求y=cos²x+2根号3sinxcosx-sin²x的最大值最小值.
人气:264 ℃ 时间:2020-03-30 01:40:58
解答
∵y
=(cosx)^2+2√3sinxcosx-(sinx)^2
=[(cosx)^2-(sinx)^2]+√3sin2x
=cos2x+√3sin2x
=2[(1/2)cos2x+(√3/2)sin2x]
=2(sin30°2cos2x+cos30°sin2x)
=2sin(30°+2x).
∴y的最大值是2,最小值是-2.
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