证明在△ABC中,(a-ccosB)/(b-ccosA)=sinB/sinA

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证明在△ABC中,(a-ccosB)/(b-ccosA)=sinB/sinA
证明在△ABC中,(a-ccosB)/(b-ccosA)=sinB/sinA

证明在△ABC中,(a-ccosB)/(b-ccosA)=sinB/sinA
由正弦定理
(a-ccosB)/(b-ccosA)
=(sinA-sinCcosB)/(sinB-sinCcosA)
=[sin(π-B-C)-sinCcosB]/[sin(π-A-C)-sinCcosA]
=[sin(B+C)-sinCcosB]/[sin(A+C)-sinCcosA]
=(sinBcosC+sinCcosB-sinCcosB)/(sinAcosC+sinCcosA-sinCcosA)
=(sinBcosC)/(sinAcosC)
=sinB/sinA