已知函数fx=根号3sinωx-cosωx(x∈R,ω>0)的最小正周期为6π1、求f(3π/2)的值2、设α、β∈[-π/2,0],f(3α+π/2)=-10/13,f(3β+2π）=6/5,求cos(α+β）的值

Question

已知函数fx=根号3sinωx-cosωx(x∈R,ω>0)的最小正周期为6π
1、求f(3π/2)的值
2、设α、β∈[-π/2,0],f(3α+π/2)=-10/13,f(3β+2π）=6/5,求cos(α+β）的值

Answer 1

已知函数fx=根号3sinωx-cosωx(x∈R,ω>0)的最小正周期为6π
1、求f(3π/2)的值
2、设α、β∈[-π/2,0],f(3α+π/2)=-10/13,f(3β+2π）=6/5,求cos(α+β）的值

(1)解析：∵函数fx=根号3sinωx-cosωx(x∈R,ω>0)的最小正周期为6π
F(x)=√3sinωx-cosωx=2sin(ωx-π/6)
T=2π/ω=6π==>ω=1/3
∴f(x)=2sin(1/3x-π/6)==>f(3π/2)=2sin(π/2-π/6)= √3
(2)解析：设α、β∈[-π/2,0],f(3α+π/2)=-10/13,f(3β+2π）=6/5
f(3α+π/2)=2sin((3α+π/2)/3-π/6)=2sin(α)= -10/13==> sin(α)=-5/13==>cos(α)=12/13
f(3β+2π)=2sin((3β+2π)/3-π/6)=2sin(β+π/2)=6/5==> cos(β)=3/5==>sin(β)=-4/5
cos(α+β）=cosαcosβ-sinαsinβ=36/65-20/65=16/65