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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?

Blue
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Beedassy Lekraj
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  • https://artofproblemsolving.com/community/c6h46202p494824 I believe it should be $$-4\sin\dfrac\pi{11}$$ – lab bhattacharjee Jun 07 '20 at 18:33
  • Thanks for comment lab bhattacharjee, but this version of the identity has no mistake ; The link you provide was already included in the previous disussion thread ! – Beedassy Lekraj Jun 07 '20 at 19:00

2 Answers2

2

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: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$

Quanto
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I don't see how it fails! The approach using the $11$th roots of unity works well here too. With $x=\exp(2i\pi/11)$ so that $x^{11} =1 $ and $x^{11/2}=-1$. So

$$4i\sin(\pi/11)= 2(x^{1/2} - x^{-1/2}) = 2(x^5-x^6)$$ $$i\tan(4\pi/11) = \frac{x^4-x^{40}}{x^4+1} = x^4-x^8+x-x^5+x^9-x^2+x^6-x^{10}+x^3-x^7$$

Adding the two: $$i\tan(4\pi/11)+i\sin(\pi/11) = \underbrace{x^9 + x^5 + x^4 + x^3 + x}_u - (\underbrace{x^{10} + x^8 + x^7 + x^6 + x^2}_v)$$

Now, $u +v = -1$ and $uv = 3$. So $u-v = i\sqrt{11}$ where we ignore the negative value since the ratios are positive.

Sahaj
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