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

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

## closed as off-topic by Xander Henderson, The Phenotype, JonMark Perry, Did, user223391 Mar 5 '18 at 23:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, The Phenotype, JonMark Perry, Did, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

• Yes, it is correct. – Netchaiev Mar 4 '18 at 16:16
• It is better, though, to use $\pi$-radiant notations : $$A+B+C=\pi$$ since in fact it is $$\sin(\pi-A)=\sin(A)$$ (180° is a notation for a given angle, but if you want to compute the sinus of it, the value to give to the sinus function (for that very angle) is $\pi$). – Netchaiev Mar 4 '18 at 16:18
• @Netchaiev I'm just going off the question I was given – Plus Twenty Mar 5 '18 at 1:19

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

• Thanks for the help, can you check over this answer : $\sin A= \sin (B+C)$ Since $A + B + C = 180$, $A = 180 - (B+C)$ And $\sin A = \sin (180 - A)$ $\sin A = \sin (180-(180-(B+C))$ $\sin A = \sin (B+C)$ – Plus Twenty Mar 3 '18 at 23:22
• @Conal This is correct – Paolo Leonetti Mar 3 '18 at 23:36
• Thanks a lot mate, appreciate it – Plus Twenty Mar 3 '18 at 23:38

HINT

Note that

$$\sin A = \sin (180°-B-C)$$

and

$$\sin (180°-x)=\sin x$$

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 $\sin D = \sin A.$