Монгол Бодлогын Сан

Заавар: $\sin\alpha\sin\beta=\dfrac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)]$ томьёог ашиглавал $4\sin 70^\circ\sin10^\circ=2[\cos 60^\circ-\cos80^\circ]=1-2\cos80^\circ$ болно.

Бодолт: $$\dfrac1{2\sin10^{\circ}}-2\sin70^{\circ}=\dfrac{1-4\sin70^\circ\sin10^\circ}{2\sin10^\circ}=$$ $$=\dfrac{1-(1-2\cos80^\circ)}{2\sin10^\circ}=\dfrac{2\cos80^\circ}{2\sin10^\circ}=\dfrac{\cos(90^\circ-80^\circ)}{\cos10^\circ}=1$$