∵最小正周期为180度=360°/ωω=2∴f(x)=√2Sin(2x-45°)∴当2x-45°=90°﹢k×360°即x=180°k﹢67.5°时Sin(2x-45°)取最大值1f(x)取最大值√2注:y=Asin﹙ωx﹢φ﹚的最小正周期为360°/ω你好!f (x)=(√2)·sin(ωx-π/4) 解:ω=2 f (x)=(√2)·sin(2x-π/4) f (x)最大=√2 , x=kπ+3π/8 (k∈z) f (x)最小=-√2 , x=kπ-π/8 (k∈z)如果对你有帮助,望采纳。f (x)=(√2)·sin(ωx-π/4) 解:ω=2 f (x)=(√2)·sin(2x-π/4) f (x)最大=√2 , x=kπ+3π/8 (k∈z) f (x)最小=-√2 , x=kπ-π/8 (k∈z)