(7) $\cos \frac{\pi}{15}-\cos \frac{2 \pi}{15}-\cos \frac{4 \pi}{15}+\cos \frac{7 \pi}{15}=$ $\qquad$
(7) $-\frac{1}{2}$ Hint:
$$
\begin{aligned}
\text { Original expression } & =\left(\cos \frac{\pi}{15}+\cos \frac{7 \pi}{15}\right)-\left(\cos \frac{2 \pi}{15}+\cos \frac{4 \pi}{15}\right) \\
& =2 \cos \frac{4 \pi}{15} \cos \frac{\pi}{5}-2 \cos \frac{\pi}{15} \cos \frac{\pi}{5} \\
& =2 \cos \frac{\pi}{5}\left(\cos \frac{4 \pi}{15}-\cos \frac{\pi}{15}\right) \\
& =-4 \cos \frac{\pi}{5} \sin \frac{\pi}{6} \sin \frac{\pi}{10} \\
& =-2 \cos \frac{\pi}{5} \sin \frac{\pi}{10}=-\frac{1}{2} .
\end{aligned}
$$
(Note: From $\sin 72^{\circ}=2 \sin 36^{\circ} \cos 36^{\circ}=4 \sin 18^{\circ} \cos 18^{\circ} \cos 36^{\circ}$, we get $\sin 18^{\circ} \cos 36^{\circ}=\frac{1}{4}$, i.e., $\cos \frac{\pi}{5} \sin \frac{\pi}{10}=\frac{1}{4}$.)