对于正弦函数y=sinx 单调递减区间为
[π/2+2kπ,3π/2+2kπ],k∈N*
又∵Y=sin(ωx+π/4) (ω＞0）
∴ωx+π/4∈[π/2+2kπ,3π/2+2kπ]
即π/2+2kπ≤ωx+π/4≤3π/2+2kπ 即π/4+2kπ≤ωx≤5π/4+2kπ
∴π/4ω+2kπ/ω≤x≤5π/4ω+2kπ/ω
∴π/4ω≤π/2且 5π/4ω≥π
∴1/2≤ω≤5/4,