y=2sin(x/2)cos(x/2)/[1+2cos²(x/2)-1]
=2sin(x/2)cos(x/2)/[2cos²(x/2)]
=sin(x/2)/cos(x/2)
=tan(x/2)
所以T=π/(1/2)=2π
函数y=sinx/1+cosx的最小正周期是?
y=2sin(x/2)cos(x/2)/[1+2cos²(x/2)-1]
=2sin(x/2)cos(x/2)/[2cos²(x/2)]
=sin(x/2)/cos(x/2)
=tan(x/2)
所以T=π/(1/2)=2π