2道高一数学证明题, 求计算过程 ,很简单的!1. 证cosA+cos(120+B)+cos(120-B)/(sinB+sin(120+A)-sin(120-A)=tan(A+B)/2 R.H.S / L.H.S = .=......= L.H.S. / R.H.S.2. 证4cosθcos[(2π/3)+θ]cos[(2π/3)-θ] = cos3θ R.H.S / L.H.S = .=......= L.H.S. / R.H.S.

2019-05-23

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2道高一数学证明题, 求计算过程 ,很简单的!
1. 证cosA+cos(120+B)+cos(120-B)/(sinB+sin(120+A)-sin(120-A)=tan(A+B)/2
R.H.S / L.H.S = .
=.
.
.
.
.
.= L.H.S. / R.H.S.
2. 证4cosθcos[(2π/3)+θ]cos[(2π/3)-θ] = cos3θ
R.H.S / L.H.S = .
=.
.
.
.
.
.= L.H.S. / R.H.S.

优质解答

1.[cosA+cos(120+B)+cos(120-B)]/[(sinB+sin(120+A)-sin(120-A)]
cosA+cos120cosB-sin120sinB+cos120cosB+sin120sinB
=------------------------------------------------------
sinB+sin120cosA+cos120sinA-sin120cosA+cos120sinA
cosA-cosB
=-----------
sinB-sinA
-2sin[(A+B)/2]sin[(A-B)/2]
=----------------------------
2cos[(A+B)/2]sin[(A-B)/2]
=tan[(A+B)/2]
2.cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
所以 cos(A+B)+cos(A-B)=2cosAcosB
所以 cosAcosB=[cos(A+B)+cos(A-B)]/2
所以
4cosθcos[(2π/3)+θ]cos[(2π/3)-θ]
=4cosθ[cos4π/3+cos2θ]/2
=2cosθ(-1/2+cos2θ)
=-cosθ+2cosθcos2θ
=-2cosθ+2[cos3θ+cosθ)/2
=cos3θ
得证 1.[cosA+cos(120+B)+cos(120-B)]/[(sinB+sin(120+A)-sin(120-A)]
cosA+cos120cosB-sin120sinB+cos120cosB+sin120sinB
=------------------------------------------------------
sinB+sin120cosA+cos120sinA-sin120cosA+cos120sinA
cosA-cosB
=-----------
sinB-sinA
-2sin[(A+B)/2]sin[(A-B)/2]
=----------------------------
2cos[(A+B)/2]sin[(A-B)/2]
=tan[(A+B)/2]
2.cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
所以 cos(A+B)+cos(A-B)=2cosAcosB
所以 cosAcosB=[cos(A+B)+cos(A-B)]/2
所以
4cosθcos[(2π/3)+θ]cos[(2π/3)-θ]
=4cosθ[cos4π/3+cos2θ]/2
=2cosθ(-1/2+cos2θ)
=-cosθ+2cosθcos2θ
=-2cosθ+2[cos3θ+cosθ)/2
=cos3θ
得证