已知函数f(x)=[2cos^4x-2cos^x+1/2]/[tan（π/4-x）sin^2（π/4+x）],求f（x）

tan（π/4-x）= (tanπ/4-tanx)/(1+tanπ/4tanx) = (1-tanx)(1+tanx) = (cosx-sinx)/(cosx+sinx)
sin^2(π/4+x) = (√2/2sinx+√2/2cosx)^2= 1/2(sinx+cosx)^2
f(x) = [2cos^4x-2cos^2x+1/2]/[tan（π/4-x）sin^2（π/4+x）]
=[2cos^4x-2cos^2x+1/2]/[ (cosx-sinx)/(cosx+sinx)*1/2(sin+cosx)^2]
=[4cos^4x-4cos^2x+1]/[ (cosx-sinx)(cosx+sinx)]
= (2cos^2x-1)^2 / (cos^2x-sin^2x)
=[cos(2x)]^2 / cos(2x)
= cos(2x) ∈【-1,1】
值域：【-1,1】