Question

函数y=cosxsin(x-π/3)+根号3/4的图像的一个对称中心是什么

Answer 1

y=cosxsin(x-π/3)+(√3)/4
y=cosx[sinxcos(π/3)+cosxsin(π/3)]+(√3)/4
y=cosx[(sinx)/2+cosx(√3)/2]+(√3)/4
y=(1/2)sinxcosx+[(√3)/2](cosx)^2+(√3)/4
y=(1/4)sin(2x)+[(√3)/2](1+cos2x)/2+(√3)/4
y=(1/4)sin(2x)+[(√3)/4]cos2x+(3√3)/4
y=(1/2){(1/2)sin(2x)+[(√3)/2]cos2x}+(3√3)/4
y=(1/2){(1/2)sin(2x)+[(√3)/2]cos2x}+(3√3)/4
y=(1/2)[cos(π/3)sin(2x)+sin(π/3)cos2x]+(3√3)/4
y=(1/2)sin(2x+π/3)+(3√3)/4
可见,y的最小周期是π,初相位是π/3
所以,y对称轴是：x=kπ+(π/3-π/4)=kπ-π/12,k=0、±1、±2、±3……
当k=0时,其对称轴是x=-π/12