5. Let $x=\cos \frac{\pi}{13}, y=\frac{3 \pi}{13}, z=\frac{9 \pi}{13}$, then $x y+y z+z x=$
Solution: $-\frac{1}{4}$

Let $\frac{\pi}{13}=\alpha$, then
$$
x=\cos \alpha, y=\cos 3 \alpha, z=\cos 9 \alpha
$$

And
$$
\begin{aligned}
x y+y z+z x= & (\cos \alpha \cos 3 \alpha+\cos 3 \alpha \cos 9 \alpha+\cos 9 \alpha \cos \alpha) \\
& =\frac{1}{2}((\cos 2 \alpha+\cos 4 \alpha)+(\cos 6 \alpha+\cos 12 \alpha)+(\cos 8 \alpha+\cos 10 \alpha)) \\
& =\frac{1}{2 \sin \alpha} \sin \alpha(\cos 2 \alpha+\cos 4 \alpha+\cdots+\cos 12 \alpha) \\
& =\frac{1}{4 \sin \alpha}(\sin 13 \alpha-\sin \alpha)=-\frac{1}{4}
\end{aligned}
$$