Answer
See proof below.
Work Step by Step
Apply the sum and difference formula:
$\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ and $\sin(A-B)=\sin(A)\cos(B) -\cos(A)\sin(B)$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$$\sin{(\alpha+\beta)}+\sin{(\alpha-\beta)}\\
=[\sin{(\alpha)} \ \cos{(\beta)}+\cos{(\alpha)}\ \sin{(\beta)}]+[\sin{(\alpha)}\ \cos{(\beta)}-\cos{(\alpha)} \ \sin{(\beta)}]\\
=[\sin{(\alpha)}\cos{(\beta)}+\sin{(\alpha)}\cos{(\beta)}+\cos{(\alpha)}]+[\sin{(\beta)}-\cos{(\alpha)}\sin{(\beta)}]\\
=2\sin{(\alpha)} \ \cos{(\beta)} \\=\text{ RHS}$$