Not as easy as it look!

I took a rather convoluted path to find this solution... there has to be a better way.

Let $t=\frac{2\pi}{7}$ throughout, and let $S$ be the sum we seek.

As we compute $S^2$ , we are in for a pleasant surprise: all the mixed terms $2\sin{pt}\sin{qt}$ cancel out (use $2\sin{a}\sin{b}=\cos(a-b)-\cos(a+b)$ ), and we are left with $S^2=(\sin{t})^2+(\sin{2t})^2+(\sin{4t})^2$ $=\frac{1}{2}\left(3-\cos{2t}-\cos{4t}-\cos{t}\right).$ Since they are the real parts of the 7th roots of unity, we have $1+\cos{t}+\cos{2t}+...+\cos{6t}=0$ and therefore $\cos{2t}+\cos{4t}+\cos{t}$ $=\cos{5t}+\cos{3t}+\cos{6t}=-\frac{1}{2}$ so that $S^2=\frac{1}{2}(3+ 1/2) = 7/4$ and $S=\boxed{\frac{\sqrt{7}}{2}}$ (We know that S is positive since $\sin{t}>1/2, \sin{2t}>0,$ and $\sin{4t}>-1/2$ )