Answer
$\dfrac{1}{2} [\cos (2 \theta) -\cos (8 \theta) ] $
Work Step by Step
Recall the Product to Sum Identities:
$a) \sin x \cos y =\dfrac{1}{2} [\sin (x-y) +\sin (x+y)] \\ b) \sin x \sin y =\dfrac{1}{2} [\cos (x-y) -\cos (x+y)] \\ c) \cos x \cos y =\dfrac{1}{2} [\cos (x-y) +\cos (x+y)]$
We also know that $\cos$ and $\sec$ are even trigonometric functions.
This implies that $f (-x)=f(x) \implies \cos(-x)=\cos(x)$
By identity $(b)$, we have:
$\sin 3 \theta \ \sin 5 \theta=\dfrac{1}{2} [\cos (3 \theta- 5 \theta) -\cos (3 \theta+5 \theta) ] \\=\dfrac{1}{2} [\cos (-2 \theta) -\cos (8 \theta) ] \\=\dfrac{1}{2} [\cos (2 \theta) -\cos (8 \theta) ] $
