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..
Best Answer
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.