仙子126
幼苗
共回答了18个问题采纳率:72.2% 举报
(1)∵函数f(x)的最大值为3,
∴A+1=3,即A=2;
∵函数图象的相邻两条对称轴之间的距离为[π/2],
∴最小正周期T=π,
∴ω=2,
∴函数f(x)的解析式为:y=2sin(2x-[π/6])+1;
(2)∵f([α/2])=2sin(α-[π/6])+1=[11/5],即sin(α-[π/6])=[3/5]>0,
∵0<α<[π/2],
∴-[π/6]<α-[π/6]<[π/3],sin(α-[π/6])=[3/5]>0,
∴0<α-[π/6]<[π/3],
∴cos(α-[π/6])=[4/5],
∴sinα=sin[(α-[π/6])+[π/6])]=sin(α-[π/6])cos[π/6]+cos(α-[π/6])sin[π/6]=[3/5]•
3
2+[4/5]•
1年前
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