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在△ABC中 ,由正弦定理asinA =bsinB =csinC =2R ,(R表示△ABC外接圆半径 )得 :a =2RsinA ,b =2RsinB ,c=2RsinC .将它们代入余弦定理的一个表达式a2 =b2 +c2 - 2bccosA .得 4R2 sin2 A =4R2 sin2 B + 4R2 si
In ΔABC, from the sine theorem asinA = bsinB = csinC = 2R, (R denotes the circumcircle radius of △ABC): a = 2RsinA, b = 2RsinB, c = 2RsinC. An expression that substitutes them into the cosine theorem a2 = B2 +c2 - 2bccosA. Get 4R2 sin2 A =4R2 sin2 B + 4R2 si