Question

已知函数f(x)=sinωxsin(ωx+π/3)+cos^2ωx(x＞0)的最小正周期为π(1)

已知函数f(x)=sinωxsin(ωx+π/3)+cos^2ωx(x＞0)的最小正周期为π(1)求ω的值(2)求函数f(x)在区间[-π/6,7π/12]的取值范围

Answer 1

已知函数f(x)=sinωxsin(ωx+π/3)+cos^2ωx(x＞0)的最小正周期为π(1)求ω的值(2)求函数f(x)在区间[-π/6,7π/12]的取值范围
(1)解析:f(x)=sinωxsin(ωx+π/3)+cos^2ωx=1/2sin^2ωx+√3/2sinωxcosωx+cos^2ωx
=1/2sin^2ωx+√3/4sin2ωx+cos^2ωx=1/2+√3/4sin2ωx+1/4(1+cos2ωx)
=3/4+√3/4sin2ωx+1/4cos2ωx=3/4+1/2sin(2ωx+π/6)
因为，f(x)的最小正周期为π
所以，2ω=2==>ω=1
(2)解析:因为，f(x)=3/4+1/2sin(2x+π/6)
单调增区间:2kπ-π/2<=2x+π/6<=2kπ+π/2==>kπ-π/3<=x<=kπ+π/6
因为，在区间[-π/6,7π/12]
f(-π/6)=3/4+1/2sin(-π/3+π/6)=1/2
f(7π/12)=3/4+1/2sin(7π/6+π/6)=3/4-√3/4
所以，函数f(x)在区间[-π/6,7π/12]的取值范围为[3/4-√3/4,5/4]