sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值?

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sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值?

sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值?
sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值?

sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值?
cos(B-C)=cosBcosC+sinBsinc
sinB+sinc=-sinA
cosB+cosC=-cosA
所以同时平方
sinB^2+sinc^2+2sinBsinc=sinA^2
cosB^2+cosc^2+2cosBcosC=cosA^2
两式相加
2(cosBcosC+sinBsinc)+2=1
cos(B-C)
=cosBcosC+sinBsinc
=-1/2