Монгол Бодлогын Сан

Заавар: $$\sin\alpha\cdot\sin\beta=\dfrac12\{\cos(\alpha-\beta)-\cos(\alpha+\beta)\}$$ $$\sin\alpha\cdot\cos\beta=\dfrac12\{\sin(\alpha-\beta)+\sin(\alpha+\beta)\}$$ болон $\sin(180^\circ-\alpha)=\sin\alpha$ томьёог ашиглаарай!

Бодолт: \begin{align*} \sin 20^\circ\cdot\sin 40^\circ&\cdot\sin 80^\circ=\dfrac12(\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ))\cdot\sin80^\circ\\ &=\dfrac12(\cos20^\circ-\cos60^\circ)\cdot\sin80^\circ\\ &=\dfrac12\left(\cos20^\circ\sin 80^\circ-\dfrac12\sin80^\circ\right)\\ &=\dfrac12\left(\dfrac12\{\sin(80^\circ-20^\circ)+\sin(80^\circ+20^\circ)\}-\dfrac12\sin80^\circ\right)\\ &=\dfrac12\left(\dfrac12\cdot\dfrac{\sqrt3}{2}+\dfrac12\cdot\sin100^\circ-\dfrac12\sin80^\circ\right)=\dfrac{\sqrt{3}}{8} \end{align*}