Question

9 класс 691. помагите срошно​

Answer 1

Ответ:

a)

$sin(\frac{\pi }{6}-\alpha )-cos(\frac{\pi }{3}+\alpha )=\\sin(\frac{\pi }{6}-\alpha )-sin(\frac{\pi }{2} -(\frac{\pi }{3}+\alpha ))=\\sin(\frac{\pi }{6}-\alpha )-sin(\frac{\pi }{6}-\alpha )= 0$

б)

$sin(\frac{\pi }{4} +\alpha )+sin(\frac{\pi }{4} -\alpha )=\\sin\frac{\pi }{4}cos\alpha +sin\alpha cos\frac{\pi }{4}+sin\frac{\pi }{4}cos\alpha-sin\alpha cos\frac{\pi }{4}=\\2sin\frac{\pi }{4}cos\alpha = 2*\frac{\sqrt{2} }{2}*cos\alpha=\\\sqrt{2}cos\alpha$

Объяснение:

$sin(\frac{\pi }{2}-\alpha ) =cos\alpha \\cos(\frac{\pi }{2}-\alpha )=sin\alpha \\sin(\alpha \pm \beta )=sin\alpha cos\beta \pm sin\beta cos\alpha \\cos(\alpha \pm \beta )=cos\alpha cos\beta \mp sin\alpha sin\beta$