Монгол Бодлогын Сан

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

Бодолт: $$\begin{align*} \text{Илэрх.}&=\sin^2\dfrac{\pi}{6}+\sin\dfrac{\pi}{3}\cdot\cos\dfrac{\pi}{6}\\ &=\sin^2\dfrac{\pi}{6}+\cos\left(\pi-\dfrac{\pi}{3}\right)\cdot\cos\dfrac{\pi}{6} & & \color{red}{\leftarrow\sin\alpha=\cos(\pi-\alpha)}\\ &=\sin^2\dfrac{\pi}{6}+\cdot\cos^2\dfrac{\pi}{6}=1 & & \color{red}{\leftarrow\sin^2\alpha+\cos^2\alpha=1} \end{align*}$$