函数y=sin(3x+π/12)sin(3x-5π/12)的最小正周期为
问题描述:
函数y=sin(3x+π/12)sin(3x-5π/12)的最小正周期为
答
y=sin(3x+π/12)sin(3x-5π/12)
=sin(π/2-3x-π/12)sin(3x-5π/12)
=cos(5π/12-3x)sin(3x-5π/12)
=cos(3x-5π/12)sin(3x-5π/12)
=1/2sin(6x-5π/6)
很明显.W=6
T=2π/W=π/3