题目

f(x)=sinx 乘cosx-根号3乘cosx的平方+2分之√3
求最小正周期及单调递增区间

提问时间：2020-10-31

答案

f(x)=sinxcosx-√3cos²x+√3/2
=sin(2x)/2-√3[cos(2x)+1]/2+√3/2
=(1/2)sin(2x)-(√3/2)cos(2x)
=sin(2x)cos(π/3)-cos(2x)sin(π/3)
=sin(2x-π/3)
最小正周期Tmin=2π/2=π
当2kπ-π/2≤2x-π/3≤2kπ+π/2 (k∈Z)时,sin(2x-π/3)单调递增,
此时x∈[kπ-π/12,kπ+5π/12] (k∈Z)
当2kπ+π/2≤2x-π/3≤2kπ+3π/2 (k∈Z)时,sin(2x-π/3)单调递减,
此时x∈[kπ+5π/12,kπ+11π/12] (k∈Z)
函数f(x)的单调递增区间为∈[kπ-π/12,kπ+5π/12] (k∈Z)；
单调递减区间为[kπ+5π/12,kπ+11π/12] (k∈Z).