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^2-2
1个回答

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