<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] You can evaluate this by using complex methods.
Let $\alpha=e^{2\pi i/7}$.  Then
$$\sin\frac{2\pi}{7}=\frac{\alpha-\alpha^{-1}}{2i}\ ,\quad
  \sin\frac{4\pi}{7}=\frac{\alpha^2-\alpha^{-2}}{2i}\ ,\quad
  \sin\frac{8\pi}{7}=\frac{\alpha^4-\alpha^{-4}}{2i}\ .$$
Now let $S$ be the sum of these three numbers.  Write $S$ in terms of powers of $\alpha$ and calculate $S^2$.  Initially it's a mess, but you can use the relations
$$\alpha^7=1\quad\hbox{and}\quad \alpha^6+\alpha^5+\cdots+\alpha=-1$$
to show that it simplifies, amazingly, to
$$S^2=\frac{7}{4}\ .$$
It's not hard to see that $S$ is positive, so
$$S=\frac{\sqrt7}{2}\ .$$
Comment.  You can also write the sum as
$$S=\frac{1}{2i}\sum_{k=1}^6 \Bigl(\frac{k}{7}\Bigr)\alpha^k\ ,$$
where $(\frac{k}{7})$ is a Legendre symbol, and this connects the sum with some very interesting and often difficult mathematics.
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