已知tanα=2,求[cos(α-x)-2cos(π/2+α)]/[sin(α-3π/2)-sinα]
[cos(α-x)-2cos(π/2+α)]/[sin(α-3π/2)-sinα]
=[cos(α-x)+2sinα]/(cosα-sinα)
=(cosαcosx+sinαsinx)/(cosα-sinα)
=(cosx+tanαsinx)/(1-tanα)
=-(cosx+2sinx)