Complenometry

Let $t=\cos(\frac{2\pi}{7})+i \sin(\frac{2\pi}{7})$ be the seventh root of unity.

Let $A=t+t^{2}+t^{4}$ and $B=t^{3}+t^{5}+t^{6}$

The given expression= $\Im(A)$

Also, $A+B=-1$ ... $(a)$

Now, $AB=(t+t^{2}+t^{4})(t^{3}+t^{5}+t^{6})$

$=t^{4}+t^{6}+t^{7}+t^{5}+t^{7}+t^{8}+t^{7}+t^{9}+t^{10}$

$=2+(1+t+t^{2}+t^{3}+t^{4}+t^{5}+t^{6})=2+0=2$

Now, $(A-B)^{2}=(A+B)^{2}-4AB=1-8=-7$

So, $(A-B)=i\sqrt{7}$ .... $(b)$

Now,one can easily see $A=\frac{1}{2}+i\frac{\sqrt{7}}{2}$ .

Hence,the given expression= $\frac{\sqrt{7}}{2}$ .

Note:The reason why i neglected the negative sign in $(b)$ is as follows-

Obviously, $\sin(\frac{2π}{7})$ and $\sin(\frac{4π}{7})$ . are both greater than zero.

Again, $\sin(\frac{8\pi}{7})=\sin(\pi-\frac{8\pi}{7})=sin(-\frac{\pi}{7})=-\sin(\frac{\pi}{7})$

Since, $\sin$ function is an increasing function in $[0,\frac{\pi}{2}]$ ,

$\sin(\frac{2\pi}{7})>\sin(\frac{\pi}{7})$ .So the given expression is greater than $0$