Question

Помогите пожалуйста.........................................

Answer 1

Ответ:

$\frac{\sqrt{2} }{2}$

Объяснение:

$cos(\alpha -\beta +\frac{\pi }{2} )+2sin(\alpha +\pi )cos(\beta -\pi )$

Формулы:

$cos(\alpha +\frac{\pi }{2} )=-sin\alpha \\\\sin(\alpha +2\pi )=sin\alpha \\sin(\alpha )cos(\beta)=\frac{1}{2} (sin(\alpha +\beta )+sin(\alpha -\beta ))$

____________

$cos(\alpha -\beta +\frac{\pi }{2} )+2sin(\alpha +\pi )cos(\beta -\pi )=-sin(\alpha -\beta )+2*\frac{1}{2} (sin(\alpha +\pi +\beta -\pi )+sin(\alpha +\pi -(\beta -\pi )))=\\=-sin(\alpha -\beta )+sin(\alpha +\beta )+sin(\alpha -\beta +2\pi )=-sin(\alpha -\beta )+sin(\alpha +\beta )+sin(\alpha -\beta )=sin(\alpha +\beta )$

Підставимо значення $\alpha =0,1\pi ;\beta =0.15\pi .$

$sin(\alpha +\beta )=sin(0.1\pi +0.15\pi )=sin(0.25\pi )=sin(\frac{1}{4} \pi )=sin\frac{\pi }{4} =\frac{\sqrt{2} }{2}$