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已知函数f(x)=cos(x-2π/3)-cosx,求函数的最小正周期及单调递增区间

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/06 13:54:37
已知函数f(x)=cos(x-2π/3)-cosx,求函数的最小正周期及单调递增区间
cos(x-2π/3)-cosx
=cosxcos(2π/3)+sinxsin(2π/3)-cosx
=-(3/2)cosx+(√3/2)sinx
=√3[sinx*cos(π/3)-cosx*sin(π/3)]
=√3sin(x-π/3)
(1)T =2π
(2) 2kπ-π/2 ≤x-π/3≤2kπ+π/2
2kπ-π/6 ≤x≤2kπ+5π/6
所以 增区间【2kπ-π/6 ,2kπ+5π/6】,k∈Z