函数y=sin(π/3-x/2)cos(π/6+x/2)的单调减区间
问题描述：
函数y=sin(π/3-x/2)cos(π/6+x/2)的单调减区间
答
y=sin(π/3-x/2)cos(π/6+x/2)
=sin(π/3-x/2)cos[π/2-(π/3-x/2)]
=sin²(π/3-x/2)
=[1-cos2(π/3-x/2)]/2
=[1-cos(2π/3-x)]/2
=[1-cos(x-2π/3)]/2
求函数的单调减区间就是求函数y=cos(x-2π/3)单调增区间
令2kπ-π＜x-2π/3＜2kπ,k∈Z
得2kπ-π/3＜x＜2kπ+2π/3,k∈Z
所以函数的单调减区间是(2kπ-π/3,2kπ+2π/3),k∈Z