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

Question

Here is the question :

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

EDIT : Here is my reviewed working :

$$ \cos B=-\cos (A+C) $$

Since $$\space A+B+C = 180^\circ, \space B =180^\circ-(A+C)$$

And $$\begin{align} -\cos B &=\cos (180+B) \\ -\cos B &=\cos(180+(180-(A+C)) \\ -\cos B &=\cos(360-(A+C)) \\ -\cos B &=\cos(A+C) \\ -\cos B &=\cos A \cos C - \sin A \sin C \\ \cos B &= \sin A \sin C - \cos A \cos C \\ \cos B &= -\cos (A+C) \end{align}$$

Can someone confirm that my working is correct?

Thanks!

Answer 1

Your first step $\cos(B) = \cos(A+C)$ is incorrect. It should be $\cos(B) = -\cos(A+C)$.

Answer 2

Your first line should be deleted, ok. But the main mistake is at the last line. You write $$ -\cos(B)=\cos(A+C), $$ which is fine. But what is $\cos(A+C)$?

Also, more directly: $\cos(x)=-\cos(180-x)$ for all $x$. So $\cos(B)=-\cos(A+C)$.