Dokaži da za svaki prirodan broj nnn djeljiv s 4 vrijedi
sin2(2πn)+sin2(2⋅2πn)+⋯+sin2((n−1)⋅2πn)+sin2(n⋅2πn)=n2.\sin^2 \left(\frac{2\pi}{n}\right) + \sin^2 \left(2 \cdot \frac{2\pi}{n}\right) + \cdots + \sin^2 \left((n - 1) \cdot \frac{2\pi}{n}\right) + \sin^2 \left(n \cdot \frac{2\pi}{n}\right) = \frac{n}{2}.sin2(n2π)+sin2(2⋅n2π)+⋯+sin2((n−1)⋅n2π)+sin2(n⋅n2π)=2n.