Question

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

Answer 1

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