已知函数f(X)=2sin(ωx-π/6)sin(ωx+π/3) (其中ω为正常数,x∈R)的最小正周期为π

问题描述:

已知函数f(X)=2sin(ωx-π/6)sin(ωx+π/3) (其中ω为正常数,x∈R)的最小正周期为π
(1)求ω的值
(2)在△ABC中,若A<B,且f(A)=f(B)=1/2,求BC/AB

(1)f(X)=2sin(ωx-π/6)sin(ωx+π/3)=-【cos(2ωx+π/6)-cos(-π/2)】=-cos(2ωx+π/6)
所以最小正周期为2π/2ω=π
ω=1
(2)f(A)=f(B)=1/2
所以cos(2x+π/6)=-1/2
可以算出A,B,C
所以BC/AB=A/C