Strange approximation to $\sqrt{\pi}$

Let $$\alpha = \sqrt{2\sin^2 1+\sqrt{2\sin^2 2 + \sqrt{2\sin^2 3 + \cdots}}} =\sqrt{3.1415...}$$ Prove that $$\alpha^2 \neq \pi$$. It is a remarkable approximation though.

• Is it till infinity? – Zenix Oct 24 at 14:47
• $\sin^2 1$, that's $1$ radian? – GEdgar Oct 24 at 14:47
• @GEdgar: yes, radians. – Vincent Granville Oct 24 at 14:58
• @Zenix: the nested radicals go indefinitely, but only the first 4 digits match those of $\pi$. – Vincent Granville Oct 24 at 15:00
• A cool find, maybe it was just random just as with $\sqrt{2}+\sqrt{3}\approx \pi$ – user712576 Oct 24 at 15:22

This is easy numerically. It is well known that $$\sqrt{2+\sqrt{2+\cdots}}=2$$ so the tail in the nested radical is always less than $$2$$. We find that $$\alpha < \sqrt{2\sin^2 1+\sqrt{2\sin^2 2 + \sqrt{2\sin^2 3 + \cdots +\sqrt{2\sin^2 13+2}}}}\approx 1.7724371077589929$$ so that $$\alpha^2 < 3.1415333009610635 < \pi$$

Here's my python script:

from math import sqrt, sin

def f(n):
for k in range(n-1,0,-1):

1.7724371077589929 3.1415333009610635

• That doesn't look $< \pi$ to me...? – Mees de Vries Oct 24 at 15:18
• However, changing 7 into 13 in your script gives $\alpha^2 < \pi$ as desired. – Mees de Vries Oct 24 at 15:19