Монгол Бодлогын Сан

Заавар:$$\sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta=\sin(\alpha-\beta)$$ $$\sin 2x=2\sin x\cos x$$

Бодолт:\begin{align*} \dfrac{\sin3x}{\sin x}&-\dfrac{\cos3x}{\cos x}=\dfrac{\sin 3x\cdot\cos x-\cos3x\cdot\sin x}{\sin x\cdot\cos x}\\ &=\dfrac{\sin(3x-x)}{\sin x\cos x}=\dfrac{\sin 2x}{\sin x\cdot\cos x}=\dfrac{2\cdot\sin x\cdot\cos x}{\sin x\cdot\cos x}=2\\ \end{align*}