Question

помогите пожалуйста. ​

Answer 1

$\displaystyle\bf\\Cos20^{0} -Cos40^{0} -Cos80^{0} =\Big(Cos20^{0} -Cos80^{0}\Big) -Cos40^{0} =\\\\\\=2Sin\frac{20^{0}+80^{0} }{2} \cdot Sin\frac{80^{0}-20^{0} }{2} -Cos40^{0} =2Sin50^{0} \cdot Sin30^{0} -Cos40^{0} =\\\\\\=2Sin50^{0} \cdot \frac{1}{2} -Cos40^{0}=Sin50^{0} -Cos40^{0} =\\\\\\=Sin(90^{0} -40^{0} )-Cos40^{0} =Cos40^{0} -Cos40^{0} =0$