<s>[INST] 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?
 [/INST] 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 $1 + 4 \sin 10^\circ \approx 1.695$ 
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