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Ответ:

Применяем формулу разности синусов  

 [tex]\bf sin\alpha -sin\beta =2\cdot sin\dfrac{\alpha -\beta }{2}\cdot cos\dfrac{\alpha +\beta }{2}[/tex]   .

[tex]\bf sin\Big(\dfrac{\pi }{6}+\alpha \Big)-sin\Big(\dfrac{5\pi }{6}-\alpha \Big)=\\\\\\=2\cdot sin\dfrac{\Big(\dfrac{\pi}{6}+\alpha+\dfrac{5\pi }{6}-\alpha \Big)}{2}\cdot cos\dfrac{\dfrac{\pi}{6}+\alpha -\dfrac{5\pi }{6}+\alpha }{2}=\\\\\\=2\cdot sin\dfrac{\pi }{2}\cdot cos\Big(-\dfrac{4\pi }{6}+\alpha \Big)=2\cdot cos\Big(\alpha -\dfrac{2\pi }{3}\Big)[/tex]