Answer
See proof below.
Work Step by Step
Apply the difference 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:
$\text{LHS } =\sin(\pi - \theta)
\\=\sin (\pi) \cos (\theta) -\cos \pi \sin(\theta)
\\=(0)( \cos(\theta)) -(-1) (\sin(\theta))
\\=\sin (\theta)\\=\text{ RHS}$