已知函数f(x)=√3sinxcosx+cos平方x-1/2,x∈R(1)求函数的最小正周期和单调增区间.
问题描述:
已知函数f(x)=√3sinxcosx+cos平方x-1/2,x∈R
(1)求函数的最小正周期和单调增区间.
答
f(x)=√3sinxcosx+(cosx)^2-1/2
=√3/2sin(2x)+1/2cos(2x)
=sin(2x+π/6)
0≤2x+π/6≤2π
-π/12≤x≤11π/12
最小正周期:[-π/12,11π/12]
单调增区间:[2kπ-5π/12,2kπ+π/12] (k∈Z)
答
f(x)=√3sinxcosx+cos²x=(√3/2)sin2x+(1+cos2x)/2-1/2=(√3/2)sin2x+(1/2)cos2x=sin2x*cos(π/6)+cos2x*sin(π/6)=sin(2x+π/6)(1)T=2π/2=π 增区间 2kπ-π/2≤2x+π/6≤2kπ+π/22kπ-2π/3≤2x≤2kπ+π/3...