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]上最的最值
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解决时间 2021-03-06 16:46
- 提问者网友：山高云阔
- 2021-03-06 06:05
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]上最的最值
最佳答案
- 二级知识专家网友：逐風
- 2021-03-06 06:19
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
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)
﹣π/12≤x≤π/12
﹣π/6≤2x≤π/6
﹣π/3≤2x-π/6≤0
当2x-π/6=0时，y最大
y(max)=0
当2x-π/6= -π/3时，y最小
y(min)= - √3/2
=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
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)
﹣π/12≤x≤π/12
﹣π/6≤2x≤π/6
﹣π/3≤2x-π/6≤0
当2x-π/6=0时，y最大
y(max)=0
当2x-π/6= -π/3时，y最小
y(min)= - √3/2
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