若函数f(x)=sinax+cosax(a>0)的最小正周期为1 日期:2011-01-08 11:11:03 人气:1 若函数f(x)=sinax+cosax(a>0)的最小正周期为1 f(x)=sinax+cosax =√2(√2/2sinax+√2/2cosax) =√2(sinaxcosπ/4+cosaxsinπ/4) =√2sin(ax+π/4)