Lời giải

cos x.cos 2x.cos 3x

\( = \frac{1}{2}\left[ {\cos \left( {x + 2x} \right) + \cos \left( {x - 2x} \right)} \right]\,.\,\cos 3x\)

\( = \frac{1}{2}\left( {\cos 3x + \cos x} \right)\,.\,\cos 3x\)

\( = \frac{{{{\cos }^2}3x}}{2} + \frac{{\cos x\,.\,\cos 3x}}{2}\)

\( = \frac{{{{\cos }^2}3x}}{2} + \frac{1}{4}\left[ {\cos \left( {x + 3x} \right) + \cos \left( {x - 3x} \right)} \right]\)

\( = \frac{{{{\cos }^2}3x}}{2} + \frac{1}{4}\left( {\cos 4x + \cos 2x} \right)\)

\( = \frac{{\cos 6x + 1}}{4} + \frac{1}{4}\left( {\cos 4x + \cos 2x} \right)\)

\( = \frac{1}{4}\left( {\cos 6x + \cos 4x + \cos 2x + 1} \right)\)