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问题 解答题
求证：在△ABC中，其中α，β，γ是三角形的内角，cosα=
sin2β+sin2γ-sin2α
2sinβ•sinγ
．
答案

证：设R为△ABC的外接圆的半径，

则由正弦定理可得，a=2Rsinα，b=2Rsinβ，c=2Rsinγ．

代入余弦定理中，则可得

cosα=

b2+c2-a2
2bc
=
4R2(sin2β+sin2γ-sin2α)
4R2•2sinβ•sinγ

=

sin2β+sin2γ-sin2α
2sinβ•sinγ
．

故cosα=

sin2β+sin2γ-sin2α
2sinβ•sinγ
成立．

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