a向量=(根号3cosx/2,2cosx/2),向量b=(2cosx/2,-sinx/2)函数f(x)=向量a·向量b1)设θ∈[-π/2,π/2],且f(θ)=根号3 +1,求θ的值2)在△ABC中,AB=1,f(C)=根号3 +1,且△ABC的面积为 根号3 /2,求sinA+sinB的值
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a向量=(根号3cosx/2,2cosx/2),向量b=(2cosx/2,-sinx/2)函数f(x)=向量a·向量b1)设θ∈[-π/2,π/2],且f(θ)=根号3 +1,求θ的值2)在△ABC中,AB=1,f(C)=根号3 +1,且△ABC的面积为 根号3 /2,求sinA+sinB的值
a向量=(根号3cosx/2,2cosx/2),向量b=(2cosx/2,-sinx/2)函数f(x)=向量a·向量b
1)设θ∈[-π/2,π/2],且f(θ)=根号3 +1,求θ的值
2)在△ABC中,AB=1,f(C)=根号3 +1,且△ABC的面积为 根号3 /2,求sinA+sinB的值
a向量=(根号3cosx/2,2cosx/2),向量b=(2cosx/2,-sinx/2)函数f(x)=向量a·向量b1)设θ∈[-π/2,π/2],且f(θ)=根号3 +1,求θ的值2)在△ABC中,AB=1,f(C)=根号3 +1,且△ABC的面积为 根号3 /2,求sinA+sinB的值
(1)f(x)=ab=2√3cosx/2*cosx/2-2sinx/2cosx/2=√3(1+cosx)-sinx=2sin(x+2π/3)+√3
f(θ)=根号3 +1
得到2sin(θ+2π/3)+√3=√3+1 得到sin(θ+2π/3)=1/2
而θ∈[-π/2,π/2] 得到θ+2π/3属于[π/6,7π/6]
故θ+2π/3=π/6或5π/6
得到θ=-π/2或π/6
(2),f(C)=根号3 +1 而C属于(0,π)
得到C+2π/3属于(2π/3,5π/3)
得到C+2π/3=5π/6
得到C=π/6
而S=1/2absinC=√3/2 得到ab=2√3
而cosC=(a*a+b*b-c*c)/2ab=[(a+b)^2-c^2]/2ab=√3/2
得到a+b=√7
由正弦定理a/sinA=b/sinB=c/sinC=2
得到sinA+sinB=(a+b)/2=√7/2
f(x)=a·b=(sqrt(3)cosx/2,2cosx/2)·(2cosx/2,-sinx/2)=2sqrt(3)cos(x/2)^2-sinx
=sqrt(3)cosx-sinx+sqrt(3)=sqrt(3)-2sin(x-π/3)
1
t表示theta:f(t)=sqrt(3)-2sin(t-π/3)=sqrt(3)+1,故:sin(t-π/3)=-1/2
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f(x)=a·b=(sqrt(3)cosx/2,2cosx/2)·(2cosx/2,-sinx/2)=2sqrt(3)cos(x/2)^2-sinx
=sqrt(3)cosx-sinx+sqrt(3)=sqrt(3)-2sin(x-π/3)
1
t表示theta:f(t)=sqrt(3)-2sin(t-π/3)=sqrt(3)+1,故:sin(t-π/3)=-1/2
-π/2≤t≤π/2,故:-5π/6≤t-π/3≤π/6,故:t-π/3=-π/6或-5π/6,即:t=π/6或-π/2
2
f(C)=sqrt(3)-2sin(C-π/3)=sqrt(3)+1,即:sin(C-π/3)=-1/2,C是内角,故:0
而:a^2+b^2-2abcosC=c^2=1,即:a^2+b^2=1+2abcosC=7
即:a^2+b^2-2ab=7-4sqrt(3),即:(a-b)^2=(2-sqrt(3))^2,即:a-b=2-sqrt(3)或sqrt(3)-2
故:a=2,b=sqrt(3)或a=sqrt(3),b=2,故△ABC是以A或B为直角的直角三角形
故:sinA+sinB=1+sin(π/3)=1+sqrt(3)/2
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