已知函数f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]

cosx=cos²(x/2)-sin²(x/2)
所以f(x)=2√2[cos²(x/2)-sin²(x/2)]/[cos(x/2)-sin(x/2)]
=2√2[cos(x/2)+sin(x/2)]
=2√2[√2*sin(x/2+π/4)]
=4sin(x/2+π/4)
sinx增区间是(2kπ-π/2,2kπ+π/2)
减区间是(2kπ+π/2,2kπ+3π/2)

2kπ-π/2<x/2+π/4<2kπ+π/2
2kπ-3π/4<x/2<2kπ+π/4
4kπ-3π/2<x<4kπ+π/2
所以增区间是(4kπ-3π/2,4kπ+π/2)
同理，减区间是(4kπ+π/2,4kπ+5π/2)

首先用积化和差公式化解f(x),得到f(x)=√ 2cotx,函数f(x)单调区间就是三角函数cotx的单调区间。

∵f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]
=(2√ 2)[cos²(x/2)-sin²(x/2)]/[cos(x/2)-sin(x/2)]
=(2√ 2)[cos(x/2)+sin(x/2)]
∴f′(x)=√ 2[cos(x/2)-sin(x/2)]
令f′(x)=0，则 cos(x/2)-sin(x/2)=0，即tan(x/2)=1
∴ x=2kπ+π/2 , (k=0,±1,±2,±3,.......).
故f(x)在区间(2kπ,2kπ+π/2)单调增加，在区间（2kπ+π/2，2kπ+π）单调减少。