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A while ago I came across the following identity in an online math forum (of which I don't remember the name): $$\tan\left(\frac{\pi}{11}\right)+4\sin\left(\frac{3\pi}{11}\right)=\sqrt{11}.$$

It is not hard to give a proof by rewriting everything in terms of $\exp(i\pi/11)$ and applying a sequence of manipulations. I am wondering where this identity is coming from. Can somebody think of a geometric interpretation? Of an algebraic one?

Edit: Here's an example of what I mean by an algebraic interpretation: The identity $$\sin\left(\frac{\pi}{7}\right)\cdot\sin\left(\frac{2\pi}{7}\right)\cdot\sin\left(\frac{3\pi}{7}\right)=\frac{\sqrt{7}}{8}$$ expresses the fact that for the Chebyshev polynomial $$T_7(x)=x(64x^6-112x^4+56x^2-7)$$ the product of the roots $\displaystyle \sin\left(\frac{k\pi}{7}\right)$, $1\leq k<7$, of the second factor is equal to the normalized constant term $\displaystyle \frac{7}{64}$.

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I could only think of a direct trigonometric interpretation of the identity.

enter image description here

The radius of the circular sector is 1. The measures of the central angles and the lengths of the line segments are:

  1. The smaller angle: $\pi/11$ rad.
  2. The bigger angle: $3\pi/11$ rad.
  3. The red line segment: $\sqrt{11}$.
  4. The vertical black line segment: $4\sin(3\pi/11)$.
  5. The vertical light red segment: $\tan(3\pi/11)$.

The red line segment is the hypotenuse of the right triangle whose catheti are the line segment with length $\sqrt{10}$ and the orthogonal unit segment. The $\sqrt{10}$ line segment is the hypotenuse of the right triangle whose catheti are the horizontal line segment with length 3 and the vertical line segment with length 1.

Edited: The angle $\pi/11=2\pi/22$ is not constructible with compass and straightedge (Wikipedia, Constructible polygon ). Therefore the figure is an impossible construction with compass and straightedge only.

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Remember the algebra form is $a+bi$. You should calculate the first: $|z| = \sqrt{a^2+b^2}$, and then consider the angle "$\Phi$" (there are about 4 cases)... I will give you a hint: Remember complex numbers, and the end is $z=|z|(\cos{\Phi}+i \sin{\Phi})$.

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    $\begingroup$ Did you read the question before posting your answer? I think @carinii states quite clearly that this computatonal approach is not the point of this question. $\endgroup$ – t.b. May 23 '11 at 19:15

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