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.

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
$$I = \sqrt{\frac18(1-\frac18)}= \frac{\sqrt7}8$$
