Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$ 
Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$

$\bf{My\; Try::}$ I have solved Using Direct formula::
$$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}$$
Now Put $n=7\;,$ We get 
 $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}\cdot \sin \frac{4\pi}{7}\cdot \sin \frac{5\pi}{7}\cdot \sin \frac{6\pi}{7}=\frac{7}{2^{7-1}}$$
So $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} =\frac{\sqrt{7}}{8}$$
Now my question is how can we solve it without using Direct Formula, Help me
Thanks

This is a Gauss sum in disguise, but it can be computed through elementary geometry, without resorting to algebraic number theory. Let us consider a regular heptagon $P_1 P_2 P_3 P_4 P_5 P_6 P_7$ with center $O$ and take $A=P_1, B=P_5, C=P_6$. The angles of $ABC$ are $\frac{\pi}{7},\frac{2\pi}{7},\frac{4\pi}{7}$. By Euler's formula and the sine theorem
$$ [ABC] = \frac{abc}{4R} = 2R^2 \sin(A)\sin(B)\sin(C) $$
but we also have
$$[ABC] = [AOB]+[BOC]-[AOC] = \frac{R^2}{2}\left(\sin\frac{4\pi}{7}+\sin\frac{2\pi}{7}-\sin\frac{6\pi}{7}\right) $$
hence the claim is equivalent to:
$$ \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = \frac{1}{2}\sqrt{7}. $$
and since the values of the sine function at $\frac{\pi}{7},\frac{2\pi}{7},\frac{4\pi}{7}$ are positive, the claim is also equivalent to:
$$ \left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{4\pi}{7}\right)\left(1-\cos\frac{8\pi}{7}\right)=\frac{7}{8}. $$
Chebyshev polynomials ensure that $\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{8\pi}{7}$ are algebraic conjugates over $\mathbb{Q}$, and they are the roots of $p(x)=8x^3+4x^2-4x-1$. By factoring such polynomial over $\mathbb{R}$ it follows that
$$ p(x) = 8x^3+4x^2-4x-1 = 8\left(x-\cos\frac{2\pi}{7}\right)\left(x-\cos\frac{4\pi}{7}\right)\left(x-\cos\frac{8\pi}{7}\right) $$
and by evaluating at $x=1$ we get
$$ 7 = p(1) = 8\left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{4\pi}{7}\right)\left(1-\cos\frac{8\pi}{7}\right) $$
QED.
