$ \cos(\hat{A})BC+ A\cos(\hat{B})C+ AB\cos(\hat{C})=\frac {A^2 + B^2 + C^2}{2} $ What more can be said about the identity derived from law of cosines (motivation below)$$ \cos(\widehat{A})BC+ A\cos(\widehat{B})C+ AB\cos(\widehat{C})+=\frac {A^2 + B^2 + C^2}{2} \tag{IV}$$
RHS seems as if operator $\cos(\widehat{\phantom{X}})$ is being applied consecutively to terms of ABC, I tried to represent it in an analogous way to the Laplacian operator convention, but maybe there are more common ways of representing RHS using some operator and sigma notation ( please let me know if there is).
My question is : Are there any identities/structures relating or looking similar to IV, I apologize if this looks like a general fishing expedition question but I can not think of anything more that I can add to this post at this stage. Thank you
Motivation for IV, 
Let $A,B,C$ be a triangle.
Let $\widehat{C}  = \widehat{AB}$ stand for the Angle opposite to side C between the sides A and B, then the law of cosines for all three sides can be written as $$ A^2 + B^2 - 2 AB\cos(\widehat{C}) = C^2 \tag{I} $$ 
$$ A^2 + C^2 - 2 AC\cos(\widehat{B}) = B^2 \tag{II} $$ 
$$ B^2 + C^2 - 2 BC\cos(\widehat{A}) = A^2 \tag{III} $$ 
Adding $I ,II,III$ and juggling the terms we get :
$$  AB\cos(\widehat{C})+AC\cos(\widehat{B})+BC\cos(\widehat{A}) =\frac {A^2 + B^2 + C^2}{2} \tag{IV} $$ 

REMARK: It seems now to me that OP is looking for generalizations of $\text{IV}$ type relations valid for quadrilaterals, pentagons, etc., and not other triangle trigonometric relations, as I exemplified below.


Notation: Consider a triangle with angles $A$, $B$, $C$ and opposite sides $a$, $b$, $c$. 
It is known that there  exists only three distinct relations between
the angles and the sides. For instance the system
$$\begin{eqnarray}
\frac{a}{\sin A} &=&\frac{b}{\sin B}   \\
\frac{a}{\sin A} &=&\frac{c}{\sin C}\tag{1}  \\
A+B+C &=&\pi;  
\end{eqnarray}$$
or this equivalent (yours (I),(II),(III))
$$\begin{eqnarray}
a^{2} &=&b^{2}+c^{2}-2bc\cos A  \\
b^{2} &=&c^{2}+a^{2}-2ac\cos B \tag{2}\\
c^{2} &=&a^{2}+b^{2}-2ab\cos C  
\end{eqnarray}$$
are two of them.  Another is 
$$\begin{eqnarray}
\frac{\tan \frac{A+B}{2}}{\tan \frac{A-B}{2}} &=&\frac{a+b}{a-b}  
\\
\frac{\tan \frac{B+C}{2}}{\tan \frac{B-C}{2}} &=&\frac{b+c}{b-c} \tag{3}\\
\frac{\tan \frac{C+A}{2}}{\tan \frac{C-A}{2}} &=&\frac{c+a}{c-a},  
\end{eqnarray}$$
from which one can derive
$$\begin{equation}
\frac{\tan \frac{A+B}{2}}{\tan \frac{A-B}{2}}+\frac{\tan \frac{B+C}{2}}{\tan 
\frac{B-C}{2}}+\frac{\tan \frac{C+A}{2}}{\tan \frac{C-A}{2}}=\frac{a+b}{a-b}+%
\frac{b+c}{b-c}+\frac{c+a}{c-a}.\tag{4}
\end{equation}$$
Also from
$$\begin{eqnarray}
\tan \frac{A-B}{2} &=&\frac{a-b}{a+b}\cot \frac{C}{2}   \\
\tan \frac{B-C}{2} &=&\frac{b-c}{b+c}\cot \frac{A}{2}\tag{5} \\
\tan \frac{C-A}{2} &=&\frac{c-a}{c+a}\cot \frac{B}{2},  
\end{eqnarray}$$
follows
$$\begin{eqnarray}
\tan \frac{A-B}{2} \tan \frac{C}{2}+\tan \frac{B-C}{2} \tan \frac{A}{2}+\tan \frac{C-A}{2}\tan \frac{B}{2} &=&\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}. \; (6)\end{eqnarray}$$
As for a reference I looked at my old trigonometry text book Compêndio de
Trigonometria (in Portuguese) by J. Jorge Calado. The system $\left(
3\right) $ is the law of tangents (Wikipedia) and $\left(
5\right) $ can be deduct from $\left( 3\right) $ by applying the relation $A+B+C=\pi $ to get 
\begin{eqnarray}
\tan \frac{A+B}{2} &=&\tan \left( \frac{\pi }{2}-\frac{C}{2}\right) =\cot 
\frac{C}{2}=\left( \tan \frac{C}{2}\right) ^{-1}\tag{7}    
\end{eqnarray}
and similar to $\tan \frac{B+C}{2}$ and $\tan \frac{C+A}{2}$.
