已知函数f(x)=sin(ωx-π/6)sin(ωx+π/3),相邻两条对称轴之间的距离为π/2,在△ABC中,abc为角ABC的对边,若A＜B且f(A)=f(B)=1/4,求c/a的值

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已知函数f(x)=sin(ωx-π/6)sin(ωx+π/3),相邻两条对称轴之间的距离为π/2,在△ABC中,abc为角ABC的对边,若A＜B且f(A)=f(B)=1/4,求c/a的值
已知函数f(x)=sin(ωx-π/6)sin(ωx+π/3),相邻两条对称轴之间的距离为π/2,
在△ABC中,abc为角ABC的对边,若A＜B且f(A)=f(B)=1/4,求c/a的值

已知函数f(x)=sin(ωx-π/6)sin(ωx+π/3),相邻两条对称轴之间的距离为π/2,在△ABC中,abc为角ABC的对边,若A＜B且f(A)=f(B)=1/4,求c/a的值
(wx+π/3)-(wx-π/6)=π/2 ==>(wx+π/3)=π/2+(wx-π/6) 两边取正弦得：
sin(wx+π/3)=sin[π/2+(wx-π/6)]=cos(wx-π/6)
f(x)=1/2sin(2wx-π/3)
因为相邻两条对称轴之间的距离为π/2
即：T/2=π/2 ==>T=π=2π/2w ==>w=1
所以：
f(x)=1/2sin(2x-π/3)
因为f(A)=f(B)=1/4,
所以：
1/2sin(2A-π/3)=1/2sin(2B-π/3)=1/4
sin(2A-π/3)=sin(2B-π/3)=1/2
2A-π/3=π/6 ; 2B-π/3=5π/6
==>A=π/4 B=7π/12 ==>C=π/6
c/a=sinC/sinA=(1/2)/(√2/2)=√2/2

等于1，过程太麻烦，懒得写了

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