f(x)=cos²x-sin²x+2sinxcosx=cos2x+sin2x=√2sin(π/4+2x)
（1）f(x)的最小正周期T＝2π/2＝π
（2）由 2kπ-π/2≤π/4+2x ≤2kπ+π/2得kπ-3π/8≤x ≤kπ+π/8
所以f(x)的单调增区间为：[kπ-3π/8,kπ+π/8]
同理：f(x)的单调减区间为：[kπ+π/8,kπ+5π/8] （k∈Z）
(3)当-π/4≤x ≤π/4时,有-π/4≤π/4+2x ≤3π/4
所以 -√2/2≤sin(π/4+2x) ≤1
即-1≤√2sin(π/4+2x) ≤√2
故,当-π/4≤x ≤π/4时,f(x)的值域为：[-1,√2]