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

  • 1
    $\begingroup$ There are no "$i$"s in the summations defining $A$ and $B$. Also, there's a close-parenthesis missing in the target expression. $\endgroup$ – Blue Aug 14 at 4:57

Simplify the question to

$$A - \cos(\pi/2018)\sin(\pi/2018)-...-\cos(\pi/2018)\sin(3027\pi/2018)+ \sin(\pi/2018)\cos(\pi/2018)+...+\sin(\pi/2018)\cos(3027\pi/2018)$$


$$=\sin(3027\pi/2018) = \sin(3π/2)=-1$$ (we used $\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$)


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