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已知函数f(x)=cos(2x-pai/3)+2sin(x-pai/4).sin(x+pai/4)求函数在区间[-pai/12,pai /12]上最大值和最小值
人气:183 ℃ 时间:2019-08-21 09:22:56
解答
f(x)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/4)cos(π/4-x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)=cos(2x-π/3)+sin(2x-π/2)=cos(2x-π/3)-cos2x=(1/2)cos2x+(√3/2)sin2x-cos2x=(...
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