If $A+B+C = 180^{\circ}$, then show that: $\sin A = \sin B \cos C + \cos B \sin C$ Here is the question :

If $A+B+C = 180^{\circ}$, then show that: $\sin A = \sin B \cos C + \cos B \sin C$.

I don't really have any clue where to start.
The only thing I would think to do would be to change $\sin A = \sin B\cos C + \cos C\sin B$ to $\sin (B+C) = \sin B\cos C + \cos C\sin B$. But I'm not sure if that is even useful.
Any suggestions appreciated.
EDIT :
Here is my working, can someone comment and confirm that this working is correct.

We want to show that
  $$
\sin A=\sin (B+C).
$$

Since 
$$
\space A+B+C = 180^\circ, \space A =180^\circ-(B+C)
$$
And 
$$
\space \sin A=\sin (180-A),
$$
then
$$
\sin A=\sin(180-(180-(B+C))
$$
$$
\sin A=\sin(180-180+(B+C))
$$
so
$$
\sin A=\sin(B+C).
$$

Hint:
$$
A+B+C=180\,\,\,\ \text{and}\,\,\,\,\sin(A)=\sin(180-A).
$$
