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

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
