Question

三角函数的题

若函数f(x)=sinωx+√3cosωx，x∈R,又f(α)=-2.f(β)=0,且|α-β|的最小值等于3π/4，则正数ω的值为多少

Answer 1

f(x)=sinωx+√3cosωx
=2(sinωxcosπ/3+cosωxsinπ/3)
=2sin(ωx+π/3)

f(α)=-2.f(β)=0
一个是最小值，一个是和x轴交点
所以最小相差1/4个周期
所以T/4=3π/4
即(2π/ω)/4=3π/4
ω=2/3

Answer 2

f(x)=sin^2ωx+√3sinωxcosωx
=（2sin^2ωx -1）/2 +（√3/2)·2sinωxcosωx +1/2
= (-1/2)·cos(2ωx) +（√3/2)·sin(2ωx) +1/2
= cos(2π/3)·cos(2ωx) + sin(2π/3)·sin(2ωx) +1/2
= cos(2ωx - 2π/3) +1/2

f(α)=-1/2
→cos(2ωα - 2π/3)=-1;
2ωα - 2π/3 = (2k1+1)π;

f(β)=1/2
→cos(2ωβ - 2π/3)=0;
2ωβ - 2π/3 = k2 π + π/2;

则
2ωα - 2ωβ = (2k1 - k2) π + π/2;
→2ω·|α-β| = (2k1 - k2) π + π/2;
|α-β|≥3π/4,则
2ω≤4[(2k1 - k2) π + π/2]/(3π)
=[4(2k1 - k2) +2]/3
ω≤[2(2k1 - k2) +1]/3
取k1 = k2 =1,
则可知ω=1.