Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x.\cos y = \frac{1}{2}\left( {\cos \left( {x + y} \right) + \cos \left( {x - y} \right)} \right)\\
\sin x.\cos y = \frac{1}{2}\left( {\sin \left( {x + y} \right) + \sin \left( {x - y} \right)} \right)\\
8\sin x.\cos x.\sin \left( {60^\circ - 2x} \right).\cos \left( {30^\circ - 2x} \right)\\
= 4.\left( {2\sin x.\cos x} \right).cos\left( {90^\circ - \left( {60^\circ - 2x} \right)} \right).\cos \left( {30^\circ - 2x} \right)\\
= 4.\sin 2x.\cos \left( {30^\circ + 2x} \right).\cos \left( {30^\circ - 2x} \right)\\
= 4.\sin 2x.\frac{1}{2}\left( {\cos \left( {30^\circ + 2x + 30^\circ - 2x} \right) + \cos \left( {30^\circ + 2x - 30^\circ + 2x} \right)} \right)\\
= 2.\sin 2x.\left( {\cos 60^\circ + \cos 4x} \right)\\
= 2.\sin 2x.\cos 60^\circ + 2.\sin 2x.\cos 4x\\
= 2.\sin 2x.\frac{1}{2} + 2\sin 2x.\cos 4x\\
= \sin 2x + 2.\frac{1}{2}.\left( {\sin \left( {2x + 4x} \right) + \sin \left( {2x - 4x} \right)} \right)\\
= \sin 2x + \left( {\sin 6x + \sin \left( { - 2x} \right)} \right)\\
= \sin 2x + \sin 6x - \sin 2x\\
= \sin 6x
\end{array}\)
