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

Question

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 $$ \sin A=\sin(B+C). $$

Answer 1

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

Answer 2

HINT

Note that

$$\sin A = \sin (180°-B-C)$$

and

$$\sin (180°-x)=\sin x$$

Answer 3

You already know that $$ \sin B \cos C + \cos B \sin C = \sin(B+C). $$ Therefore to prove that the left side of that identity is equal to $\sin A,$ it is enough to prove that the right side is equal to $\sin A.$ Thus the problem is to prove that $$ \sin(B+C) = \sin A $$ when you know that $A+B+C=180^\circ.$

Suppose the problem had been phrased as follows:

Prove that if $A+B+C= 180^\circ$ then $\sin(B+C) = \sin A.$

In that case, neither $B$ nor $C$ is mentioned except in the sum $B+C$. Thus one can just let $D=B+C$ and forget about the fact that it's a sum, so the problem becomes this:

Prove that if $A+D=180^\circ$ then $\sin D = \sin A.$