Answer
See proof below.
Work Step by Step
Apply the sum formula:
$\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B)$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$\text{LHS } =\cos(\dfrac{3 \pi}{2}+\theta)
\\=\cos(\dfrac{3 \pi}{2})\cos(\theta) -\sin (\theta)\sin(\dfrac{3 \pi}{2})
\\=(0) \cos(\theta) -(-1) \sin(\theta)
\\=\sin(\theta)+0
\\=\sin{\theta} \\=\text{ RHS}$