<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] Let $\Phi_7(x)$ the minimal polynomial of $\xi=\exp\left(\frac{2\pi i}{7}\right)$. Its Galois group over $\mathbb{Q}$ is cyclic and generated by $\xi\mapsto\xi^3$, since $3$ is a generator of $\mathbb{Z}/(7\mathbb{Z})^*$. The quadratic residues $\!\!\pmod{7}$ are $1,2,4$ and the quadratic non-residues are $3,5,6$, hence:
$$ \alpha = \xi+\xi^2+\xi^4,\qquad \beta=\xi^3+\xi^5+\xi^6 $$
are two conjugated algebraic numbers with degree $\frac{6}{3}=2$. In particular, both $\alpha+\beta$ and $\alpha\beta$ are rational numbers:
$$ \alpha+\beta=\sum_{k=1}^{6}\xi^k=-1,\qquad \alpha\beta=3+\sum_{k=1}^{6}\xi^k=2 $$
and the minimal polynomial of $\alpha$ and $\beta$ is given by $x^2+x+2$.
It is straightforward to check that $\text{Im}(\alpha)>0$ and $\text{Im}(\beta)<0$, hence by computing the discriminant of  $x^2+x+2$ we have:
$$ \alpha-\beta = \sqrt{-7} $$
and the claim follows by De Moivre's formula, since, for instance, 
$$ \xi-\xi^6 = \exp\left(\frac{2\pi i}{7}\right)-\exp\left(-\frac{2\pi i}{7}\right)=2i\sin\left(\frac{2\pi}{7}\right).$$
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