已知函数f(x)=2sin(wx-π/6)•sin(wx+π/3)(其中w>0,x∈R的最小正周期为π).问:(1)求W的值.(2已知函数f(x)=2sin(wx-π/6)•sin(wx+π/3)(其中w>0,x∈R的最小正周期为π).问:(1)求w的值.(2)在三角形ABC中,若A〈B,且f(A)=f(B)=1/2,求BC/AB

问题描述:

已知函数f(x)=2sin(wx-π/6)•sin(wx+π/3)(其中w>0,x∈R的最小正周期为π).问:(1)求W的值.(2
已知函数f(x)=2sin(wx-π/6)•sin(wx+π/3)(其中w>0,x∈R的最小正周期为π).问:(1)求w的值.(2)在三角形ABC中,若A〈B,且f(A)=f(B)=1/2,求BC/AB

f(x)=2sin(wx-π/6)•sin(wx+π/3)=2cos(wx+π/3)•sin(wx+π/3)=sin(2wx+2π/3)
由于周期为π,则w=1
f(x)=sin(2x+2π/3)
在三角形ABC中,若f(x)=sin(2x+2π/3)=1/2,则2x+2π/3=5π/6,或2x+2π/3=13π/6
解得x=π/12或x=3π/4,即A=π/12,B=3π/4,所以C=π/6
BC/AB=sinA/sinC=(sinπ/12)/(sinπ/6)=2sinπ/12=(根号6-根号2)/2

解:f(x)=2sin(wx-丌/6)sin(wx+兀/3)=2sin(wx-兀/6)cos(wx-兀/6)=Sin(2wx-兀/3).因为f(x)最小正周期为兀且W>O,所以2兀/2w=兀→w=|

(1) f(x)=2sin(wx-π/6)•sin(wx+π/2-π/6)
=2sin[π/2+(wx-π/6)]•sin(wx-π/6)
=2cos(wx-π/6)•sin(wx-π/6)
=sin(2wx-π/3)
因周期T=2π/2w=π,则w=1
所以f(x)=sin(2x-π/3)
在三角形ABC中,若A〈B,且f(A)=f(B)=1/2
则由f(x)=sin(2x-π/3)=1/2,
知2A-π/3=π/6,A=π/4
2x-π/3=π-π/6,B=7π/12
所以C=π-A-B=π-π/4-7π/12=π/6
由正弦定理BC/AB=sinA/sinC
=sin(π/4)/sin(π/6)
=(√2/2)/(1/2)
=√2

因为f(x)=2sin(wx-π/6)•sin(wx+π/3),
所以f(x)=cos(wx-π/6-wx-π/3)-cos(wx-π/6+wx+π/3)=-cos(2wx+π/6)
因为最小正周期为π,所以w=1
所以f(x)=-cos(2x+π/6)
f(A)=f(B)=1/2,所以-cos(2A+π/6)=-cos(2B+π/6)=1/2
因为cos(2π/3)=cos(4π/3)=-1/2
又因为A〈B,所以A=π/4,B=7π/12
所以BC/AB=sinA/sinC=sin(π/4)/sin(π-π/4-7π/12)=sin(π/4)/sin(π/6)=根号2