用正弦定理证明余弦定理
由正弦定理:a/sinA=b/sinB=c/sinC=2R,得:a/(2R)=sinA,b/(2R)=sinB,c/(2R)=sinC.
进而得:(a^2+b^2-2ab×cosC)/(2R)^2=(sinA)^2+(sinB)^2-2sinAsinBcosC
=(sinA)^2+(sinB)^2-2sinAsinBcos(180°-A-B)
=(sinA)^2+(sinB)^2+2sinAsinBcos(A+B)
=(sinA)^2+(sinB)^2+2sinAsinB(cosAcosB-sinAsinB)
=(sinA)^2+(sinB)^2+2sinAsinBcosAcosB-2(sinAsinB)^2
=[(sinA)^2-(sinAsinB)^2]+[(sinB)^2-(sinAsinB)^2]+2sinAcosBcosAsinB
=(sinA)^2[1-(sinB)^2]+(sinB)^2[1-(sinA)^2]+2sinAcosBcosAsinB
=(sinAcosB)^2+(cosAsinB)^2+2sinAcosBcosAsinB
=(sinAcosB+cosAsinB)^2
=[sin(A+B)]^2
=[sin(180°-C)]^2
=(sinC)^2
=c^2/(2R)^2
两边同时乘以(2R)^2,得:a^2+b^2-2ab×cosC=c^2
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