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?
 A: Actually you have written the identity incorrectly. The original, by Ramanujan, stated:
$$ 1 + 4\sin 10^\circ = \sqrt{11 - 2\sqrt{11 + 2\sqrt{11 - 2(1 + 4\sin 10^\circ) \dots }}} $$
Edit: note the period of signs is $-, +, -, -, +, -, -, +, -, \dots$
or letting $ x = 1 + 4\sin 10^\circ $
$$ x = \sqrt{11 - 2\sqrt{11 + 2\sqrt{11 - 2x \dots }}} $$
According to WA, we have $x \approx 1.695$ and again $1 + 4 \sin 10^\circ \approx 1.695$ 
A: Let $x=\sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}$ and $ y=\sqrt{11+2\sqrt{11-2\sqrt{11+2\sqrt{11-\cdots}}}}$ then
\begin{eqnarray*}
x=\sqrt{11-2y} \\
y=\sqrt{11+2x}
\end{eqnarray*}
so $x^2=11-2y$ & $y^2=11+2x$ ... after a little algebra ...
\begin{eqnarray*}
x^4-22x^2-8x+77=0 \\
(x^2+2x-7)(x^2+2x-11)=0
\end{eqnarray*}
So we have the possible solutions $x=-1 \pm 2 \sqrt{2}$ & $x=1 \pm 2 \sqrt{3}$.
One can verify that 
\begin{eqnarray*}
\sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=\color{red}{2 \sqrt{2}-1}.
\end{eqnarray*}
If the $+$ and $-$'s alternate then the above value is correct ... Ramanujan actually does the $(-,+,-)$ repeating every $3$ times ... then the result is 
\begin{eqnarray*}
\sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11\color{red}{-}\cdots}}}}=1+ \sin(10^o).
\end{eqnarray*}
A: 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
