Cho $cos2x = \frac{1}{4}$. Tính: $A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right)$; $B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)$.

Ta có:

$A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right)$

    $= \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]$

    $= \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right]$

    $= \frac{1}{2}\left[ {\frac{1}{4} + \frac{1}{2}} \right] = \frac{3}{8}$.

$B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)$

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

    $= - \frac{1}{2}\left[ {\cos 2x - \cos \frac{{2\pi }}{3}} \right]$

    $= - \frac{1}{2}\left[ {\frac{1}{4} - \left( { - \frac{1}{2}} \right)} \right] = - \frac{3}{8}$.

Vậy $A = \frac{3}{8},B = - \frac{3}{8}$.