求函数y=sin(x+pai/3)sin(x+pai/2)的周期.谢谢一楼的老师回答,但请问=-[cos(x+π/3+x+π/2)-cos(x+π/3-x-π/2)]/2 是怎么出来的呢?是的,积化和差没学过啊
问题描述：
求函数y=sin(x+pai/3)sin(x+pai/2)的周期.
谢谢一楼的老师回答,但请问=-[cos(x+π/3+x+π/2)-cos(x+π/3-x-π/2)]/2
是怎么出来的呢?
是的,积化和差没学过啊
答
y=sin(x+π/3)sin(x+π/2)
=-[cos(x+π/3+x+π/2)-cos(x+π/3-x-π/2)]/2
=[cosπ/6-cos(2x+5π/6)]/2
=-cos(2x+5π/6)/2+√3/4
所以最小正周期T=2π/2=π
积化和差，还没学过么。
答
y=sin(x+π/3)sin(x+π/2) =sin(x+π/3)cosx=(sinxcosπ/3+cosxsinπ/3)cosx=1/2sinxcosx+√3/2cos^2(x)[cos^2(x)指cosx的平方)=1/2(1/2sin2x+√3/2cos2x)+√3/4 =1/2(2x+π/3)+√3/4所以最小正周期T=2π/w=2π/2=π...