Giải thích các bước giải:
$VT=\cos x.\cos 5x+\sin x.\sin3x\\
=\dfrac{1}{2}.\left [ \cos(x-5x)+\cos(x+5x) \right ]+\dfrac{1}{2}\left [ \cos(x-3x)-\cos(x+3x) \right ]\\
=\dfrac{1}{2}.\left [ \cos4x+\cos6x \right ]+\dfrac{1}{2}\left [ \cos2x-\cos4x \right ]\\
=\dfrac{1}{2}\left [ \cos4x+\cos6x+\cos2x-\cos4x \right ]\\
=\dfrac{1}{2}(\cos6x+\cos2x)\\
=\dfrac{1}{2}.2.\cos\dfrac{6x+2x}{2}\cos\dfrac{6x-2x}{2}\\
=\cos4x.\cos2x=VP$