If $A+B+C+D=2\pi$, prove that: If $A+B+C+D=2\pi$, prove that: $$\cos A+\cos B+\cos C+\cos D=4\cos\frac {A+B}{2}\cdot\cos\frac {A+C}{2}\cdot\cos\frac {B+C}{2}$$.
My Approach:
Here,
$$A+B+C+D=2\pi$$
$$A+B=2\pi - (C+D)$$
$$ \sin(A+B)=\sin(2\pi-(C+D))$$
$$\sin(A+B)=-\sin(C+D)$$
Again,
$$\cos(A+B)=\cos(2\pi-(C+D))$$
$$\cos(A+B)=\cos(C+D)$$
Now,
$$L.H.S=\cos A+\cos B+\cos C+\cos D$$
$$=2 \cos\frac {A+B}{2}\cdot\cos\frac {A-B}{2} + 2 \cos\frac {C+D}{2}\cdot \cos\frac {C-D}{2}$$.
I got stuck at here. Please help me to complete the proof.

As $D=2\pi-(A+B+C),\cos D=\cos(A+B+C)$
$$\implies\cos A+\cos B+\cos C+\cos(A+B+C)$$
$$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2+2\cos\dfrac{A+B+C-C}2\cos\dfrac{A+B+C+C}2$$  (using Prosthaphaeresis Formulas)
$$=2\cos\dfrac{A+B}2\left(\cos\dfrac{A-B}2+\cos\dfrac{A+B+2cc}2\right)$$
Apply Prosthaphaeresis Formula on $$\cos\dfrac{A-B}2+\cos\dfrac{A+B+2C}2$$
