Question

已知三角形ABC的三个内角A,B,C的对边分别为a,b,c

求证(a^2-b^2)/(cos A+ cos B)+(b^2-c^2)/(cos B+cos C)+(c^2-a^2)/(cos A+cos C)=0

Answer 1

证明：
利用正弦定理a/(sina)=b/(sinb)=c/(sinc)=2R, 就有：
a^2=4R^2sin^2A
b^2=4R^2sin^2B
c^2=4r^2sin^2C
(a^2-b^2)=4R^2(sin^2A-sin^2B)
=4R^2(1-cos^2A-1+cos^2B)
=4R^2(cos^2B-cos^2A)
=4R^2(cosA+cosB)(cosB-cosA)……（1）式
同理，可得
(b^2-c^2)=4R^2(sin^2B-sin^2C)
=4R^2(cosB+cosC)(cosC-cosB)………（2）式
(C^2-a^2)=4R^2(sin^2C-sin^2A)
=4R^2(cosC+cosA)(cosA-cosC）…………（3）式
(a^2-b^2)/(cosA+cosB)+(b^2-c^2)/(cosB+cosC)+(c^2-a^2)/(cosC+cosA)
=4R^2(cosB-cosA)+4R^2(cosC-cosB)+4R^2(cosA-cosC)
=0
得证