What's wrong with my proof?

Prove $\cos A + \cos B + \cos C -1=4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ when $A+B+C=180^\circ$

$\cos A + \cos B + \cos C -1= 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}-2\sin^2\frac{C}{2}$

$=2\sin\frac {C}{2}\cos \frac {A-B}{2}-2\sin^2 \frac{C}{2}$

$2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\sin \frac {C}{2} )$

$2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})$

$2\sin \frac{C}{2}(-2\sin\frac {A}{2}\sin \frac {B}{2})$

$-4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

I had the same steps up towards $2\sin\frac {C}{2} ( \cos \frac{A-B}{2}-\cos \frac{A+B}{2})$ , but then the negative sign just disappeared from the next step. Isn't the formula for $\cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}$ ?