3. 16 Evaluate
(1) $\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cdots+\cos \frac{2(n-1) \pi}{n}$;
(2) $\sin \frac{2 \pi}{n}+\sin \frac{4 \pi}{n}+\cdots+\sin \frac{2(n-1) \pi}{n}$.
[Solution] Let $i=\sqrt{-1}$,
By De Moivre's theorem, we have:
$$
\begin{array}{l}
\left.\left(\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}\right)+\cdots+\cos \frac{2(n-1) \pi}{n}\right) \\
+i\left(\sin \frac{2 \pi}{n}+\sin \frac{4 \pi}{n}+\cdots+\sin \frac{2(n-1) \pi}{n}\right) \\
=\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)+\left(\cos \frac{4 \pi}{n}+i \sin \frac{4 \pi}{n}\right)+\cdots \\
+\left(\cos \frac{2(n-1) \pi}{n}+i \sin \frac{2(n-1) \pi}{n}\right) \\
=\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)+\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)^{2}+\cdots \\
+\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)^{n-1} \\
=\frac{\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)-\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)^{n}}{1-\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)} \\
=\frac{\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)-(\cos 2 \pi+i \sin 2 \pi)}{1-\left(\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\right)}=-1 . \\
\end{array}
$$

Therefore, $\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cdots+\cos \frac{2(n-1) \pi}{n}=-1$,
$$
\sin \frac{2 \pi}{n}+\sin \frac{4 \pi}{n}+\cdots+\sin \frac{2(n-1) \pi}{n}=0 .
$$