Answer
$\dfrac{1}{2} [\sin (6 \theta) +\sin (2 \theta) ] $
Work Step by Step
Recall the Product to Sum Identities:
$a) \sin x \cos y =\dfrac{1}{2} [\sin (x-y) +\sin (x+y)] \\ b) \sin x \sin y =\dfrac{1}{2} [\cos (x-y) -\cos (x+y)] \\ c) \cos x \cos y =\dfrac{1}{2} [\cos (x-y) +\cos (x+y)]$
By identity $(a)$, we have:
$\sin 4 \theta \ \cos 2 \theta=\dfrac{1}{2} [\sin (4 \theta +2 \theta) +\sin (4 \theta -2 \theta) ] \\=\dfrac{1}{2} [\sin (6 \theta) +\sin (2 \theta) ] $
