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求函数y=2sin²x+2根号3 sinxcosx-2的周期
人气:189 ℃ 时间:2020-05-09 21:11:14
解答
y=2sin²x+2√3sinxcosx-2
=(1-cos2x)+√3sin2x-2
=√3sin2x-cos2x-1
=2[sin2x*(√3/2)-cos2x(1/2)]-1
=2[sin2xcos(π/6)-cos2xsin(π/6)]-1
=2sin(2x-π/6)-1
T=2π/2=π
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