化简sina^2+sinb^2+2sinasinbcos(a+b)
5个回答

不需要“积化和差”公式的证法:

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.