Answer
$- 2 \sin (\dfrac{ \theta}{2}) \cos \theta $
Work Step by Step
Recall the Sum to Product Identities:
$a) \sin x +\sin y =2 \sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2} \\ b) \sin x - \sin y =2 \sin \dfrac{x -y}{2} \cos \dfrac{x +y}{2} \\ c) \cos x +\cos y =2 \cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2} \\ d) \cos x - \cos y = -2 \sin \dfrac{x+y}{2} \sin \dfrac{x-y}{2}$
We know that $\sin$, $\csc$, and $\tan$ are odd trigonometric functions. This implies that $f (-x)=f(x) \implies \sin(-x)=- \sin(x)$
By identity $(b)$, we have:
$\sin (\dfrac{ \theta}{2}) - \sin (\dfrac{3\theta}{2}) = 2 \sin \dfrac{\dfrac{ \theta}{2} - \dfrac{3 \theta}{2} }{2} \ \cos \dfrac{ \dfrac{ \theta}{2} + \dfrac{3 \theta}{2} }{2} \\= 2 \sin (\dfrac{-\theta}{2} ) \cos \theta \\=- 2 \sin (\dfrac{ \theta}{2}) \cos \theta $
