<s>[INST] 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.
 [/INST] $$cos\frac{C+D}{2}=cos\frac{2\pi-(A+B)}{2}=-cos\frac{A+B}{2}$$
$$cos\frac{C-D}{2}=cos\frac{C+A+B+C-2\pi}{2}=-cos\frac{A+C+B+C}{2}$$
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 
</s>