Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2\sin x.\sin y = - \left[ {\cos \left( {x + y} \right) - \cos \left( {x - y} \right)} \right]\\
2\sin x.\cos y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)\\
4\sin \alpha .\sin \left( {\dfrac{\pi }{3} - \alpha } \right).\sin \left( {\dfrac{\pi }{3} + \alpha } \right)\\
= 2\sin \alpha .\left[ {2.\sin \left( {\dfrac{\pi }{3} + \alpha } \right).\sin \left( {\dfrac{\pi }{3} - \alpha } \right)} \right]\\
= 2\sin \alpha .\left[ { - \left( {\cos \left( {\dfrac{\pi }{3} + \alpha + \dfrac{\pi }{3} - \alpha } \right) - \cos \left( {\dfrac{\pi }{3} + \alpha - \dfrac{\pi }{3} + \alpha } \right)} \right)} \right]\\
= 2\sin \alpha .\left[ { - \left( {\cos \dfrac{{2\pi }}{3} - \cos 2\alpha } \right)} \right]\\
= 2\sin \alpha .\left( {\dfrac{1}{2} + \cos 2\alpha } \right)\\
= \sin \alpha + 2\sin \alpha .\cos 2\alpha \\
= \sin \alpha + \sin \left( {\alpha + 2\alpha } \right) + \sin \left( {\alpha - 2\alpha } \right)\\
= \sin \alpha + sin3\alpha + sin\left( { - \alpha } \right)\\
= \sin \alpha + \sin 3\alpha - \sin \alpha \\
= \sin 3\alpha
\end{array}\)
