函数y=sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)cos(x+π/3)的图象的一条对称轴的方

问题描述:

函数y=sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)cos(x+π/3)的图象的一条对称轴的方

y=sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)cos(x+π/3)
=sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)sin[π/2-(x+π/3)]
=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/2)
=cos2x
其中一条对称轴的方程为x=π/2

cos(x+π/3)=-sin(x-π/6)
∴sin(3x+π/3)cos(x-π/6)+cos(3x+π/3)cos(x+π/3)
=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/2)
=cos(2x)
∴对称轴为2x=kπ,即x=kπ/2,k∈Z
若取k=0,得x=0

cos(x+π/3)
=sin[π/2-(x+π/3)]
=sin(π/6-x)
=-sin(x-π/6)
所以y=sin(3x+π/3)cos(x-π/6)-cos(3x+π/3)sin(x-π/6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/2)
=cos2x
cosx的对称轴用就是取最大或最小的地方,即x=kπ
所以此处是2x=kπ
x=kπ/2