用正弦定理证明余弦定理

1个回答

  • 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