<s>[INST] Geometric proof of $\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$ 
Prove:
$$\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$$
Geometrically.

I found the algebraic solutions here but I want to figure out how to calculate the above using geometrical arguments of complex numbers/ vectors.

Some hints of geometry I figured out already:

*

*$\frac{A+B}{2}$ is equidistant in angle mangitude from $A$ and $B$, that is $|\frac{A+B}{2} -A| = | \frac{A+B}{2} - B|$, similar results hold for the other angle averages.

*We may think of the above as summing the projections of three unit vectors onto the x-axis: $ \left( \tau(A) + \tau(B) + \tau(C) + \tau( A+B+C)\right) \cdot \hat{i}$, where $\tau(\phi)$ is the unit vector making angle $\phi$ with the x-axis

Related , related
 [/INST] Let
$$\,\bar e=\dbinom01,\;\bar a=\dbinom{\sin A}{\cos A},\;
\bar b=\dbinom{\sin(A+B)}{\cos(A+B)},\;
\bar c=\dbinom{\sin(A+B+C)}{\cos(A+B+C)},\tag1$$
where $\dbinom yx$ means the radius-vector with the cartesian coordinates $\,x,\,y.$

Then the scalar productions are
$$\dbinom{\sin\alpha}{\cos\alpha}\cdot\dbinom{\sin\beta}{\cos\beta}
=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta),\tag2$$
$$\bar e\cdot\bar a=\cos A,\quad
\bar a\cdot\bar b=\cos B,\quad
\bar b\cdot\bar c=\cos C,\quad
\bar e\cdot\bar c=\cos(A+B+C).\tag3$$
Also, are known the identities
$$\sin x +\sin y = 2\sin\dfrac{x+y}2\,\cos\dfrac{x-y}2,\tag4$$
$$\cos x +\cos y = 2\cos\dfrac{x+y}2\,\cos\dfrac{x-y}2.\tag5$$
From $(1)-(5)$ should
$$\cos A+\cos B+\cos C+\cos(A+B+C)=\bar e\cdot\bar a+\bar a\cdot\bar b+
\bar b\cdot\bar c+\bar e\cdot\bar c=(\bar e+\bar b)\cdot(\bar a+\bar c)$$
$$=\dbinom{\sin(A+B)}{1+\cos(A+B)}\cdot\dbinom{\sin A+\sin(A+B+C)}{\cos A+\cos(A+B+C)}$$
$$=2\cos\dfrac{A+B}2\,
\begin{pmatrix}\sin\dfrac{A+B}2\\ \cos\dfrac{A+B}2\end{pmatrix}
\cdot2\cos\dfrac{B+C}2
\begin{pmatrix}\sin\dfrac{2A+B+C}2\\ \cos\dfrac{2A+B+C}2\end{pmatrix}$$ $$\color{green}{\mathbf{=4\cos\dfrac{A+B}2\,\cos\dfrac{B+C}2\,\cos\dfrac{C+A}2.}}$$
At the same time,
$$\cos x \cos y = \dfrac12(\cos(x-y)+\cos(x+y)),$$
and the approach
$$4\cos\dfrac{A+B}2\cos\dfrac{B+C}2\cos\dfrac{C+A}2
=2\left(\cos\dfrac{A-C}2+\cos\dfrac{A+2B+C}2\right)\cos\dfrac{C+A}2$$
$$=\cos C+\cos A+\cos B+\cos(A+B+C)$$
looks the best.
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