[1+2cosx/2*sinx/2+cos(x/2)^2-sin（x/2)^2 ][sin（x/2)-cos(x/2)]/根号下的[2*2cos(x/2)^2]
=2cos(x/2)[sin（x/2)+cos(x/2)][sin（x/2)-cos(x/2)]/2cos(x/2)
=-cosx