Vieta's Formula for Pi

Theorem

 * $$\pi = 2 \times \frac{2} {\sqrt{2}} \times \frac{2} {\sqrt{2 + \sqrt{2}}} \times \frac{2}{\sqrt {2 + \sqrt{2 + \sqrt{2}}}} \times \frac{2} {\sqrt{2 + \sqrt{2 + \sqrt{2 +\sqrt{2}}}}} \times \cdots$$

Proof
$$ $$ $$ $$ $$ $$

Thus:

$$ $$

Then we have from the Half Angle Formulas for Sine and Cosine that:

$$ $$ $$

So we can replace all the instances of $$\cos \frac \pi 4$$, $$\cos \frac \pi 8$$, etc. with their expansions in square roots of $$2$$.

Finally, we note that from Limit of Sine of X over X we have:
 * $$\lim_{\theta \to 0} \frac {\sin \theta} {\theta} = 1$$

As $$n \to \infty$$, then, we have that $$\frac \pi {2^n} \to 0$$, and so:
 * $$\lim_{n \to \infty} \frac {\sin \left({\pi / 2^n}\right)} {\pi / 2^n} = 1$$

The result follows after some algebra.