Explorations in math and programming
David Lowry-Duda

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

Multiply our expression by $ \sin(\dfrac{\xi}{2^n})$. Then we have $$ \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) \cdots \cos(\dfrac{\xi}{2^n})\sin(\dfrac{\xi}{2^n})$$ Using the double angle identity, we can reduce this: $$ = \dfrac{1}{2} \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) ... \cos(\dfrac{\xi}{2^{n-1}})sin(\dfrac{\xi}{2^{n-1}}) =$$ $$ = \dfrac{1}{4} \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) ... \cos(\dfrac{\xi}{2^{n-2}})\sin(\dfrac{\xi}{2^{n-2}}) =$$ $$ \ldots $$ $$ = \dfrac{1}{2^{n-1}}\cos(\xi / 2)\sin(\xi / 2) = \dfrac{1}{2^n}\sin(\xi)$$ 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{\sin x}{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? $$ \dfrac{\pi}{2} = \dfrac{1}{\sqrt{1/2} \sqrt{ 1/2 + 1/2 \sqrt{1/2} } \sqrt{1/2 + 1/2 \sqrt{ 1/2 + 1/2 \sqrt{1/2} \cdots}}}$$ 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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