作业帮已知函数fx=cos(2x-π/3)+2sin(x-4)cos(x-π/4),x∈R若对任意x∈[-π/12,π/2]
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已知函数fx=cos(2x-π/3)+2sin(x-4)cos(x-π/4),x∈R若对任意x∈[-π/12,π/2]
都有fx≥a成立,求a的取值范围
都有fx≥a成立,求a的取值范围
f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4),
=1/2cos2x+√3/2sin2x+sin(2x-π/2)
=1/2cos2x+√3/2sin2x-cos2x
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)
因,f(x)≥a
sin(2x-π/6)≥a
x∈[-π/12,π/2]
2x-π/6∈[-π/3,5π/6]
a
=1/2cos2x+√3/2sin2x+sin(2x-π/2)
=1/2cos2x+√3/2sin2x-cos2x
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)
因,f(x)≥a
sin(2x-π/6)≥a
x∈[-π/12,π/2]
2x-π/6∈[-π/3,5π/6]
a
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