已知函數(shù)f(x)=(√3)sin(ωx+φ)-cos(ωx+φ)(o0)為偶函數(shù) 且函數(shù)y=f(x)圖像的兩相鄰對(duì)稱軸間的距離為π/2.求f(π/8)的值;還有一問是：將函數(shù)y=f(x)的圖像向右平移π/6個(gè)單位后,得到函數(shù)y=g(x)的圖像,求g(x)的單調(diào)遞減區(qū)間.
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)是偶函數(shù),0＜φ＜π,∴φ=π-π/3=2π/3
故f(x)=2cos(ωx)
∵函數(shù)y=f(x)圖像的兩相鄰對(duì)稱軸間的距離為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)的單調(diào)遞減區(qū)間為(kπ+π/6,kπ+2π/3)