Find $\sin(2π/7)+\sin(4π/7)+ \sin(8π/7)$

Question

Find $$\sin\left(\frac{2π}{7}\right)+\sin\left(\frac{4π}{7}\right)+\sin\left(\frac{8π}{7}\right).$$

I already know of two methods. The first is in which we let $2π=7\theta$ and proceed as such-

$\begin{align} 2\pi-3\theta=4\theta\\ \implies\sin(3\theta)+\sin(4\theta)=0\\ \implies-4\sin^3\theta+3\sin\theta+ 4\sin\theta\cos\theta(1-2\sin^2\theta)=0 \end{align}$

On solving further we get a cubic polynomial in $\sin^2\theta$. We can then find the required sum from the sum of roots and some algebra.

I also know of the method that involves complex numbers and the roots of unity.

What I am interested in is a method involving trigonometric identities. I tried a few things. In particular I noticed one fact that might be of significance, that is the sum of the angles $\frac{2π}7, \frac{4π}7$ and $\frac{8π}7$ is precisely $=2π$. However, I was not able to make much progress. Can someone please help me with this?

Answer 1

Since $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}>0,$ we obtain:

$$\begin{aligned} \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} &=\sqrt{\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\right)^2} \\ &=\sqrt{\tfrac{1-\cos\tfrac{4\pi}{7}}{2}+\tfrac{1-\cos\tfrac{8\pi}{7}}{2}+\tfrac{1-\cos\frac{16\pi}{7}}{2}+\cos\tfrac{2\pi}{7}-\cos\tfrac{6\pi}{7}+\cos\tfrac{6\pi}{7}-\cos\tfrac{10\pi}{7}+\cos\tfrac{4\pi}{7}-\cos\tfrac{12\pi}{7}} \\ &=\sqrt{\frac{3}{2}-\frac{1}{2}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)} \\ &=\sqrt{\frac{3}{2}-\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{4\sin\frac{\pi}{7}}} \\ &=\sqrt{\frac{3}{2}-\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{4\sin\frac{\pi}{7}}} \\ &=\frac{\sqrt7}{2} \end{aligned}$$