单选题设f0(x)＝sinx，f1(x)＝f′0(x)，f2(x)＝f′1(x)，…f

单选题 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

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．

我好好复习下