设f0(x)=sinx,f1(x)=f′0(x),f2(x)=f′1(x),…fn(x)=fn-1′(x),n∈N,则f2 005(x)=A.sinxB.-sinxC.cosxD.-cosx
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C解析(sinx)′=cosx,(cosx)′=-sinx,(-sinx)′=-cosx,(-cosx)′=sinx,由此可知,其周期为4,故可得fn+4(x)=…=…=fn(x)故猜测fn(x)是以4为周期的函数,有f4n+1(x)=f(1)=cosxf4n+2(x)=-sinxf4n+3(x)=-cosx,f4n+4(x)=f(4)=sinx.