已知函数f(x)=tan(3x+π/4)(1)求f(π/9)的值(2)若f( α/3 π/4)=2,

问题描述:

已知函数f(x)=tan(3x+π/4)(1)求f(π/9)的值(2)若f( α/3 π/4)=2,

tan(3x+π/4)
=[tan3x+1]/(1-tan3x)
=(1+tan3x)/(1-tan3x)
f(α/3+π/4)=2,求cos2α
f(α/3+π/4)=tan(a+3π/4+π/4)=tana=2
所以
tan²a=4
sin²a=4cos²a
5cos²=1 cos²=1/5 sin²a=4/5
cos2a=cos²a-sin²a=1/5-4/5=-3/5

1.f(π/9)=tan(π/3+π/4)=(tanπ/3+tanπ/4)/(1-tanπ/3tanπ/4)=(√3+1)/(1-√3)=-√3-22.∵f(a/3+π/4)=2 ∴tan(a+3π/4+π/4)= tan(π+a)=tana=2sina/cosa=2sina=2cosa代入sin²a+cos²a=1cos²a=1/...