依题意,得
f(x)＝-(cosx)^2+sinxcosx＝-cos(2x)/2+sin(2x)/2-1/2＝√2[√2sin(2x)/2-√2cos(2x)/2]/2
-1/2＝√2[sin(2x-π/4)]/2-1/2
故f(x)√2[sin(2x-π/4)]/2-1/2
令-π/2+2kπ≤2x-π/4≤π/2+2kπ (k∈Z)
即-π/8+kπ≤x≤3π/8+kπ (k∈Z)
此时f(x)单调递增,故f(x)的单调递增区间为{x|-π/8+kπ≤x≤3π/8+kπ (k∈Z)}
注：仅供参考!