sinA=sinBcosB+sinBcosC+sinCcosB+sinCcosC
=sin(180-(B+C))=sin(B+C)=sinBcosC+cosBsinC
由
sinBcosC+cosBsinC=sinBcosB+sinBcosC+sinCcosB+sinCcosC
得:sinBcosB+sinCcosC=0
1/2sin2B+1/2sin2C=0
sin2B+sin2C=0
sinA=sinBcosB+sinBcosC+sinCcosB+sinCcosC
=sin(180-(B+C))=sin(B+C)=sinBcosC+cosBsinC
由
sinBcosC+cosBsinC=sinBcosB+sinBcosC+sinCcosB+sinCcosC
得:sinBcosB+sinCcosC=0
1/2sin2B+1/2sin2C=0
sin2B+sin2C=0