设f0(x)=cosx,f1(x)=f′0(x),f2(x)=f′1(x),…,fn+1(x)=f′n(x),n∈N*,则函数y=|4f2

2025-03-06 22:14:59
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回答1:

解答:解:f0(x)=cosx,
f1(x)=f′0(x)=-sinx,
f2(x)=f′1(x)=-cosx
f3(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)=-sinx
y=|-4cosxsinx-1|=|1+2sin2x|,
结合图象可知T=π,
故答案为π