IMO 1963 - Problem 5

Problem: Prove that $cos(\frac{\pi}{7})-cos(\frac{2\pi}{7})+cos(\frac{3\pi}{7})=\frac{1}{2}$.
Solution: The LHS of the desired idetity equals $S=cos(\frac{\pi}{7})+cos(\frac{3\pi}{7})+\frac{5\pi}{7}$. Now
$S.sin(\frac{\pi}{7})=\frac{sin\frac{2\pi}{7}}{2}+\frac{sin\frac{4\pi}{7}-sin\frac{2\pi}{7}}{2}+\frac{sin\frac{6\pi}{7}-sin\frac{4\pi}{7}}{2}=\frac{sin\frac{6\pi}{7}}{2}\implies S=\frac{1}{2}$.