Proof: Let $A=\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}$, and construct its complementary conjugate $B=\sin \frac{\pi}{7}-\sin \frac{2 \pi}{7}+$ $\sin \frac{3 \pi}{7}$

Then $A^{2}+B^{2}=3-4 \cos \frac{\pi}{7}+2 \cos \frac{2 \pi}{7}$,
$A^{2}-B^{2}=-\cos \frac{\pi}{7}+3 \cos \frac{2 \pi}{7}-5 \cos \frac{3 \pi}{7}$.
(1) + (2) gives $2 A^{2}=3-5 A$.

Solving for $A$ yields $A=\frac{1}{2}$ or $A=-3$ (discard the latter).