Y=sin(x+兀/6)cos(兀/3-x)的最小正周期怎么算啊
问题描述:
Y=sin(x+兀/6)cos(兀/3-x)的最小正周期怎么算啊
答
Y=sin(x+兀/6)cos(兀/3-x)
=sin(x+兀/6)cos(x+兀/6)
=1/2sin(2x+π/3)
所以最小正周期2π/2=π