f(x)=(cosx,sinx)*(cosx-3,sinx-3) =cosx(cosx-3)+sinx(sinx-3) =cosx)^2-3cosx+sinx)^2-3sinx =1-3(sinx+cosx) =1-3√2sin(x+pi/4) f'(x)=1'-(3√2sin(x+pi/4))' =0-3√2cos(x+pi/4)*1 =-3√2cos(x+pi/4)>=0 cos(x+pi/4)<=0 pi/2+2kpi<=x+pi/4<=3pi/2+2kpi pi/4+2kpi<=x<=5pi/4+2kpi 所以x在[pi/4+2kpi,5pi/4+2kpi](其中k属于Z)上单调递增 x在[-3pi/4+2kpi,pi/4+2kpi](其中k属于Z)上单调递减
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