解答:解:f0(x)=cosx,f1(x)=f′0(x)=-sinx,f2(x)=f′1(x)=-cosxf3(x)=f′2(x)=sinx,f4(x)=f′3(x)=cosx=f0(x)…可知周期T=4,∴f2008(x)=f0(x)=cosx,f2009(x)=f1(x)=-sinxy=|-4cosxsinx-1|=|1+2sin2x|,结合图象可知T=π,故答案为π