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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?

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  • $\begingroup$ $\sin(\color{red}{45}^o)=\frac{1}{\sqrt{2}}$ ? $\endgroup$ Commented Jun 10, 2017 at 13:50
  • $\begingroup$ I think you could show how you obtained $1+2\sqrt{2}$ for the left hand side since that seems to be the issue $\endgroup$
    – user12345
    Commented Jun 10, 2017 at 13:55
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    $\begingroup$ I got $2 \sqrt{2} - 1$ for the LHS. $\endgroup$
    – Dando18
    Commented Jun 10, 2017 at 13:58
  • $\begingroup$ @Dando18 Even W|A says that. $\endgroup$ Commented Jun 10, 2017 at 14:03
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    $\begingroup$ It is a more long for comment but I write it: $$2\sqrt(2)-1=\sqrt{(2\sqrt(2)-1)^2}=\sqrt{9-4\sqrt(2)}=\sqrt {11-2(2\sqrt(2)+1)}$$ $$=\sqrt {11-2\sqrt {(2\sqrt(2)+1)^2}}=\sqrt {11-2\sqrt {9+4\sqrt(2)}}=\sqrt {11-2\sqrt {11+2(2\sqrt(2)-1)}}$$ $$ =\sqrt {11-2\sqrt {11+2(\sqrt{(2\sqrt(2)-1)^2}}}=\sqrt {11-2\sqrt {11+2\sqrt {11-2\sqrt{\cdots}}}} $$ Maybe useful. $\endgroup$
    – Amin235
    Commented Jun 10, 2017 at 15:04

3 Answers 3

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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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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*}

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  • $\begingroup$ From your solution it reveals that $\sin 10^o = \frac {\sqrt {2} -1} {2} \approx 0.20711$ where as the actual value of $\sin 10^o$ is approximately equal to $0.1736$.Why is this happening? $\endgroup$ Commented Jun 10, 2017 at 14:35
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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

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