作业帮函数 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
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/12 04:38:51
![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](/uploads/image/z/5920720-16-0.jpg?t=%E5%87%BD%E6%95%B0+y%3D%5Bcos%28x%2B2%2F5%CF%80%29-1%5D%2Fsin%28x%2B2%2F5%CF%80%29%E7%9A%84%E9%80%92%E5%87%8F%E5%8C%BA%E9%97%B4%E6%98%AF%E8%BE%93%E9%94%99%E4%BA%86%21%E5%87%BD%E6%95%B0+y%3D%5Bcos%28x%2B2%CF%80%2F5%29-1%5D%2Fsin%28x%2B2%CF%80%2F5%29%E7%9A%84%E9%80%92%E5%87%8F%E5%8C%BA%E9%97%B4%E6%98%AF)
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
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
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
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 kπ/2
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)...
全部展开
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
收起