Given the function f (x) = [2cos ^ 4x-2cos ^ x + 1 / 2] / [Tan (π / 4-x) sin ^ 2 (π / 4 + x)], find the range of F (x)

Given the function f (x) = [2cos ^ 4x-2cos ^ x + 1 / 2] / [Tan (π / 4-x) sin ^ 2 (π / 4 + x)], find the range of 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】
Range: [- 1,1]

Let F1 (x) = 2 / (1 + x), define f (n + 1) (x) = F1 [FN (x)], an = [FN (0) - 1] / [FN (0) + 2], then a (2007) is equal to

f1(x)=2/(1+x),
f(n+1)(x)=f1[fn(x)]=2/[1+fn(x)]
f(n+1)(x)=2/[1+fn(x)]
f(n+1)(x)-1=2/[1+fn(x)]-1=[1-fn(x)]/[1+fn(x)]
f(n+1)(x)+2=2/[1+fn(x)]+2=2[2+fn(x)]/[1+fn(x)]
Division of two equations
2[f(n+1)(x)-1]/[f(n+1)(x)+2]=[1-fn(x)]/[2+fn(x)]=-[fn(x)-1]/[2+fn(x)]
When x = 0, 2 [f (n + 1) (0) - 1] / [f (n + 1) (0) + 2] = - [FN (0) - 1] / [2 + FN (0)]
2a(n+1)=-an
an=[(-1/2)^(n-1)]a1
And A1 = [F1 (0) - 1] / [F1 (0) + 2]
=1/4
an=[(-1/2)^(n+1)]
a2007=[(-1/2)^2008=1/2^2008