题目

求函数y=2cosxsinx(x+π/3)-根号3sin^2x+sinxcosx的周期,最值和单调区间

提问时间：2021-03-07

答案

解
原式=2cosx(sinx*1/2+cosx√3/2)-√3sin²x+sinxcosx
=cosxsinx+√3cos²x-√3sin²x+sinxcosx
=2sinxcosx+√3(cosx²-sin²x)
=sin2x+√3cos2x
=2sin(2x+π/3)
周期为2π/2=π
当sin(2x+π/3)=1时,有最大值2
当sin(2x+π/3)=-1时,有最小值-2
sinx在x∈[2kπ-π/2,2kπ+π/2]上单增
所以sin(2x+π/3)在x∈[kπ-π/12,kπ+π/12]上单调递增
sinx在x∈[2kπ+π/2,2kπ+3π/2]上单减
所以sin(2x+π/3)在x∈[kπ+π/12,kπ+7π/12]上单调递减