Let $$A = \sum_{i=1}^{3027} \sin(\frac{\pi i}{2018})$$ $$B = \sum_{i=1}^{3027} \cos(\frac{\pi i}{2018})$$
Evaluate $$A(1-\cos(\frac{\pi}{2018}) + B(\sin(\frac{\pi}{2018}))$$
Dividing the summation into 3 parts, $$A = \sum_{i=1}^{1008} \sin(\frac{\pi i}{2018}) +\sum_{i=1010}^{2017} \sin(\frac{\pi i}{2018}) +\sum_{i=2018}^{3026} \sin(\frac{\pi i}{2018}) + \sin(\frac{1009\pi}{2018}) +\sin(\frac{2018\pi}{2018}) +\sin(\frac{3027\pi}{2018})$$
Using Trig properties, we have $$A = \sum_{i=1}^{1008} \sin(\frac{\pi i}{2018})$$
Doing the same logic for B
$$B = \sum_{i=1}^{1008} \cos(\frac{\pi i}{2018}) - 1$$ which is $$B = -\sum_{i=1}^{1008} \sin(\frac{\pi i}{2018})-1$$
After this, I cannot find a solution for the remaining expression
Simplify the question to
$$A - \cos(\pi/2018)\sin(\pi/2018)-...-\cos(\pi/2018)\sin(3027\pi/2018)+ \sin(\pi/2018)\cos(\pi/2018)+...+\sin(\pi/2018)\cos(3027\pi/2018)$$
$$=A+(\sin(-\pi/2018)+...+\sin(-3026\pi/2018))$$
$$=\sin(3027\pi/2018) = \sin(3π/2)=-1$$ (we used $\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$)