已知函数f(x)=(cos4x-1)/(2cosπ/2+2x)+cos∧2x-sin∧2x.

f(x)=(cos4x-1)/(2cos(π/2+2x))+cos²x-sin²x
=(cos4x-1)/(-2sin2x)+cos2x
=-2sin²2x/(-2sin2x)+cos2x
=sin2x+cos2x=√2sin(2x+π/4),
函数最小正周期为2π/2=π.
2kπ+π/2≤2x+π/4≤2kπ+3π/2,k∈Z.
kπ+π/8≤x≤kπ+5π/8,k∈Z.
所以函数的单调递减区间是[kπ+π/8,kπ+5π/8],k∈Z.