已知函数f(x)=(3x+π/4) (1)求f(π/9)的值 (2)若f(a/3+π/4)=2 求cos2a的值

前提是不是错误的啊？

（1）求f(π/9)就是把f(x)=(3x+π/4) 里的x换成π/9
（2）f(a/3+π/4)=2 就是把x=a/3+π/4代入原式求出a的值 再求出cos2a
求的f(π/9)=（7π/12） 求的a= 2-π 代入cos2a

(1)f(π/9)=π/3+π/4=7π/12
(2)f(a/3+π/4)=2 =a+π a=2-π
cos2a=cos(4-2π)=cos4

f(x)=这里是什么？(3x+π/4)
题目打错了

f(x)=tan(3x+π/4)
1.f(π/9)=tan(π/3+π/4)
=(tanπ/3+tanπ/4)/(1-tanπ/3tanπ/4)
=(√3+1)/(1-√3)
=-√3-2
2.
∵f(a/3+π/4)=2
∴tan(a+3π/4+π/4)
= tan(π+a)=tana=2
sina/cosa=2
sina=2cosa代入sin²a+cos²a=1
cos²a=1/5
∴cos2a=2cos²a-1=-3/5

1. 3*π/9+π/4=7π/12
2.3*（a/3+π/4)+π/4=2 可以解得a= 2-π
cos（2A）=cos（4-2π）=cos4

函数f(x)=tan(3x+4分之派) 求：
1。f(9分之派)=tan(3分之派+4分之派)=[tanπ/3+tanπ/4]/(1-tanπ/3*tanπ/3) =(√3+1) /(1-√3)
=(√3+1)^2/(1-3)
=-2-√3
2. f(a/3+派/4)==tan(a+π/4)=(tana+1)/(1-tana)=2
tana=1/3
tana=sina/cosa=1/3 cosa=3sina
sin^2a+cos^2a =1 sin^2a=1/10
cos2a=1-2sin^2a=1-2*1/10=4/5