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.
 A: We will use $2\cos X \cos Y = \cos(X+Y) + \cos(X-Y)$.
\begin{align*}
4\cos \frac{A+B}{2}\cos \frac{A+C}{2} \cos\frac{B+C}{2} &= 2\left(\cos\left(A + \frac{B+C}{2}\right) + \cos\frac{B-C}{2}\right)\cos\frac{B+C}{2}\\
&= 2\cos\left(A + \frac{B+C}{2}\right)\cos\frac{B+C}{2} + 2 \cos\frac{B-C}{2}\cos\frac{B+C}{2}\\
&= \cos(A+B+C) + \cos A + \cos B + \cos C \\
&= \cos(2\pi - D) + \cos A + \cos B + \cos C \\
&= \cos A + \cos B + \cos C + \cos D
\end{align*}
A: $$cos\frac{C+D}{2}=cos\frac{2\pi-(A+B)}{2}=-cos\frac{A+B}{2}$$
$$cos\frac{C-D}{2}=cos\frac{C+A+B+C-2\pi}{2}=-cos\frac{A+C+B+C}{2}$$
Now go to your last line
$$2cos\frac{A+B}{2}(cos\frac{A-B}{2}+cos\frac{A+C+B+C}{2})$$
Use the rule to turn an addition of cosinuses to a multiplication and you will get your proof 
A: 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 $$\cos\dfrac{A-B}2+\cos\dfrac{A+B+2C}2$$
