{\xi}{2^{n-1}}) =$
$= \dfrac{1}{4} cos( \dfrac{\xi}{2})cos(\dfrac{\xi}{4}) ... cos(\dfrac{\xi}{2^{n-2}})sin(\dfrac{\xi}{2^{n-2}}) =$
$...$
$= \dfrac{1}{2^{n-1}}cos(\xi / 2)sin(\xi / 2) = \dfrac{1}{2^n}sin(\xi)$

So we can rewrite this as

$cos( \dfrac{\xi}{2})cos(\dfrac{\xi}{4}) ... cos(\dfrac{\xi}{2^n}) = \dfrac{sin \xi}{2^n sin( \dfrac{\xi}{2^n} )}$ for $\xi \not = k \pi$

Because we know that $lim_{x \to \infty} \dfrac{sinx}{x} = 1$, we see that $lim_{n \to \infty} \dfrac{\xi / 2^n}{sin(\xi / 2^n)} = 1$. So we see that

$cos( \dfrac{\xi}{2})cos(\dfrac{\xi}{4}) ... = \dfrac{\xi}{\xi}$
$\xi = \dfrac{sin( \xi)}{cos( \dfrac{\xi}{2})cos(\dfrac{\xi}{4}) ...}$

Now we set $\xi := \pi /2$. Also recalling that $cos(\xi / 2 ) = \sqrt{ 1/2 + 1/2 cos \xi}$. What do we get?

$\dfrac{\pi}{2} = \dfrac{1}{\sqrt{1/2} \sqrt{ 1/2 + 1/2 \sqrt{1/2} } \sqrt{1/2 + 1/2 \sqrt{ 1/2 + 1/2 \sqrt{1/2} ...}}}$

This is pretty cool. It’s called Vieta’s Formula for $\dfrac{\pi}{2}$. It’s also one of the oldest infinite products.