trigonometry - If the sides $a$, $b$ and $c$ of $triangle ABC$ are in Arithmetic Progression

If the sides $a$, $b$ and $c$ of $\triangle ABC$ are in Arithmetic Progression, then prove that:
$$\cos (\dfrac {B-C}{2})=2\sin (\dfrac {A}{2})$$

My Attempt:

Since, $a,b,c$ are in AP
$$2b=a+c$$
$$\sin A+\sin C=2\sin B$$

$$2\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2\sin B$$
$$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=\sin B$$
$$\sin (\dfrac {A+C}{2}).\cos (\dfrac {A-C}{2})=2.\sin (\dfrac {A+C}{2}).\cos (\dfrac {A+C}{2})$$
$$2\cos (\dfrac {A+C}{2})=\cos (\dfrac {A-C}{2})$$