<s>[INST] 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..
 [/INST] 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.
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