已知f(x)=4cos^4x-2cosx-1/tan(π/4+x)sin²(π/4-x)
(1),若tan(π/4+x)=3,
(tanπ/4+tanx)/(1-tanx*tanπ/4)=(1+tanx)/(1-tanx)=3
tanx=2
1+tan²x=1/cos²x
cos²x=1/5 cos^4x=1/25 cosx=1/√5
sin²x=4/5 sinx=2/√5
sin²(π/4-x)=(sinπ/4cosx-cosπ/4*sinx)²=［1/√2(cosx-sinx)］² =1/10
f(x)=4*1/25-2/√5-1/3*(1/10)=4/25-2/√5-10/3=1/75*(238-30√2)