In a similar vein as $\tan\frac{3\pi}{11} + 4\sin\frac{2\pi}{11} = \sqrt{11}$ discussed in this question is this identity:

$$\tan\frac{4\pi}{11} + 4\sin\frac{\pi}{11} = \sqrt{11}$$

Trying to adopt a method on the same line however lamentably fails. I wonder if the arguement of $11$th roots of unity can still be effectively employed in this case. Is there a way to adapt it or there could be possibly an easier way out to prove the result?


Let $a=\frac\pi{11}$ to evaluate

\begin{align} & 4 (\sin 2a -\sin a)-(\tan 4a - \tan3a )\\ = & 4\sin a (2\cos a -1)-\frac{\sin a}{\cos 3a\cos4a} =\frac{4\sin a}{\cos 7a+\cos a}\cdot A\tag1 \end{align}

where \begin{align} A = & 2\cos a( \cos7a + \cos a) - (\cos7a + \cos a)-\frac12 \\ = & \cos10a+ \cos8a+ \cos 6a +\cos 4a +\cos2a +\frac12\\ = & \frac12 \sum_{k=0}^{10} e^{i 2ka}=0 \\ \end{align}

Substitute $A=0$ into (1) to obtain

$$\tan\frac{4π}{11} + 4\sin\frac{π}{11} = \tan\frac{3π}{11} + 4\sin\frac{2π}{11}= \sqrt{11}$$

where How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$


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