已知向量m=(a+1,sinx)(a∈R,且a为常数),n=(1,4cos(x+π/6)),设g(x)=m*n1、求g(x)的最小正周期2、若g(x)在[0,π/3)上的最大值与最小值之和为7,求a的值.
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已知向量m=(a+1,sinx)(a∈R,且a为常数),n=(1,4cos(x+π/6)),设g(x)=m*n1、求g(x)的最小正周期2、若g(x)在[0,π/3)上的最大值与最小值之和为7,求a的值.
已知向量m=(a+1,sinx)(a∈R,且a为常数),n=(1,4cos(x+π/6)),设g(x)=m*n
1、求g(x)的最小正周期
2、若g(x)在[0,π/3)上的最大值与最小值之和为7,求a的值.
已知向量m=(a+1,sinx)(a∈R,且a为常数),n=(1,4cos(x+π/6)),设g(x)=m*n1、求g(x)的最小正周期2、若g(x)在[0,π/3)上的最大值与最小值之和为7,求a的值.
(1)g(x)=m*n=a+1+4sinxcos(x+π/6)=a+1+4sinx(cosxcosπ/6-sinxsinπ/6)=1+2√3sinxcosx-2sin²x+a
=cos²x-sin²x+2√3sinxcosx+a=cos2x+√3sin2x+a=2sin(2x+π/6)+a
则求g(x)的最小正周期
T=2π/2=π
(2)g(x)在[0,π/3)得π/6
因为向量m=(a+1,sinx)(a∈R,且a为常数),向量n=(1,4cos(x+π/6))
于是g(x)=m•n
=a+1+4 sinx cos(x+π/6)
= a+1+4 sinx(cosx cosπ/6-sinxsinπ/6)
= a+1+2 (根号3)sinxcosx-2(sinx) ^2
= a+1+ (根号3)sin2x+cos2...
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因为向量m=(a+1,sinx)(a∈R,且a为常数),向量n=(1,4cos(x+π/6))
于是g(x)=m•n
=a+1+4 sinx cos(x+π/6)
= a+1+4 sinx(cosx cosπ/6-sinxsinπ/6)
= a+1+2 (根号3)sinxcosx-2(sinx) ^2
= a+1+ (根号3)sin2x+cos2x-1
=a+2sin(2x+π/6)
最小正周期T=π
因为x∈[0,π/3)
所以2x+π/6 ∈[π/6,5π/6)
则2sin(2x+π/6) ∈〔1,2〕
g(x)max=2+a, g(x)min=1+a
于是3+2a=7
即a=2
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