sin(x + π/6）cos（π/3 - x）的最小正周期怎么算?
问题描述：
sin(x + π/6）cos（π/3 - x）的最小正周期怎么算?
答
cos[π/2-(π/3+x)]=cos(π/6-x)=sin(π/3+x)
y=sin(x+π/3)cos(π/6-x)
=sin(x+π/3)sin(π/3+x)
=sin²(π/3+x)
=[1-cos(2π/3+2x)]/2
=1/2-1/2cos(2π/3+2x)
T=2π/2=π