函数f(x)=3sin(wx+π/4)(w>0,x∈R),的最小周期为2π/31.求f(x)的解析式2.已知f(2a/3+π/12)=-3/2*√2,0

问题描述:

函数f(x)=3sin(wx+π/4)(w>0,x∈R),的最小周期为2π/3
1.求f(x)的解析式
2.已知f(2a/3+π/12)=-3/2*√2,0

(1)、因为函数f(x)=3sin(wx+π/4)(w>0,x∈R),的最小周期为2π/3,
所以2π/w=2π/3,即:w=3,
所以f(x)的解析式是:f(x)=3sin(3x+π/4).
(2)、因为f(2a/3+π/12)=-3/2*√2,
所以3sin【3(2a/3+π/12)+π/4】=-3/2*√2
3sin(2a+π/2)=-3/2*√2
sin(2a+π/2)=-√2/2
cos2a=-√2/2
因为 0