Question

已知函数f(x)=cosx＾2+ 根号3sinxcosx+1,求函数f(x)的最小正周期和单调增区间

高一数学题，详解

Answer 1

f(x)=cosx＾2+ √3*sinxcosx+1
=0.5（2cosx＾2-1）+√3*sinxcosx+1.5
=0.5cos2X+0.5√3sin2X+1.5
=sin（2X+π/6）+1.5
因此最小正周期为π，单调递增区间为-π/2+kπ <= 2X+π/6 <= π/2+kπ
即[-7π/12+kπ/2 ，5π/12+kπ/2]

Answer 2

解：f(x)=cosx＾2+ 根号3sinxcosx+1
=（1/2）cos2x+(根号3/2)sin2x+3/2
=sin(2x+π/6)+3/2
所以函数的周期为π，
单调增区间为（）