y=1－(sin2x＋cos2x)研究下：sin2x＋cos2x=√2[(√2/2)sin2x＋(√2/2)cos2x]=√2[sin2xcos(π/4)＋cos2xsin(π/4)]=√2sin(2x＋π/4)则：y=1－√2sin(2x＋π/4)最小正周期是2π/2=π增区间是：2kπ＋π/2≤2x＋π/4...]=√2[sin2xcos(π/4)＋cos2xsin(π/4)]=√2sin(2x＋π/4)这步是怎么的来的1、sin(a＋b)=sinacosb＋cosasinb2、sin2x＋cos2x=√2[(√2/2)sin2x＋(√2/2)cos2x]=√2[sin2xcos(π/4)＋cos2xsin(π/4)]=√2sin(2x＋π/4)=√2[(√2/2)sin2x＋(√2/2)cos2x]=√2[sin2xcos(π/4)＋cos2xsin(π/4)]老师我想问的是这步逆用公式：sin(2x＋π/4)=sin2xcos(π/4)＋cos2xsin(π/4)=(√2/2)sin2x＋(√2/2)cos2x那(√2/2)sin2x用什么定理等于(√2/2)sin2x＋(√2/2)cos2x的(√2/2)sin2x是不会等于(√2/2)sin2x＋(√2/2)cos2x的。你这问法有问题。。 sin45°=cos45°=√2/2则：(√2/2)sin2x＋(√2/2)cos2x=sin2xcos45°＋cos2xsin45°=√2sin(2x＋45°)