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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 :

$$ AB\cos(\widehat{C})+AC\cos(\widehat{B})+BC\cos(\widehat{A}) =\frac {A^2 + B^2 + C^2}{2} \tag{IV} $$

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An attractively symmetrical formula! – André Nicolas Jul 3 '11 at 14:28
@user6312: I agree. @Arjang: I'd prove this without appealing to the law of cosines. Indeed, in the usual trigonometric proof of the law of cosines you start by writing $c = a \cos{\beta} + b \cos{\alpha}$ and multiply this by $c$. Now do this for each side, add the three resulting equalities up and divide by two to get (IV). In order to get the law of cosines you write $a^2 + b^2 - c^2$ using these expressions and solve for $c^2$. – t.b. Jul 3 '11 at 18:01
Oddly enough, this looks (to me) quite related to my answer here… (this would be the case for N=3, for N>3 we have the inequality) – leonbloy Jul 3 '11 at 18:59
@Theo : tx for the tip, I will be using it for revision. – Arjang Jul 5 '11 at 10:02
Do you want to generalize $IV$ for quadrilaterals, pentagons, etc.? – Américo Tavares Aug 2 '11 at 14:40

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.

enter image description here

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

$$\begin{eqnarray} \tan \frac{A-B}{2} &=&\frac{a-b}{a+b}\cot \frac{C}{2} \\ \tan \frac{B-C}{2} &=&\frac{b-c}{b+c}\cot \frac{A}{2}\tag{5} \\ \tan \frac{C-A}{2} &=&\frac{c-a}{c+a}\cot \frac{B}{2}, \end{eqnarray}$$


$$\begin{eqnarray} \tan \frac{A-B}{2} \tan \frac{C}{2}+\tan \frac{B-C}{2} \tan \frac{A}{2}+\tan \frac{C-A}{2}\tan \frac{B}{2} &=&\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}. \; (6)\end{eqnarray}$$

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 $\tan \frac{C+A}{2}$.

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