Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach : 
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) $
I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks..

Here's my approach, it only use basic trigonometry identities (i.e, it requires no complex numbers), so it's quite lengthy. I would be glad if you guys can help me shorten it a bit.


*

*Use Product to Sum Formula to prove that:


$\displaystyle \begin{align*}& \ \cos\left(\frac{2\pi}{7} \right) \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{2\pi}{7} \right)\cos\left(\frac{8\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right)\cos\left(\frac{8\pi}{7} \right)\\
= & \ \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{6\pi}{7} \right) \\
= & \ \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{8\pi}{7} \right)\end{align*}$


*

*Let $\displaystyle A = \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{8\pi}{7} \right)$. Prove that $A < 0$, then square $A$ and using (1) along with Power Reduction Formula to prove that $A$ is a solution to the equation: $\displaystyle X^2 - \frac{5}{2}X - \frac{3}{2} = 0$. Hence $\displaystyle A = -\frac{1}{2}$.

*Again use the Product to Sum Formula to prove that $\displaystyle \sin\left(\frac{2\pi}{7} \right) \sin\left(\frac{4\pi}{7} \right) + \sin\left(\frac{2\pi}{7} \right)\sin\left(\frac{8\pi}{7} \right) + \sin\left(\frac{4\pi}{7} \right)\sin\left(\frac{8\pi}{7} \right) = 0$.

*Let $\displaystyle B = \sin\left(\frac{2\pi}{7} \right) + \sin\left(\frac{4\pi}{7} \right) + \sin\left(\frac{8\pi}{7} \right)$. Prove that $B > 0$. Using the result in (3), we have $\displaystyle B^2 = \sin^2\left(\frac{2\pi}{7} \right) + \sin^2\left(\frac{4\pi}{7} \right) + \sin^2\left(\frac{8\pi}{7} \right)$, using Power Reduction Formula, and (2), you'll get $\displaystyle B^2 = \frac{3}{2} - \frac{1}{2} A = \frac{7}{4}$. From the fact that $B > 0$, we'll have $\displaystyle B = \frac{\sqrt{7}}{2}$.
The above are just kind of hints, and you'll have to apply some trig formulae (identities) to get the result.
