已知函数f(x)=sin(ωx+π/3),cos(ωx-π/6)(ω>0),f(x)的最小正周期为π.(1)求f(x)的解析式 (2)求f(x)的单调增区间

问题描述:

已知函数f(x)=sin(ωx+π/3),cos(ωx-π/6)(ω>0),f(x)的最小正周期为π.(1)求f(x)的解析式 (2)求f(x)的单调增区间

f(x)=sin(ωx+π/3)cos(ωx-π/6) =sin(wx+π/3)sin(wx+π/3) =1-cos^2(wx+π/3) =1-(1+cos(2wx+2π/3) =cos(2wx+2π/3) T=2π/2w=π w=1 (1)f(x)=cos(2x+2π/3) (2) 2x+2π/3属于 【π+2kπ,2π+k...