Question

cos(7п/5+a)cos(2п/5+a)+sin(7п/5+a)sin(2п/5+a) пж пж пж

Answer 1

Объяснение:

Пусть:   $\frac{7\pi }{5} +a=\alpha \ \ \ \ \ \frac{2\pi }{5} +a=\beta \ \ \ \ \Rightarrow$

$cos\alpha *cos\beta +sin\alpha *sin\beta =\\=\frac{1}{2}*(cos(\alpha -\beta )+cos(\alpha +\beta ))+\frac{1}{2}*(cos(\alpha -\beta )-cos(\alpha +\beta ))=\\=\frac{1}{2}*(cos(\alpha -\beta )+cos(\alpha +\beta )+(cos(\alpha -\beta )-cos(\alpha +\beta ))=\\=\frac{1}{2}*2*cos(\alpha -\beta )=cos(\alpha -\beta )=cos(\frac{7\pi }{5}-\frac{2\pi }{5})=cos(\frac{5\pi }{5} )=cos\pi =-1.$