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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!

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    $\begingroup$ $\cos(B)$ does not equal $\cos(A+C)$. You know $B=180^{\circ}-A-C$ and you can verify $\cos(180^{\circ}-\theta)=-\cos(\theta)$ by looking at a circle, so you're missing a sign at the beginning. $\endgroup$ – anon Mar 4 '18 at 0:16
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    $\begingroup$ @anon right, but he never used the first "identity" $\endgroup$ – Paolo Leonetti Mar 4 '18 at 0:22
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    $\begingroup$ @PaoloLeonetti OP began with $\cos(B)=\cos(A+C)$ in the original version of the question. $\endgroup$ – anon Mar 4 '18 at 5:22
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Your first step $\cos(B) = \cos(A+C)$ is incorrect. It should be $\cos(B) = -\cos(A+C)$.

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  • $\begingroup$ Ah thanks for pointing this out to me I just realised my mistake $\endgroup$ – Plus Twenty Mar 4 '18 at 0:19
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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 $\cos(B)=-\cos(A+C)$.

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  • $\begingroup$ Can you check over my working please I just edited my post $\endgroup$ – Plus Twenty Mar 4 '18 at 0:22
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    $\begingroup$ @Conal Everything correct now $\endgroup$ – Paolo Leonetti Mar 4 '18 at 0:23

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