求证a cos A + b cos B = c cos (A-B )
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求证acosA+bcosB=ccos(A-B)求证acosA+bcosB=ccos(A-B)求证acosA+bcosB=ccos(A-B)设t=a/sinA=b/sinB=c/sinCa=tsinAb
求证a cos A + b cos B = c cos (A-B )
求证a cos A + b cos B = c cos (A-B )
求证a cos A + b cos B = c cos (A-B )
设t=a/sinA=b/sinB=c/sinC a=tsinA b=tsinB c=tsinC acosA+bcosB-ccos(A-B) =t(sinAcosA+sinBcosB-sinCcos(A-B)) =t(sinAcosA+sinBcosB-sin(A+B)cos(A-B)) =t/2 * (2sinAcosA+2sinBcosB-2sin(A+B)cos(A-B)) =t/2 * (sin2A + sin2B - (sin2A+sin2B)) =0 所以a cos A + b cos B = c cos (A-B )
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