求函数y=sin(x+1/3兀)sin(x+1/2兀)的周期

网友回答

y=sin(x+1/3兀)sin(x+1/2兀)
=-1/2[cos(2x+5π/6)-cos(-π/6)]
所以T=2π/w=2π/2=π
======以下答案可供参考======
供参考答案1:
周期是π供参考答案2:
y=sin(x+1/3兀)sin(x+1/2兀)
=[sinxcos(π/3)+cosxsin(π/3)][sinxcosπ/2+cosxsin(π/2)]
=[(sinx)/2+(√3/2)cosx]cosx
=(1/2)sinxcosx+(√3/2)cos²x
=(1/4)sin(2x)+(√3/4)(cos2x+1)
=(1/2)[(1/2)sin2x+(√3/2)cos2x]+√3/4
=(1/2)sin(2x+π/3)+√3/4
所以 函数的周期为 2π/2=π
供参考答案3:
sinα·sinβ=-(1/2)[cos(α+β)-cos(α-β)]
y=sin(x+1/3兀)sin(x+1/2兀)
=(-1/2)[cos(2x+5π/6)-cos(π/6)]
=(-1/2)cos(2x+5π/6)+√3/4
所以，函数的周期为kπ，k∈Z
供参考答案4:
y=[(1/2)sinx+√3/2cosx]cosx
=(1/4)sin2x+√3/4(1+cos2x)
=(1/2)[(1/2)sin2x+√3/2cos2x]+√3/4
=(1/2)sin(2x+π/3)+√3/4
周期为π。