f(x)=2cosx/2×(√2sin(x/2+π/4）+ tan(x/2+π/4)×tan(x/2－π/4))
=√2[sin(x+π/4)+sin(π/4)] + [1+tan(x/2)]/[1－tan(x/2)]×[tan(x/2)－1]/[1+tan(x/2)]
=√2sin(x+π/4)
最大值=√2
最小正周期=2π
sinx的增区间是：-π/2+2kπ≤x≤π/2+2kπ
带入-π/2+2kπ≤x+π/4≤π/2+2kπ
所以增区间-3π/4+2kπ≤x≤π/4+2kπ
减区间同理