For triangular trigonometric identity, it is typically proved by combining the first two angles and replacing $C$ with $(\pi - A - B)$. Thus \begin{align} \cos A + \cos B + \cos C &= (\cos A + \cos B) - cos (A+B)\\ &= 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} -( 2\cos^2\frac{A+B}{2} -1) \\ &= 1+ 2\cos\frac{A+B}{2}\big(\cos\frac{A-B}{2} - \cos\frac{A+B}{2}\big)\\ &= 1 + 2\sin\frac{C}{2}\cdot 2 \sin\frac{A}{2}\sin\frac{B}{2}\\ &= 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \end{align}