Монгол Бодлогын Сан

Заавар: $$\begin{gather} \sin\alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}\\ \cos\alpha+\cos\beta=2\cos\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2} \end{gather}$$

Бодолт: $$\dfrac{\sin20^\circ-\sin30^\circ+\sin40^\circ}{\cos20^\circ-\cos30^\circ+\cos40^\circ}=\dfrac{(\sin20^\circ+\sin40^\circ)-\sin30^\circ}{(\cos20^\circ+\cos40^\circ)-\cos30^\circ}=$$ $$=\dfrac{\frac12\sin30^\circ\cos10^\circ-\sin30^\circ}{\frac12\cos30^\circ\cos10^\circ-\cos30^\circ}=\dfrac{\sin30^\circ}{\cos30^\circ}=\dfrac{1}{\sqrt3}$$