Монгол Бодлогын Сан

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

Бодолт: $$\begin{align*} \lim\limits_{x\to \frac{\pi}{3}}\frac{\sin(x-\frac{\pi}{3})}{1-2\cos x}&=\lim\limits_{x\to \frac{\pi}{3}}\dfrac{2\sin(\frac{x}{2}-\frac{\pi}{6})\cos(\frac{x}{2}-\frac{\pi}{6})}{2\cos\frac{\pi}{3}-2\cos x}\\ &=\lim\limits_{x\to \frac{\pi}{3}}\dfrac{\sin(\frac{x}{2}-\frac{\pi}{6})\cos(\frac{x}{2}-\frac{\pi}{6})}{2\sin\frac{\frac\pi3+x}{2}\sin\frac{x-\frac\pi3}{2}}\\ &=\lim\limits_{x\to \frac{\pi}{3}}\frac{\cos\Bigl(\frac{x}{2}-\frac{\pi}{6}\Bigr)} {2\sin\Bigl(\frac{x}{2}+\frac{\pi}{6}\Bigr)}\\ &=\frac{\cos 0}{2\cdot\sin\frac{\pi}{3}}=\dfrac{1}{\sqrt3} \end{align*}$$