Let $$A=\sum_{n=1}^{3027}\sin\frac{n\pi}{2018} \qquad B=\sum_{n=1}^{3027}\cos\frac{n\pi}{2018}$$ Evaluate $$A\left(1-\cos\frac{\pi}{2018}\right)+B\sin\frac{\pi}{2018}$$
I don't know how to do it. I used a smaller case. I changed $3027 \to 3$ and changed the denominator by $2$, and I noticed that the answer is $-1$. I don't know how to solve the bigger case.
I'll generalize the problem a bit, writing $m$ for $3027$ and $2\theta$ for $\pi/2018$.
Adapting the identities proven here, we can write $$\begin{align} A &:= \sum_{n=1}^{m}\sin 2n\theta = \frac{\sin(m+1)\theta}{\sin\theta}\;\sin m\theta \\[4pt] B &:= \sum_{n=1}^{m}\cos 2n\theta = \frac{\sin(m+1)\theta}{\sin\theta}\;\cos m\theta - 1 \end{align}$$ where we subtract $\cos 0=1$ from the cosine sum (and $\sin 0=0$ from the sine sum!) because our summations start at index $1$ instead of $0$. Then we have $$\begin{align} A \left(1-\cos 2\theta\right) + B\sin 2\theta &= \phantom{+}A\cdot 2 \sin^2\theta + B\cdot 2\sin\theta\cos\theta \tag{1}\\[4pt] &= \phantom{+}2\sin\theta\;\sin(m+1)\theta\;\sin m\theta \\ &\phantom{=}+ 2 \cos\theta\;\sin(m+1)\theta\;\cos m\theta - 2\sin\theta\cos\theta \tag{2}\\[4pt] &= \phantom{+}2\sin(m+1)\theta\;\left(\sin\theta\sin m\theta+\cos\theta\cos m\theta\right)-\sin 2\theta \tag{3}\\[4pt] &= \phantom{+}2\sin(m+1)\theta\;\cos(m-1)\theta -\sin 2\theta \tag{4} \\[4pt] &= \phantom{+}\left( \sin 2m\theta + \sin 2\theta \right) - \sin 2\theta \tag{5} \\[4pt] &= \phantom{+} \sin 2m\theta \tag{6} \end{align}$$ where $(1)$ uses the double-angle identities, $(4)$ uses the angle-difference identity for cosine, and $(5)$ invokes a somewhat-lesser-known product-to-sum identity. Restoring the specific values for the problem at hand, $$2\theta = \frac{\pi}{2018} \qquad m = 3027 \qquad\to\qquad 2m\theta = \frac{3\pi}{2} \qquad\to\qquad \sin 2m\theta = -1 \tag{$\star$}$$
It's a bit of an identity slog, but there it is. $\square$
