关于余弦定理的证明题!
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/27 08:25:16
关于余弦定理的证明题!
三角形ABC中,abc为ABC所对的边,a^2+b^2-2abcos(C+D)=b^2+c^2-2bccos(A+D)
D属于[90,120],求证:三角形ABC为等腰三角形
三角形ABC中,abc为ABC所对的边,a^2+b^2-2abcos(C+D)=b^2+c^2-2bccos(A+D)
D属于[90,120],求证:三角形ABC为等腰三角形
证明:
将已知等式化简为:
a^-c^=2b[a*cos(C+D)-c*cos(A+D)] ①
在△ABC中,由正弦定理可得:a/sinA=b/sinB=c/sinC=2R
故:a=2R*sinA,b=2R*sinB,c=2R*sinC
代入已知等式中,可化简得:
sin^A-sin^C=2sinB*[sinA*cos(C+D)-sinC*cos(A+D)]
=2sinB*[sinA*(cosC*cosD-sinC*sinD)-sinC*(cosA*cosD-sinA*sinD)]
=2sinB*cosD*(sinA*cosC-cosA*sinC)
=2sinB*cosD*sin(A-C)
∵A+B+C=180°,∴sin(A+C)=sin(180°-B)=sinB
∴sin^A-sin^C
=2sin(A+C)*sin(A-C)*cosD
={cos[(A+C)-(A-C)]-cos[(A+C)+(A-C)]}*cosD
=(cos2C-cos2A)*cosD
=[1-2sin^C-(1-2sin^A)]*cosD
=2(sin^A-sin^C)*cosD
∴(sin^A-sin^C)*(2cosD-1)=0
∵D∈[90°,120°]
∴cosD∈[-1/2,0]
∴2cosD-1≠0
∴sin^A-sin^C=0
(sinA+sinC)*(sinA-sinC)=0
∵A,B,C∈(0,180°)
∴00
sinA=sinC
A=C (A+C=180°显然不成立,舍去)
因此,△ABC为等腰三角形
将已知等式化简为:
a^-c^=2b[a*cos(C+D)-c*cos(A+D)] ①
在△ABC中,由正弦定理可得:a/sinA=b/sinB=c/sinC=2R
故:a=2R*sinA,b=2R*sinB,c=2R*sinC
代入已知等式中,可化简得:
sin^A-sin^C=2sinB*[sinA*cos(C+D)-sinC*cos(A+D)]
=2sinB*[sinA*(cosC*cosD-sinC*sinD)-sinC*(cosA*cosD-sinA*sinD)]
=2sinB*cosD*(sinA*cosC-cosA*sinC)
=2sinB*cosD*sin(A-C)
∵A+B+C=180°,∴sin(A+C)=sin(180°-B)=sinB
∴sin^A-sin^C
=2sin(A+C)*sin(A-C)*cosD
={cos[(A+C)-(A-C)]-cos[(A+C)+(A-C)]}*cosD
=(cos2C-cos2A)*cosD
=[1-2sin^C-(1-2sin^A)]*cosD
=2(sin^A-sin^C)*cosD
∴(sin^A-sin^C)*(2cosD-1)=0
∵D∈[90°,120°]
∴cosD∈[-1/2,0]
∴2cosD-1≠0
∴sin^A-sin^C=0
(sinA+sinC)*(sinA-sinC)=0
∵A,B,C∈(0,180°)
∴00
sinA=sinC
A=C (A+C=180°显然不成立,舍去)
因此,△ABC为等腰三角形