Монгол Бодлогын Сан

Заавар:$$\sin\alpha+\sin\beta=2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}$$ $$\cos\alpha+\cos\beta=2\cos\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}$$

Бодолт:\begin{align*} \text{Илэрх.}&=\dfrac{\sin2\alpha+\sin6\alpha+\sin10\alpha}{\cos2\alpha+\cos6\alpha+\cos10\alpha}=\dfrac{\sin6\alpha+\sin2\alpha+\sin10\alpha}{\cos6\alpha+\cos2\alpha+\cos10\alpha}\\ &=\dfrac{\sin6\alpha+2\sin6\alpha\cos(-4\alpha)}{\cos6\alpha+2\cos6\alpha\cos(-4\alpha)}=\dfrac{\sin6\alpha(1+2\cos4\alpha)}{\cos6\alpha(1+2\cos4\alpha)}=\tg6\alpha \end{align*}