Ta đi chứng minh đẳng thức ${\sin ^2}C - {\sin ^2}A = \sin \left( {C + A} \right)\sin \left( {C - A} \right)$
$\begin{array}{l} \sin \left( {C + A} \right)\sin \left( {C - A} \right) \\= \dfrac{1}{2}\left[ {\cos \left( {C + A - C + A} \right) - \cos \left( {C + A + C - A} \right)} \right]\\ = \dfrac{1}{2}\left( {\cos 2A - \cos 2C} \right) = \dfrac{1}{2}\left( {1 - 2{{\sin }^2}A - 1 + 2{{\sin }^2}C} \right) = {\sin ^2}C - {\sin ^2}A \end{array}$
$\begin{array}{l} {\sin ^2}B + {\sin ^2}C - {\sin ^2}A\\ = {\sin ^2}B + \sin \left( {C - A} \right)\sin \left( {C + A} \right) = {\sin ^2}B + \sin \left( {\pi - B} \right)\sin \left( {C - A} \right)\\ = {\sin ^2}B + \sin B\sin \left( {C - A} \right)\\ = \sin B\left[ {\sin B + \sin \left( {C - A} \right)} \right]\\ = 2\sin B\sin \left( {\dfrac{{B + C - A}}{2}} \right)\cos \left( {\dfrac{{B - C + A}}{2}} \right)\\ = 2\sin B\sin \left( {\dfrac{{B + C + A}}{2} - A} \right)\cos \left( {\dfrac{{B + C + A}}{2} - C} \right)\\ = 2\sin B\sin \left( {\dfrac{\pi }{2} - A} \right)\cos \left( {\dfrac{\pi }{2} - C} \right)\\ = 2\sin B.\cos A.\sin C \end{array}$
