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