Given $\cos C(\sin A +\sin B)=\sin C\cos(A-B)$, prove that triangle $ABC$ is equilateral.
My attempt:
By applying some trigonometric identities, I have: $$\cos C(\sin A +\sin B)=\sin C \cos(A-B)\\ \iff \cos C \cdot2\sin \frac{A+B}{2} \cos \frac{A-B}{2}=\sin \frac{C}{2}\cos \frac{C}{2}\cos(A-B)\\ \iff \cos C \cos \frac{A-B}{2}=\sin \frac{C}{2}\cos(A-B)\\ \iff \left( 1-2\sin^2 \frac{C}{2}\right)\cos \frac{A-B}{2}=\sin \frac{C}{2}\left( 2\cos^2 \frac{A-B}{2}-1\right)\\ \iff \cos \frac{A-B}{2}+\sin \frac{C}{2}=2\sin \frac{C}{2}\cos \frac{A-B}{2}\left(\sin \frac{C}{2}+\cos \frac{A-B}{2} \right)\\ \implies 2\sin \frac{C}{2}\cos \frac{A-B}{2}=1$$ or $$2\cos \frac{A+B}{2}\cos \frac{A-B}{2}=1$$ or $$\cos A +\cos B=1.$$
And that's where I'm stuck.
I got the same result!
Now take $A=30^{\circ}$, for example, and we see that our triangle is not equilateral.