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:
$\text{LHS } =\sin(\dfrac{\pi}{2}+\theta)
\\=\sin(\dfrac{\pi}{2})\cos(\theta)+\sin(\theta)\cos(\dfrac{\pi}{2})
\\=1\cdot \cos(\theta)+0\cdot \sin(\theta)
\\=\cos(\theta)+0
\\=\cos{\theta} \\=\text{ RHS}$