Strange approximation to $\sqrt{\pi}$

Question

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.

Answer 1

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):
    answer = 2+sqrt(2*sin(n)**2)
    for k in range(n-1,0,-1):
        answer = sqrt(2*(sin(k)**2)+answer)
    return answer

alpha = f(13)
print(alpha, alpha**2)

This prints

1.7724371077589929 3.1415333009610635