在三角形ABC中，求证(sinA)^2+(sinB)^2+2sinAsinBcos(A+B)=[sin(A+B)]^2

sina^2+sinb^2+2sinasinbcos(a+b)
=sina^2+sinb^2+2sinasinb(cosacosb-sinasinb)
=sina^2+sinb^2+2sinasinbcosacosb-2(sina)^2(sinb)^2
=[sina^2-(sina)^2(sinb)^2]+[sinb^2-(sina)^2(sinb)^2+2sinasinbcosacosb
=(sina)^2[1-(sinb)^2]+(sinb)^2[1-(sina)^2]+2sinasinbcosacosb
=(sina)^2(cosb)^2+(sinb)^2(cosa)^2+2sinasinbcosacosb
=(sinacosb+cosasinb)^2
=[sin(a+b)]^2

(sinA)^2+(sinB)^2+2sinAsinBcos(A+B)
=(sinA)^2+(sinB)^2+2sinAsinB(cosAcosB-sinAsinB)
=(sinA)^2+(sinB)^2+2sinAsinBcosAcosB-2sinAsinBsinAsinB

[sin(A+B)]^2
=(sinAcosB)^2+(cosAsinB)^2+2sinAsinBcosAcosB

两式相减有:
(sinA)^2+(sinB)^2+2sinAsinBcos(A+B)-[sin(A+B)]^2
=(sinA)^2+(sinB)^2-2sinAsinBsinAsinB-(sinAcosB)^2-(cosAsinB)^2
=(sinA)^2-(sinAcosB)^2+(sinB)^2-(cosAsinB)^2-2sinAsinBsinAsinB
=(sinA)^2(1-(cosB)^2)+(sinB)^2(1-(cosA)^2)-2sinAsinBsinAsinB
=(sinA)^2(sinB)^2+(sinB)^2(sinA)^2)-sinAsinBsinAsinB
=0
得证