<s>[INST] 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.
 [/INST] 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: $$2^6\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{3\pi}7\sin\frac{4\pi}{7}\sin\frac{5\pi}7\sin\frac{6\pi}{7}=\left(8\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{4\pi}7\right)^2=7.$$
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