设f0(x)=cosx,f1(x)f0'(x),f2(x)=f1'(x),...,fn+1(x)=fn'(x),n属于正整数,则f2008

问题描述:

设f0(x)=cosx,f1(x)f0'(x),f2(x)=f1'(x),...,fn+1(x)=fn'(x),n属于正整数,则f2008

f0(x)=cosx
所以f1(x)=-sinx
f2(x)=-cosx
f3(x)=sinx
f4(x)=cosx=f0(x)
所以是4个一循环
2008÷4余0
所以f2008(x)=f0(x)=cosx