試題分析:利用函數的周期性及x∈[3,5]時的表達式f(x)=2-|x-4|,可求得x∈[-1,1]時的表達式,從而可判斷逐個選項的正誤。解:∵f(x+2)=f(x),∴函數f(x)是周期為2的周期函數,又當x∈[3,5]時f(x)=2-|x-4|,∴當-1≤x≤1時,x+4∈[3,5],∴f(x)=f(x+4)=2-|x|,∴f(sin

))=f(

)=

-

=f(cos

)),排除A, f(sin1)=2-sin1<2-cos1=f(cos1)排除B, f(sin

))=2-

<2-

=f(cos

))=f(cos

),D正確; f(sin2)=2-sin2<2-(-cos2)=f(cos2)排除C.故選D
點評:本題考查函數的周期性,難點在于求x∈[-1,1]時的表達式,屬于中檔題.