Монгол Бодлогын Сан

Заавар: $$1+\cos2\alpha=2\cos^2\alpha$$ $$\sin2\alpha=2\sin\alpha\cos\alpha$$

Бодолт: \begin{align*} A&=\dfrac{1+\cos40^{\circ}+\cos80^{\circ}}{\sin 80^{\circ}+\sin 40^{\circ}}\cdot \tg 40^{\circ}\\ &=\dfrac{(1+\cos80^\circ)+\cos40^\circ}{\sin(2\cdot 40^\circ)+\sin40^\circ}\cdot\tg40^\circ\\ &=\dfrac{2\cos^2\alpha+\cos\alpha}{2\sin40^\circ\cos40^\circ+\sin40^\circ}\cdot\tg40^\circ\\ &=\dfrac{\cos40^\circ(2\cos40^\circ+1)}{\sin40^\circ(2\cos40^\circ+1)}\cdot\tg40^\circ\\ &=\dfrac{\cos40^\circ}{\sin40^\circ}\cdot\dfrac{\sin40^\circ}{\cos40^\circ}=1 \end{align*}