Монгол Бодлогын Сан

Заавар:$$\sin\alpha\cdot\cos\beta=\dfrac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]$$

Бодолт:\begin{align*} \dfrac{\tg60^{\circ}}{\sin40^{\circ}}&+4\cos100^{\circ} = \dfrac{\sqrt3+4\cos100^\circ\cdot\sin40^\circ}{\sin 40^\circ}\\ & = \dfrac{\sqrt3+2(\sin140^\circ-\sin 60^\circ)}{\sin 40^\circ}=\dfrac{2\sin (180^\circ-40^\circ)}{\sin 40^\circ}=2 \end{align*}