We have

$y=\cos x \cdot \cos 2x \cdot \cos 3x$

$\therefore y = \dfrac {1}{2} \cdot \cos x \cdot 2\cos 2x \cdot \cos 3x = \dfrac {1}{2} \cdot \cos x \cdot \big [ \cos 5x + \cos x \big ]$

$\therefore y = \dfrac {1}{2} \cdot \cos x \cdot \cos 5x + \dfrac {1}{5} \cdot \cos x \cdot \cos x$

$\therefore y = \dfrac {1}{4} \big [ \cos 6x + \cos 4x + 1 + \cos 2x \big ] $

Usign formula,

$y = \cos (ax + b) \text { then }y_n = a^n \cos \left ( ax + b + \dfrac {n \pi}{2} \right )$

$\therefore y_n = \dfrac {1}{4} \left [ 6^n \cos \left ( 6x + \dfrac {n\pi}{2} \right ) + 4^n \cos \left ( 4x + \dfrac {n \pi}{2} \right ) + 2^n \cos \left ( 2x + \dfrac {n\pi}{2} \right ) \right ]$