有关正弦定理在三角形ABC中,（b²－c²）／a² - 知识问答

有关正弦定理在三角形ABC中,（b²－c²）／a² ·sin2A＋（c²－a&s

1个回答

由正弦定理
a/sinA=b/sinB=c/sinC
(b^2-c^2)/a^2*sin2A
=[(sinB)^2-(sinC)^2]/(sinA)^2*2sinAcosA
由(sinB)^2-(sinC)^2
=(1-cos2B)/2-(1-cos2C)/2
=(cos2C-cos2B)/2
=[2sin(B+C)sin(B-C)]/2
=sin(B+C)sin(B-C)
=sinAsin(B-C)
(b^2-c^2)/a^2*sin2A
=2sin(B-C)cosA
=2sin(B-C)cos(B+C)
=sin2B-sin2C
可知原式=sin2B-sin2C+sin2C-sin2A+sin2A-sin2B=0

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