TITLE: Find $\cos{A}+\cos{B}$
QUESTION [5 upvotes]: In $\Delta ABC$,if 
  $$\cos{C}\cdot(\sin{A}+\sin{B})=\sin{C}\cdot\cos{(A-B)}$$
  Find $\cos{A}+\cos{B}$

Thus 
$$\sin{A}+\sin{B}=\tan{C}\cdot\cos{(A-B)}=\dfrac{\sin{(A+B)}}{\cos{(A+B)}}\cos{(A-B)}$$
$$(\sin{A}+\sin{B})\cos{(A+B)}=\sin{(A+B)}\cos{(A-B)}$$
follow  I can't figure it out.

REPLY [3 votes]: Let me try. $\cos C (\sin A +\sin B)=\sin C \cos(A-B)$ 
$\implies2\left(1-2\sin^2\dfrac C2\right)\sin\dfrac{A+B}2\cos\dfrac{A-B}2 = 2\sin \dfrac C2\cos\dfrac C2\left(2\cos^2\dfrac{A-B}2-1\right)\ \ \ \  (1)$
Now as $\sin\dfrac{A+B}2=\sin\left(\dfrac\pi2-\dfrac C2\right)=\cos\dfrac C2$
and $\cos\dfrac C2=0\implies\dfrac C2=(2n+1)\dfrac\pi2\iff C=(2n+1)\pi$ where $n$ is any integer
But $0<C<\pi\implies\sin\dfrac{A+B}2=\cos\dfrac C2\ne0$
$(1)\implies$
$$\left(1-2\cos^2\dfrac{A+B}2\right)\cos\dfrac{A-B}2 = \cos\dfrac{A+B}2\left(2\cos^2\dfrac{A-B}2-1\right)$$
$$\implies\left(\cos\dfrac{A-B}2+\cos\dfrac{A+B}2\right)\left(1-2\cos\dfrac{A-B}2\cos\dfrac{A+B}2\right)=0$$
Now, $\cos\dfrac{A-B}2+\cos\dfrac{A+B}2=2\cos\dfrac A2\cos\dfrac B2$ which can not be zero due to the reason mentioned above.
$$\implies1-2\cos\dfrac{A-B}2\cos\dfrac{A+B}2=0$$
But $2\cos\dfrac{A-B}2\cos\dfrac{A+B}2=\cos A+\cos B$