函数 y=[cos(x+2/5π)-1]/sin(x+2/5π)的递减区间是输错了!函数 y=[cos(x+2π/5)-1]/sin(x+2π/5)的递减区间是
问题描述：
函数 y=[cos(x+2/5π)-1]/sin(x+2/5π)的递减区间是
输错了!函数 y=[cos(x+2π/5)-1]/sin(x+2π/5)的递减区间是
答
楼主所给函数中的“2/5π”，应该是“2π/5”吧？
如果是的话：
y=[cos(x+2π/5)-1]/sin(x+2π/5)
y‘={-[sin(x+2π/5)]^2-[cos(x+2π/5)-1]cos(x+2π/5)}/{[sin(x+2π/5)]^2}
y‘={-[sin(x+2π/5)]^2-[cos(x+2π/5)]^2-cos(x+2π/5)}/{[sin(x+2π/5)]^2}
y‘=-[1+cos(x+2π/5)]/{[sin(x+2π/5)]^2}
令：y'≤0，即：-[1+cos(x+2π/5)]/{[sin(x+2π/5)]^2}≤0
整理，有：1+cos(x+2π/5)≥0
即：cos(x+2π/5)≥-1
2kπ≤x+2π/5≤2kπ+2π，其中k∈Z
2kπ-2π/5≤x≤2kπ+8π/5
即：f(x)的递减区间是x∈[2kπ-2π/5，2kπ+8π/5]，其中k∈Z
答
y=[cos(x+2π/5)-1]/sin(x+2π/5) =[1-2sin²(x/2+π/5)-1]/[2sin(x/2+π/5)cos(x/2+π/5)] =-sin(x/2+π/5)/cos(x/2+π/5) =-tan(x/2+π/5)定义域sin(x/2+π/5)≠0,cos(x/2+π/5)≠0∴x/2+π/5≠kπ/2,k∈Z由 ...