已知函数f(x)=(√3)sin(ωx+φ)-cos(ωx+φ)(o0)为偶函数 且函数y=f(x)图像的两相邻对称轴间的距离为π/2.求f(π/8)的值;还有一问是：将函数y=f(x)的图像向右平移π/6个单位后,得到函数y=g(x)的图像,求g(x)的单调递减区间.
f(x)=(√3)sin(ωx+φ)-cos(ωx+φ)
=-2[(1/2)cos(ωx+φ)-(√3/2)sin(ωx+φ)]
=-2[cos(π/3)cos(ωx+φ)-sin(π/3)sin(ωx+φ)]
=-2cos(ωx+φ+π/3)
∵f(x)是偶函数,0＜φ＜π,∴φ=π-π/3=2π/3
故f(x)=2cos(ωx)
∵函数y=f(x)图像的两相邻对称轴间的距离为π/2.∴ω=2.
故f(x)=2cos2x.
于是f(π/8)=2cos(π/4)=√2.
向右平移π/6得g(x)=2cos[2(x-π/6)]=2cos(2x-π/3)
g(x)的单调递减区间为(kπ+π/6,kπ+2π/3)