4yg0rbk5
幼苗
共回答了20个问题采纳率:100% 举报
f(x)=sinxcosx-√3sin²x
=1/2 sin2x-√3/2 (1-cos2x)
=cosπ/3 sin2x+sinπ/3 cos2x -√3/2
=sin(2x+π/3)-√3/2
(1)最小正周期T=2π/2=π
(2)因为x∈(0,π/2)
所以 0<2x<π
π/3<2x+π/3<4π/3
所以sin(2x+π/3)∈(-√3/2,1)
所以f(x)=sin(2x+π/3)-√3/2 ∈(-√3,1-√3/2)
1年前
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