Монгол Бодлогын Сан

Заавар:$$\cos2\alpha=1-2\sin^2\alpha$$ $$\ctg\alpha=\tg(90^\circ-\alpha)$$ $$\cos(180^\circ+\alpha)=-\cos\alpha$$

Бодолт:\begin{align*} A&=\dfrac{2\sin^220^{\circ}-1}{3\ctg 65^{\circ}\cdot \cos^2205^{\circ}}\\ &=-\dfrac{\cos 40^{\circ}}{3\tg 25^{\circ}\cdot (-\cos25^{\circ})^2} & \color{red}{\leftarrow} & \color{red}{\tg\alpha=\dfrac{\sin\alpha}{\cos\alpha}}\\ &=-\dfrac{\cos40^\circ}{3\cdot\dfrac{\sin25^\circ}{\cos25^\circ}\cdot\cos^225^\circ}\\ &=-\dfrac{\cos40^\circ}{\dfrac{3}{2}\cdot(2\cos25^\circ\sin25^\circ)} & \color{red}{\leftarrow} & \color{red}{\sin2\alpha=2\sin\alpha\cos\alpha}\\ &=-\dfrac{2\cdot\cos40^\circ}{3\cdot\sin 50^\circ}=-\dfrac23 & \color{red}{\leftarrow} & \color{red}{\cos40^\circ=\sin 50^\circ} \end{align*}