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
 A: $$
\begin{align}
\prod_{k=1}^3\sin\left(\frac{k\pi}7\right)^2
&=\prod_{k=1}^6\sin\left(\frac{k\pi}7\right)\tag{1}\\
&=-\frac1{64}\prod_{k=1}^6\left(e^{ik\pi/7}-e^{-ik\pi/7}\right)\tag{2}\\
&=\frac1{64}\prod_{k=1}^6\left(1-e^{-i2k\pi/7}\right)\tag{3}\\
&=\frac1{64}\lim_{z\to1}\prod_{k=1}^6\left(z-e^{-i2k\pi/7}\right)\tag{4}\\
&=\frac1{64}\lim_{z\to1}\frac{z^7-1}{z-1}\tag{5}\\[6pt]
&=\frac7{64}\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\sin(x)=\sin(\pi-x)$
$(2)$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$(3)$: pull $\prod\limits_{k=1}^6e^{ik\pi/7}=-1$ out of the product
$(4)$: $1=\lim\limits_{z\to1}z$
$(5)$: $\prod\limits_{k=1}^6\left(z-e^{-i2k\pi/7}\right)=\frac{z^7-1}{z-1}$
$(6)$: evaluate limit
Therefore, taking the square root of $(6)$, we get
$$
\prod_{k=1}^3\sin\left(\frac{k\pi}7\right)=\frac{\sqrt7}8\tag{7}
$$
A: Using $2\sin a\sin b=\cos(a-b)-\cos(a+b)$ and $2\sin a\cos b=\sin(a+b)+\sin(a-b)$, write
$$\sin \frac{\pi}7\cdot\sin \frac{2\pi}7\cdot \sin \frac{3\pi}7 = \frac12\left(\cos\frac{\pi}7-\cos\frac{3\pi}7\right)\sin\frac{3\pi}7=\frac14\left(\sin\frac{4\pi}7+\sin\frac{2\pi}7-\sin\frac{\pi}7\right)\\=\frac14\left(\sin\frac{2\pi}7+\sin\frac{4\pi}7+\sin\frac{8\pi}7\right)$$
Then have a look at this question: Trigo Problem : Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
A: 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.
A: Hint;
Assume
$$ \theta = (n\pi)/7$$
$$\implies 4\theta = n\pi -3\theta$$
Take the sine on both the sides
An apply the required manipulations
You should get a cubic in $\sin^2 \theta$
Since we have defined out $\theta$, we are quite aware what the roots are going to be
So the numerical part in the end divided by the coefficient of $x^3$ gives you
$$ {{[\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}}]}^2 = 7/64$$
That should be it
