5. Given $\alpha=\frac{2 \pi}{1999}$, find the value: $\cos \alpha \cos 2 \alpha \cos 3 \alpha \cdots \cos 999 \alpha$.
5. Let $P=\cos \alpha \cos 2 \alpha \cos 3 \alpha \cdots \cos 999 \alpha, Q=\sin \alpha \sin 2 \alpha \sin 3 \alpha \cdots \sin 999 \alpha$.

Thus, $2^{999} P Q=(2 \cos \alpha \sin \alpha)(2 \cos 2 \alpha \sin 2 \alpha)(2 \cos 3 \alpha \sin 3 \alpha) \cdots(2 \cos 999 \alpha \cdot \sin 999 \alpha)$
$$
\begin{array}{l}
=\sin 2 \alpha \sin 4 \alpha \sin 6 \alpha \cdots \sin 1998 \alpha[-\sin (2 \pi-1000 \alpha)][-\sin (2 \pi-1002 \alpha)] \cdots[-\sin (2 \pi-1998 \alpha)] \\
=\sin 2 \alpha \sin 4 \alpha \cdots \sin 998 \alpha \sin 999 \alpha \sin 997 \alpha \cdots \sin \alpha=Q
\end{array}
$$

Since $Q \neq 0$, we have $P=\frac{1}{2^{999}}$