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三角形ABC中,求证(a2-b2/cosA+cosB)+(b2-c2/cosB+cosC)+(c2-a2/cosC+cosA)=0
人气:470 ℃ 时间:2019-10-19 13:31:41
解答
证明:利用正弦定理a/(sina)=b/(sinb)=c/(sinc)=2R,就有:a^2=4R^2sin^2Ab^2=4R^2sin^2Bc^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)……...
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