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.

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