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
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![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](/uploads/image/z/1626141-21-1.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%E6%B1%82%E8%AF%81%28a%5E2-b%5E2%29%2F%28cos+A%2B+cos+B%29%2B%28b%5E2-c%5E2%29%2F%28cos+B%2Bcos+C%29%2B%28c%5E2-a%5E2%29%2F%28cos+A%2Bcos+C%29%3D0)
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
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
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
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