由于3π/2<α<2π,故cosα>0,
3π/4<α/2<π,故cos(α/2)<0.
cos2α=2(cosα)^2-1,
故cos2α+1=2(cosα)^2,
故(1/2)(1+cos2α)=(cosα)^2,
故√[(1/2)(1+cos2α)]=cosα,
故原式=√[(1/2)(1+cosα)].
而cosα=2[cos(α/2)]^2-1,
故1+cosα=2[cos(α/2)]^2,
故原式=√[cos(α/2)]^2=-cos(α/2).