求函数y=sin(x+п/6)cos(п/3-x)的最小正周期是多少?
问题描述：
求函数y=sin(x+п/6)cos(п/3-x)的最小正周期是多少?
答
y=sin(x+π/6)sin[π/2-(π/3-x)]
=sin²(x+π/6)
={1-cos[2(x+π/6)]/2
=-cos(2x+π/3)/2+1/2
所以T=2π/2=π
答
2sinacosb=sin(a+b)+sin(a-b)
sin(x+6/π)cos(x-3/π)=1/2( sin(2x+6/π) + sin(9/π))
sin(2x+6/π)的周期同sin(2x)的正周期一样是π,所以最小正周期是π