Với tam giác \(ABC\). Hãy chứng minh rằng \(\sin 2A + \sin 2B + \sin 2C \)\(= 4\sin A\sin B\sin C\).

Ta có: \(\sin 2A + \sin 2B + \sin 2C\)

\(\begin{array}{l} = 2\sin \frac{{2A + 2B}}{2}\cos \frac{{2A - 2B}}{2} + \sin 2C\\ = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + 2\sin C\cos C\end{array}\)

\( = 2\sin \left( {{{180}^0} - \left( {A + B} \right)} \right)\cos \left( {A - B} \right)\)\( + 2\sin C\cos C\)

\( = 2\sin C.\cos \left( {A - B} \right)\) \( + 2\sin C\cos C\)

\( = 2\sin C\left( {\cos \left( {A - B} \right) + \cos C} \right)\)

\( = 2\sin C.2\cos \frac{{A - B + C}}{2}.\cos \frac{{A - B - C}}{2}\)

\( = 4\sin C.\cos \frac{{A + C - B}}{2}.\cos \frac{{B + C - A}}{2}\)

Ta có: \(\cos \frac{{A + C - B}}{2} = \sin \left( {{{90}^0} - \frac{{A + C - B}}{2}} \right)\)  \( = \sin \left( {\frac{{A + B + C}}{2} - \frac{{A + C - B}}{2}} \right)\)\( = \sin B\) (do \(A + B + C = {180^0}\) )

\(\cos \frac{{B + C - A}}{2} = \sin \left( {{{90}^0} - \frac{{B + C - A}}{2}} \right)\)  \( = \sin \left( {\frac{{A + B + C}}{2} - \frac{{B + C - A}}{2}} \right)\)\( = \sin A\) (do \(A + B + C = {180^0}\) )

Từ đó ta có: \(4\sin C.\cos \frac{{A + C - B}}{2}.\cos \frac{{B + C - A}}{2}\)\( = 4\sin A\sin B\sin C\)

Hay  \(\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C\).