解:f(x)=(cosx-sinx)(-cosx)=sinxcosx-cox^2x=1/2sin2x-1/2(2cos^2x-1+1)=1/2sin2x-1/2(cos2x+1)=1/2sin2x-1/2cos2x-1/2=根号2/2sin(2x-π/4)-1/2所以最小正周期T=2π/2=π单调递增区间-π/2+2kπ<=2x-π/4<=π/2+2kπ[-π/8+kπ<=x<=3π/8+kπ]
