Ramanujan and his problem Following was proposed by Ramanujan:
$ \sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=1+4\sin(10^o)$
Working on this I got the radical on the left equal to $(1+2\sqrt{2})$ implying that $\sin(10^o)=1/\sqrt{2}$
How is this possible? What is wrong here?

Inspired by Ramanujan's above formula, I added some similar results here:
$ $
$ 4 \cos \left(\frac{4 \pi }{9}\right)+1=\sqrt{11-2 \sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{4 \pi }{9}\right)+1\right)}}} $
$ 4 \cos \left(\frac{2 \pi }{9}\right)+1=\sqrt{11+2 \sqrt{11-2 \sqrt{11-2 \left(4 \cos \left(\frac{2 \pi }{9}\right)+1\right)}}} $
$ 4 \cos \left(\frac{\pi }{9}\right)-1=\sqrt{11-2 \sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{9}\right)-1\right)}}} $
$  $
$ 4 \cos \left(\frac{\pi }{4}\right)+1=\sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{\pi }{4}\right)+1\right)}} $
$ 4 \cos \left(\frac{\pi }{4}\right)-1=\sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{4}\right)-1\right)}} $
$  $
$ 4 \cos \left(\frac{\pi }{6}\right)+1=\sqrt{11+2 \left(4 \cos \left(\frac{\pi }{6}\right)+1\right)} $
$ 4 \cos \left(\frac{\pi }{6}\right)-1=\sqrt{11-2 \left(4 \cos \left(\frac{\pi }{6}\right)-1\right)} $
$ $
$ 4 \cos \left(\frac{\pi }{4}\right)+1=\sqrt{7+2 \left(4 \cos \left(\frac{\pi }{4}\right)+1\right)} $
$ 4 \cos \left(\frac{\pi }{4}\right)-1=\sqrt{7-2 \left(4 \cos \left(\frac{\pi }{4}\right)-1\right)} $
$ $
Other similar results can be found here:
https://math.stackexchange.com/a/4233146/954936
https://math.stackexchange.com/a/4232525/954936
http://eslpower.org
http://eslpower.org/Notebook.htm
