函数y=sin(2x+派/6)cos(2x+派/6)的最小正周期为
问题描述:
函数y=sin(2x+派/6)cos(2x+派/6)的最小正周期为
答
y=sin(2x+π/6)cos(2x+π/6)
=1/2·(2sin(2x+π/6)cos(2x+π/6))
=1/2·sin(2×(2X+π/6))
=1/2·sin(4X+π/3)
T=2π/4=π/2
答
利用二倍角公式sin2x=2sinx*cosx可得 y=sin(2x-π/6)*cos(2x-π/6) =0.5sin(4x-π/3) 因为y=Asin(ax+b)得最小正周期是2π/a 所以该函数最小正周期是2π/4=π/2 格式这样写:∵y=sin(2x+π/6)cos(2x+π/6) =1...