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?

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*}
