已知sinA+sinB=sinC,cosA+cosB=cosC,求cos(A-B)的值
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已知sinA+sinB=sinC,cosA+cosB=cosC,求cos(A-B)的值
sinA+sinB=sinC,cosA+cosB=cosC
(sinA)^2+(sinB)^2+2sinAsinB=(sinC)^2
(cosA)^2+(cosB)^2+2cosAcosB=(cosC)^2
(sinA)^2+(cosA)^2+(sinB)^2+(cosB)^2+2sinAsinB+2cosAcosB=(sinC)^2+(cosC)^2
1+1+2sinAsinB+2cosAcosB=1
sinAsinB+cosAcosB=-1/2
cos(A-B)=-1/2
(sinA)^2+(sinB)^2+2sinAsinB=(sinC)^2
(cosA)^2+(cosB)^2+2cosAcosB=(cosC)^2
(sinA)^2+(cosA)^2+(sinB)^2+(cosB)^2+2sinAsinB+2cosAcosB=(sinC)^2+(cosC)^2
1+1+2sinAsinB+2cosAcosB=1
sinAsinB+cosAcosB=-1/2
cos(A-B)=-1/2
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