I need to position each textbox like the following image
code
\documentclass[10pt,twoside]{article}
\usepackage[spanish,es-lcroman, es-tabla, es-noshorthands]{babel}
\usepackage[paperwidth=8.8cm, paperheight=11.1cm, top=0.6cm, left=0.5cm, right=0.8cm, bottom=0.4cm,nomarginpar]{geometry}
% COMANDOS PERSONALES ----------------------------------------------
\newcommand{\sen}{\mathop{\rm sen}\nolimits}
\newcommand{\tg}{\mathop{\rm tg}\nolimits}
\newcommand{\ctg}{\mathop{\rm ctg}\nolimits}
% nuevo
\usepackage{fancyhdr}
\usepackage{tikz,amsmath,colortbl}
\usepackage{fancyhdr}
\usepackage{calc}
\usepackage[lining,tabular]{fbb}
\parindent=0cm
\usetikzlibrary{matrix,arrows, positioning,shadows,shadings,backgrounds,
calc, shapes, tikzmark}
\fancyhf{}
\newcommand{\margenes}{%
\begin{tikzpicture}[remember picture, overlay]
\draw [line width=0.5pt,blue]
($ (current page.south west) + (0.5cm,0.4cm) $)
rectangle
($ (current page.north east) + (-0.8cm, -0.6cm)$);
\end{tikzpicture}}
\newcommand{\mydos}{%
\begin{tikzpicture}[remember picture, overlay]
\def\A##1{($ (current page.south west) + (0.5cm,0.4cm) + ##1 $)};
\def\B##1{($ (current page.north east) + (-0.8cm, -0.6cm) + ##1 $)};
\draw [line width=0.5pt,blue] \A{(0cm,0cm)} rectangle \B{(0cm,0cm)};
\draw [line width=0.5pt,blue] \A{(0cm,4.4cm)} -- \A{(3.4cm,4.4cm)};
\draw [line width=0.5pt,blue] \A{(0cm,7.8cm)} -- \A{(7.5cm,7.8cm)};
\draw [line width=0.5pt,blue] \A{(3.4cm,7.8cm)} -- \A{(3.4cm,0cm)} ;
\draw [line width=0.5pt,blue] \A{(3.4cm,6.2cm)} -- \A{(7.5cm,6.2cm)};
\draw [line width=0.5pt,blue] \A{(3.4cm,3.8cm)} -- \A{(7.5cm,3.8cm)};
\draw [line width=0.5pt,blue] \A{(3.4cm,1.8cm)} -- \A{(7.5cm,1.8cm)};
\draw [line width=0.5pt,blue] \A{(5.8cm,0cm)} -- \A{(5.8cm,1.1cm)};
\end{tikzpicture}}
\begin{document}
\fontsize{5}{5}\selectfont
\mydos
\put(2,-107){
\begin{minipage}[c]{3.4cm}
%\centering
IDENTIDADES TRIGONOMÉTRICAS \
$\sen^2A+\cos^2A = 1$ \
$1 + \tg^2A = \sec^2A$ \
$1 + \ctg^2A = \csc^2A$ \
$\tg A = \sen A/\cos A$ \
$\ctg A = \cos A/\sen A$ \
$\sec A = 1/\cos A$ \
$\csc A = 1/\sen A$ \
$\ctg A = 1/\tg A$ \
\end{minipage}}\
\put(2,-127){
\begin{minipage}[c]{3.4cm}
%\centering
ÁNGULO DOBLE \
$\sen 2A =2\sen A\ \cos A $ \
$\cos 2A\ =\cos^2 A-\ \sen^2 A $ \
$\hspace*{0.625cm} =1-2\sen^2 A $ \
$\tg 2A=\dfrac{2\tg A}{ 1-\tg^2 A} $\
ÁNGULO MITAD\
$\sen \dfrac{A}{2} = \sqrt{\dfrac{1-\cos A}{2}}$\
$\sen^2x = \dfrac{1}{2}(1-\cos 2x)$\
$\cos \dfrac{A}{2} = \sqrt{\dfrac{1+\cos A}{2}}$\
$\cos^2x = \dfrac{1}{2}(1+\cos 2x)$
\end{minipage}}\
\newpage
\margenes
\end{document}



