Монгол Бодлогын Сан

Заавар: $$\sin2\alpha=2\cos\alpha\sin\alpha$$ $$\sin(90^\circ-\alpha)=\cos\alpha$$ $$\sin\alpha=\sin(180^\circ-\alpha)$$

Бодолт: \begin{align*} \text{Илэрх.}&=\sin70^{\circ}\cdot\sin50^{\circ}\cdot\sin10^{\circ}\\ &=\dfrac{\sin70^{\circ}\cdot\sin50^{\circ}\cdot(2\cos10^\circ\sin10^{\circ})}{2\cos10^\circ}\\ &=\dfrac{\sin70^{\circ}\cdot\sin50^{\circ}\cdot\sin20^\circ}{2\cos10^\circ}\\ &=\dfrac{\sin50^\circ\cdot(2\cdot \sin70^\circ\cdot\cos70^\circ)}{4\cos10^\circ}\\ &=\dfrac{\sin50^\circ\cdot\sin140^\circ}{4\cos10^\circ}=\dfrac{\sin50^\circ\cdot\sin40^\circ}{4\cos10^\circ}\\ &=\dfrac{2\sin50^\circ\cos50^\circ}{8\cos10^\circ}=\dfrac{\sin100^\circ}{8\cos10^\circ}\\ &=\dfrac{\sin80^\circ}{8\cos10^\circ}=\dfrac{\cos10^\circ}{8\cos10^\circ}=\dfrac18 \end{align*}