Viete's Formula Proof

Because the entire circle contains $2^{n+1}$ angles equal to $\theta$ , $\theta=\frac{\pi}{2^n}$ . Thus we can see that $S(n)=2\sin\frac{\pi}{2^n}$ . Using the double angle identity for $\sin x$ tells us that $\frac{S(n+1)}{S(n)}=\frac 12 \sec\frac{\pi}{2^{n+1}}$

Bonus: Notice that an infinity-gon is basically a circle. In equation form, we have $(\text{perimeter of an inscribed square}) \frac{2P(3)}{P(2)} \frac{2P(4)}{P(3)} \frac{2P(5)}{P(4)} \cdots = (\text{circumference of the circle}) \implies 4\sqrt2 \sec\frac{\pi}{8} \sec\frac{\pi}{16} \sec\frac{\pi}{32} \cdots = 2\pi \implies \cos\frac{\pi}{4} \cos\frac{\pi}{8} \cos\frac{\pi}{16} \cdots = \frac{2}{\pi}$ Starting with $\cos\frac{\pi}{4} = \frac{\sqrt2}{2}$ and using the identity $\cos\frac{\theta}{2}=\sqrt\frac{1+\cos\theta}{2}$ for each successive term gives us the identity $\frac{2}{\pi}=\frac{\sqrt2}{2} \frac{\sqrt{2+\sqrt2}}{2} \frac{\sqrt{2+\sqrt{2+\sqrt2}}}{2} \cdots$ which is better known as Viete's Formula.