Is there an interpretation for this trigonometric identity?

Question

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 $\displaystyle \frac{7}{64}$.

Answer 1

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

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.

Answer 2

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 $z=|z|(\cos{\Phi}+i \sin{\Phi})$.