y=sinx+sin(x+π/2)＝sinx＋cosx＝√2sin﹙x＋π／4﹚
∵x∈(0,π﹚
∴x＋π／4∈﹙π／4,5π／4﹚
∴sin﹙x＋π／4﹚∈﹙﹣√2／2,1]
∴y=sinx+sin(x+π/2)
＝sinx＋cosx
＝√2sin﹙x＋π／4﹚∈﹙﹣√2,2]＝√2sin﹙x＋π／4﹚∈﹙﹣√2,2] 值域是不是算错了?不好意思！大意了∴y=sinx+sin(x+π/2)＝sinx＋cosx＝√2sin﹙x＋π／4﹚∈﹙﹣1,√2]y=2sinx(sinx+cosx)那个这道题能不能帮忙算下?y=2sinx(sinx+cosx)＝2sin²x＋2sinxcosx＝1－cos2x＋sin2x＝sin2x－cos2x＋1＝√2sin﹙2x－π／4﹚＋1∵sin﹙2x－π／4﹚∈[﹣1,1]∴√2sin﹙2x－π／4﹚∈[﹣√2,√2]∴√2sin﹙2x－π／4﹚＋1∈[1－√2,1＋√2]∴y=2sinx(sinx+cosx)∈[1－√2,1＋√2]记得采纳哦