Giải thích các bước giải:
$VT=\sin4x\cos2x-\sin3x\cos x\\
=\frac{1}{2}\left (\sin(4x-2x)+\sin(4x+2x) \right )-\frac{1}{2}\left ( \sin(3x-x)+\sin(3x+x) \right )\\
=\frac{1}{2}\left ( \sin2x+\sin6x-\sin2x-\sin4x \right )\\
=\frac{1}{2}\left (\sin6x-\sin4x \right )\\
=\frac{1}{2}.2\cos\frac{6x+4x}{2}\sin\frac{6x-4x}{2}\\
=\cos5x\sin x=VP\Rightarrow ĐPCM