The area of a triangle determined by two diagonals at a vertex of a regular heptagon 
In a circle of diameter 7, a regular heptagon is drawn inside of it. Then, we shade a triangular region as shown:

What’s the exact value of the shaded region, without using trigonometric constants?

My attempt
I tried to solve it with the circumradius theorem : $A=(abc)/(4R)$, where $a$, $b$, and $c$ are the three sides, and $R$ is the circumradius of the triangle. However, I needed to find the exact value of $\cos(5\pi/14)$, $\cos(4\pi/7)$, and $\sin(5\pi/14)$. Finally, I found an explicit formula for this particular triangle, but the proof was missing.
You can find the formula in Wikipedia's "Heptagonal triangle" entry.
 A: Expressing the sides $a,b,c$ via the sines of corresponding central angles one obtains:
$$
A=\frac{abc}{4R}=2R^2\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{4\pi}7.\tag1
$$
For a product of sines we have the following theorem:
$$
\prod_{0<m_i<n}2\sin\frac{\pi m_i}{n}=n.
\tag2
$$
Therefore: $$2^6\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{3\pi}7\sin\frac{4\pi}{7}\sin\frac{5\pi}7\sin\frac{6\pi}{7}=\left(8\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{4\pi}7\right)^2=7.$$
A: Here is a proof using complex numbers geometry : Let 
$$\zeta=\exp(2i \pi/7).$$
For the sake of simplicity, we will work in a heptagon inscribed in a unit circle.
The vertices of the triangle can be considered as being 
$$1, \ \ \ \zeta^2, \ \ \ \zeta^3$$
Using the (oriented) area formula one can find in one of my recent answers, we get :
$$A = -\frac{1}{4i}\begin{vmatrix}1 & 1 & \bar{1} \\ 1 & \zeta^2 & \bar{\zeta}^2 \\ 1 & \zeta^3 & \bar{\zeta}^3 \end{vmatrix}$$
$$=-\frac{1}{4i}\left(-(\zeta-\bar{\zeta})-(\zeta^2-\bar{\zeta}^{-2})+(\zeta^3-\bar{\zeta}^{-3})\right)$$
$$=-\frac{1}{4i} \ 2i \ \left(-\sin \frac{2\pi}{7}-\sin \frac{4\pi}{7}+\sin \frac{6\pi}{7}\right)=-\frac12 \left(-\frac{\sqrt{7}}{2}\right)$$
(see for example here), or alternatively take the imaginary part of relationship
$$1+2(1+\zeta+\zeta^2+\zeta^4)=\sqrt{7}i$$
see lemma 1 p. 128 of this very interesting document, involving Gauss sums. This document is especially interesting because it displays many sides of heptagonal triangles using complex geometry. For Gauss sums, see as well this question and its answer.
A: It boils down to
$$I= \sin\frac\pi7  \sin\frac{2\pi}7 \sin\frac{3\pi}7= \frac{\sqrt7}8
$$
Let $a =\frac\pi7$ and evaluate
\begin{align}
I^2 & = (1-\cos^2 a )  (1-\cos^2 2a )  (1-\cos^2 3a )\\
 & =\frac18 (1-\cos 2a )  (1-\cos 4a )  (1-\cos 8a )\\
 & =\frac18 (1-\cos 2a \cos 4a \cos 8a )\tag1\\
\end{align}
where the following equality via $2\cos x\cos y= \cos(x+y)+ \cos(x-y)$ is used
\begin{align}
\cos 2a \cos 4a + \cos 4a \cos 8a+ \cos 8a \cos 2a
 =\cos 2a +\cos 4a + \cos 8a\\
\end{align}
Note that 
$$\cos 2a \cos 4a \cos 8a= \frac{\sin4a \cos 4a \cos 8a}{2\sin 2a}
= \frac{\sin16a}{8\sin 2a}=\frac18
$$
Thus, (1) yields
$$I = \sqrt{\frac18(1-\frac18)}= \frac{\sqrt7}8$$
