题目

函数f(x)＝acosωx＋bsinωx(ω＞0)的最小正周期为π/2,当x＝π╱6时,有最大值4.
(1)求a.b.ω的值
(2)若0＜x＜π/4,且f(x)＝4/3.求f(x－π/8)的值

提问时间：2021-02-22

答案

1)求a.b.ω的值
T=π/2=2π/ω ,ω=4
f(x)＝acosωx＋bsinωx= acos4x＋bsin4x=A sin(4x+arctanb/a ),x＝π╱6时,有最大值4.A=4
2π/3+arctanb/a=π/2 ,arctanb/a=-π/6 ,,cos(-π/6)=a/A ,a=2√3,sin(-π/6)=b/A ,b=-2
2)若0＜x＜π/4,且f(x)＝4/3.求f(x－π/8)的值 ,sin(4x-π/6)=3/16 ,cos(4x-π/6)=√247/16
f(x)=4sin(4x-π/6)
f(x-π/8)=4sin(4x-π/2-π/6)=-4sin[π/2-(4x-π/6)]=-4cos(4x-π/6)=-√247/4