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y=2sinxcosx_根号3cos2x+1的最小正周期和最值
最小值是?过程
人气:120 ℃ 时间:2020-05-08 22:29:16
解答
y=2sinxcosx-√3cos2x+1
=2(1/2sin2x-√3/2cos2x)+1
=2(sin2x*cosπ/3-cos2xsinπ/3)+1
=2sin(2x+π/3)+1
T=π
当2x+π/3=π/2+2kπ,即x=π/12+kπ,k∈Z时,y取最大值为3
x=-5π/12+kπ,k∈Z时,y取最小值为-1
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