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
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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
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
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
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
f(x)=这里是什么?(3x+π/4)
题目打错了
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
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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
收起
(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
(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
前提是不是错误的啊?