作业帮已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
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已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
由积化和差公式可得：
f(x)=2[sin(2x+π/6)+sin(π/6)]-1=2sin(2x+π/6)
故 f的最小正周期T=π
因为π/6
f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
f(x)最小正周期T=2π/2=π
(2)
∵x∈[-π/6,π/4]
∴2x∈[-π/3...
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f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
f(x)最小正周期T=2π/2=π
(2)
∵x∈[-π/6,π/4]
∴2x∈[-π/3,π/2]
∴2x+π/6∈[-π/6,2π/3]
∴2x+π/6=π/2时，f(x)max=2
2x+π/6=-π/6时，f(x)min=-1
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f(x)=4cosxsin(x+π/6)-1=4cosx(sinx*√3/2+cosx*1/2)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2*(sin2x*√3/2+cos2x*1/2)
=2sin(2x+π/6) ，
因此，最小正周期为 2π/2=π ，
当 -π/6<=x<=π/4 时，-π/6<=2...
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f(x)=4cosxsin(x+π/6)-1=4cosx(sinx*√3/2+cosx*1/2)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2*(sin2x*√3/2+cos2x*1/2)
=2sin(2x+π/6) ，
因此，最小正周期为 2π/2=π ，
当 -π/6<=x<=π/4 时，-π/6<=2x+π/6<=2π/3 ，
因此 -1/2<=sin(2x+π/6)<=1 ，
即函数在区间 [-π/6，π/4] 上的最大值为 2 ，最小值为 -1 。
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