高一数学题···求详解f(x) = cos^4x - 2sinx*cosx - sin^4x

f(x)=cos^4x-2sinx*cosx-sin^4x
=(cos²x+sin²x)(cos²x-sin²x)-sin2x
=cos2x-sin2x
=(√2)cos(2x+π/4)
(1)∵-1≤cos(2x+π/4)≤1
∴值域为[-√2,√2]
(2)最小正周期T=2π/2=π
(3)令-π+2kπ≤2x+π/4≤2kπ
即-5π/8+kπ≤x≤-π/8+kπ
∴f(x)的单调递增区间[-5π/8+kπ,-π/8+kπ](k∈Z)
(4)令2x+π/4=kπ,即x=-π/8+kπ/2,k∈Z
∴f(x)图像的对称轴方程为x=-π/8+kπ/2,k∈Z