函数y=sin(π/12-π)cos(π/12-π)的单调减区间
问题描述:
函数y=sin(π/12-π)cos(π/12-π)的单调减区间
答
y=sin(π/12-x)cos(π/12-x)=-1/2sin(2x-π/6)2kπ-π/2≤2x-π/6≤2kπ+π/2kπ-π/6≤x≤kπ+π/3单减区间:[kπ-π/6,kπ+π/3]2kπ+π/2≤2x-π/6≤2kπ+3π/2kπ+π/3≤x≤kπ+5π/6单增区间:[kπ+π/3,kπ+5...