题目

求f(x)=-cos²x+根号3sinxcosx的最小正周期和单调增区间

提问时间：2021-04-06

答案

f(x)=-cos²x+√3sinxcosx
=-1/2(1+cos2x)+√3/2sin2x
=√3/2sin2x-1/2cos2x-1/2
=sin(2x-π/6)-1/2
f(x)最小正周期T=2π/2=π
由2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z
得kπ-π/6≤x≤kπ+π/3,k∈Z
f(x)递增区间为[kπ-π/6,kπ+π/3],k∈Z
由2kπ+π/2≤2x-π/6≤2kπ+3π/2,k∈Z
得kπ+π/3≤x≤kπ+5π/6,k∈Z
f(x)递减区间为[kπ+π/3,kπ+5π/6],k∈Z