Thu gọn \(A = {\sin ^2}\alpha  + {\sin ^2}\beta  + 2\sin \alpha \sin \beta .\cos \left( {\alpha  + \beta } \right)\) ta được:

$A = {\sin ^2}\alpha  + {\sin ^2}\beta  $ $+ 2\sin \alpha \sin \beta .\cos \left( {\alpha  + \beta } \right)$

$= {\sin ^2}\alpha  + {\sin ^2}\beta  $ $+ 2\sin \alpha \sin \beta .\left( {\cos \alpha .\cos\beta  - \sin \alpha .\sin\beta } \right)$

$= {\sin ^2}\alpha  + {\sin ^2}\beta  - 2{\sin ^2}\alpha {\sin ^2}\beta $ $ + 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$

$= {\sin ^2}\alpha \left( {1 - {{\sin }^2}\beta } \right) + {\sin ^2}\beta \left( {1 - {{\sin }^2}\alpha } \right) + 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$

$= {\sin ^2}\alpha {\cos ^2}\beta  + {\sin ^2}\beta {\cos ^2}\alpha  $ $+ 2\sin \alpha \sin \beta \cos \alpha .\cos\beta$

$= {\left( {\sin \alpha \cos \beta  + \sin \beta \cos \alpha } \right)^2} = {\sin ^2}\left( {\alpha  + \beta } \right)$