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已知函数f(x)=2√3sinxcosx+2cos²x-1(x∈R)
1.求函数f(x)的最小正周期及在区间[0,π/2]的最大和最小值 2.若f(x0)=6/5,x0∈[π/4,π/2],求cos2x0的值
人气:476 ℃ 时间:2020-03-24 17:34:35
解答
f(x)=2√3sinxcosx+2cos²x-1(x∈R) f(x) = 2√3sinxcosx+2cos²x-1= √3 * ( 2sinxcosx) + (2cos²x-1)= 根号3 sin2x + cos2x= 2 (sin2xcosπ/6+cos2xsinπ/6)=2 sin(2x+π/6)最小正周期:2π/2 = ...
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