ae4hm
幼苗
共回答了19个问题采纳率:94.7% 举报
(1)∵f(x)=cos(2x-[π/3])+2sin(x-[π/4])cos(x-[π/4])
=[1/2]cos2x+
3
2sin2x+sin(2x-[π/2])
=
3
2sin2x+[1/2]cos2x-cos2x
=
3
2sin2x-[1/2]cos2x
=sin(2x-[π/6]),
∴f(x)的最小正周期为T=[2π/2]=π;
(2)由(1)知,f(x)=sin(2x-[π/6]),
∵x∈[-[π/12],[π/2]],
∴2x-[π/6]∈[-[π/3],[5π/6]],
∴f(x)=sin(2x-[π/6])在区间[-[π/12],
1年前
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