Монгол Бодлогын Сан

Заавар: $$\cos(\pi-\alpha)=-\cos\alpha$$

Бодолт: \begin{align*} \text{Илэрх.}&=\cos^2\dfrac{\pi}{3}-\sqrt3\sin\dfrac{\pi}{6}\cdot\cos\dfrac{5\pi}{6}\\ &=\cos^2\dfrac{\pi}{3}-\sqrt3\sin\dfrac{\pi}{6}\cdot\left(-\cos\left(\pi-\dfrac{5\pi}{6}\right)\right)\\ &=\cos^2\dfrac{\pi}{3}-\sqrt3\sin\dfrac{\pi}{6}\cdot\left(-\cos\dfrac{\pi}{6}\right)\\ &=\cos^2\dfrac{\pi}{3}+\sqrt3\sin\dfrac{\pi}{6}\cdot\cos\dfrac{\pi}{6}\\ &=\left(\dfrac{1}{2}\right)^2+\sqrt3\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt3}{2}\\ &=\dfrac14+\dfrac34=1 \end{align*}