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.

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*}
