Question

(2cos(a-2pi)-sin(a-pi/2)):(cos(9pi/2-a)+2sin(2pi-a))
упростить, пожалуйста.

Answer 1

(2cos(a-2π)-sin(a-π/2))/(cos(9π/2-a)+2sin(2π-a))=
=(2cos(2π-a)+sin(π/2-a))/(cos(9π/2-a)+2sin(2π-a))=
=(2cosα+cosα)/(cos(4π+(π/2-α))-2sinα=3cosα/(cos(π/2-α)-2sinα)=
=3cosα/(sinα-2sinα)=3cosα/-sinα=-3ctgα.