Question

设函数f（x）=2sin（ω x+π/3）,ω＞0,x∈R,且以3π为最小正周期.

Answer 1

最小正周期T=2π/ω=2π/3,

——》ω=3,

f(π/12)=Asin(3*π/12+π/4)=2,

——》A=2,

——》f(x)=2sin(3x+π/4),

1、若是f(2/3*a-π/12)=6/5

——》f(2/3*a-π/12)=2sin2a=6/5,

——》sin2a=3/5,cos2a=4/5,

——》sin(2a-π/3)=3/5*1/2-4/5*√3/2=(3-4√3)/10,

2、若是f(2/3*a+π/12)=6/5,

——》f(2/3*a+π/12)=2sin(2a+π/2)=6/5,

——》cos2a=3/5,sin2a=4/5,

——》sin(2a-π/3)=4/5*1/2-3/5*√3/2=(4-3√3)/10.