What is the value of the expression $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of 
$\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$ 
Do I just use 
$\cos(a-b)-\cos(a+b) = 2\sin(a)\sin(b)$
 A: You can use complex numbers: let $x=2\pi/7$, so
$$
\sin x=\frac{e^{ix}-e^{-ix}}{2i}
$$
and similarly for the other terms:
$$
\frac{(e^{ix}-e^{-ix})(e^{2ix}-e^{-2ix})}{-4}
+
\frac{(e^{2ix}-e^{-2ix})(e^{4ix}-e^{-4ix})}{-4}
+
\frac{(e^{4ix}-e^{-4ix})(e^{ix}-e^{-ix})}{-4}
$$
Expanding the numerator we find
$$
e^{3ix}-e^{ix}-e^{-ix}+e^{-3ix}+
e^{6ix}-e^{2ix}-e^{-2ix}+e^{-6ix}+
e^{5ix}-e^{-3ix}-e^{3ix}+e^{-5ix}
$$
which simplifies to
$$
e^{6ix}+e^{5ix}-e^{2ix}-e^{ix}+
e^{-6ix}+e^{-5ix}-e^{-2ix}-e^{-ix}
$$
Now use the fact that $e^{6ix}=e^{-ix}$ and $e^{5ix}=e^{-2ix}$
A: $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}$
$\sin \frac{8\pi}{7} = \sin \frac{-6 \pi}{7}$
-> $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{-6\pi}{7}$

$\sin a \sin (b-c) + \sin b \sin (c-a)+ \sin c \sin (a-b)$ is always equal to "$0$".

Can you see it?
A: As you observed, 
$$\sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]$$
Applying that observation here yields
\begin{align*}
\sin\left(\frac{2\pi}{7}\right)&\sin\left(\frac{4\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right)\sin\left(\frac{8\pi}{7}\right) + \sin\left(\frac{8\pi}{7}\right)\sin\left(\frac{2\pi}{7}\right)\\  
& = \frac{1}{2}\left[\cos\left(-\frac{2\pi}{7}\right) - \cos\left(\frac{6\pi}{7}\right) + \cos\left(-\frac{4\pi}{7}\right) - \cos\left(\frac{12\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) - \cos\left(\frac{10\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(-\frac{2\pi}{7}\right) + \cos\left(-\frac{4\pi}{7}\right) - \cos\left(\frac{12\pi}{7}\right) - \cos\left(\frac{10\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) - \cos\left(\frac{12\pi}{7}\right) - \cos\left(\frac{10\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\pi - \frac{3\pi}{7}\right) - \cos\left(2\pi - \frac{2\pi}{7}\right) - \cos\left(\pi + \frac{3\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(\frac{2\pi}{7}\right) - \cos\left(\frac{3\pi}{7}\right) - \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{3\pi}{7}\right)\right]\\
& = 0
\end{align*}
where we have used the identities
\begin{align*}
\cos(-\theta) & = \cos\theta\\
\cos(\pi - \theta) & = -\cos\theta\\
\cos(\pi + \theta) & = -\cos\theta\\
\cos(2\pi - \theta) & = \cos\theta
\end{align*}
