我需要将每个文本框定位如下图所示
代码
\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}
答案1
此解决方案基于xcoffin
软件包。无需迷你页面或tikz
。
简单来说,就是将内容物装入盒子,然后通过角落将一个盒子与另一个盒子连接起来。
例如将和\JoinCoffins\Framex[l,t]\titulo[l,t]
的左上角连接起来。\Framex
\titulo
\Framex
将收集所有其他框,包括分隔线和边距线,并将结果排版在文本区域的第一行。
一旦定义了两列的宽度,就无需猜测距离。框将根据其内容调整其高度,其他所有内容都将自动根据它们计算。
\documentclass[10pt,twoside]{article}
\usepackage[spanish,es-lcroman, es-tabla, es-noshorthands]{babel}
\usepackage[paperwidth=8.8cm, paperheight=11.1cm, top=0.2cm, left=0.8cm, right=0.8cm, bottom=0.4cm,nomarginpar]{geometry}
\setlength{\parindent}{0cm}
% COMANDOS PERSONALES ----------------------------------------------
\newcommand{\sen}{\mathop{\rm sen}\nolimits}
\newcommand{\tg}{\mathop{\rm tg}\nolimits}
\newcommand{\ctg}{\mathop{\rm ctg}\nolimits}
\newcommand{\lgo}{\mathop{\rm Log}\nolimits}
% nuevo
\usepackage{fancyhdr}
\usepackage{amsmath,colortbl}
\usepackage{calc}
\setlength{\parindent}{0cm}
\fancyhf{}
\usepackage{newtxtext,newtxmath} %better fonts for smaller size <<<<<<<<<<<<<<<
\usepackage{xcoffins} % added <<<<<<<<<<<
% allocate
\NewCoffin\titulo
\NewCoffin\Framex
\NewCoffin\identidades
\NewCoffin\angdoble
\NewCoffin\sumdif
\NewCoffin\transprod
\NewCoffin\transsum
\NewCoffin\logaritmo
\NewCoffin\hrulex
\NewCoffin\vrulex
\NewCoffin\vrulexx
\newlength{\leftcolwidth}
\newlength{\rightcolwidth}
\setlength{\leftcolwidth}{2.8cm} % adjust columns width <<<<<
\setlength{\rightcolwidth}{4.5cm}
\begin{document}
\setlength{\abovedisplayskip}{0pt} % reduce vertical space in align*
\setlength{\belowdisplayskip}{2pt}
\setlength{\abovedisplayshortskip}{0pt}
\setlength{\belowdisplayshortskip}{0pt}
\SetVerticalCoffin\titulo{\leftcolwidth+\rightcolwidth}{%
\vspace*{15pt}
\centering \sffamily \Large Title
\vspace*{15pt}% reserved for the title
}
\SetVerticalCoffin\identidades{\leftcolwidth}{%
\setlength{\parindent}{1ex} \tiny \smallskip
IDENTIDADES
TRIGONOMÉTRICAS\smallskip
$\sen^2A+\cos^2A = 1$ \smallskip
\begin{align*}
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{align*}
\color{blue}\rule{\leftcolwidth}{0.5pt}
}
\SetVerticalCoffin\angdoble{\leftcolwidth}{%
\setlength{\parindent}{1ex} \tiny \smallskip
ÁNGULO DOBLE \smallskip
\begin{align*}
\sen 2A &=2\sen A \cdot \cos A \\
\cos 2A &=\cos^2 A - \sen^2 A \\
&=1-2\sen^2 A \\
\tg 2A &=\dfrac{2\tg A}{1-\tg^2 A} \\
\end{align*}
ÁNGULO MITAD\smallskip
\begin{align*}
\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{align*}
}
\SetVerticalCoffin\sumdif{\rightcolwidth}{%
\setlength{\parindent}{1ex} \tiny \smallskip
SUMA Y DIFERENCIA DE ÁNGULOS\smallskip
\begin{align*}
\sen \left(A \pm B \right) &= \sen A \cdot \cos B \pm \cos A \cdot \sen B\\
\cos \left(A \pm B \right) &= \cos A \cdot \cos B \mp \sen A \cdot \sen B\\
\tg \left(A \pm B \right) &= \dfrac{\tg A \pm \tg B}{1 \mp \tg A \cdot \tg B }
\end{align*}
\color{blue}\noindent\rule{\rightcolwidth}{0.5pt}
}
\SetVerticalCoffin\transprod{\rightcolwidth}{%
\setlength{\parindent}{1ex} \tiny \smallskip
TRANSFORMACION EN PRODUCTO\smallskip
\begin{align*}
\sen A + \sen B &= 2 \sen \dfrac{A + B}{2}\cos \dfrac{A - B}{2} \\
\sen A - \sen B &= 2 \cos \dfrac{A + B}{2}\sen \dfrac{A - B}{2} \\
\cos A + \cos B &= 2 \cos \dfrac{A + B}{2}\cos \dfrac{A - B}{2} \\
\cos A - \cos B &= -2 \sen \dfrac{A + B}{2}\sen \dfrac{A - B}{2}
\end{align*}
\color{blue}\noindent\rule{\rightcolwidth}{0.5pt}
}
\SetVerticalCoffin\transsum{\rightcolwidth}{%
\setlength{\parindent}{1ex} \tiny \smallskip
TRANSFORMACION DE PRODUCTO EN SUMA\smallskip
\begin{align*}
\sen A \cdot \cos B &= \dfrac{1}{2}\left[\sen (A + B) + \sen(A - B)\right]\\
\sen A \cdot \sen B &= \dfrac{1}{2}\left[\cos (A - B) + \cos(A + B)\right]\\
\cos A \cdot\cos B &= \dfrac{1}{2}\left[\cos (A + B) + \cos(A - B)\right]
\end{align*}
\color{blue}\noindent\rule{\rightcolwidth}{0.5pt}
}
\SetVerticalCoffin\logaritmo{\rightcolwidth}{%
\setlength{\parindent}{1ex} \tiny \smallskip
LOGARITMO: $\lgo_{b}{N} = x \Leftrightarrow N = b^x$
\phantom{LOGARITMO:} $ b > 0, b \neq 1, N > 0$ \medskip
PROPIEDADES: \smallskip
\begin{align*}
\lgo{(A \cdot B)} &= \lgo{A} + \lgo{B}; &\lgo_{b}{b}&= 1 \\
\lgo{(A/B)} &= \lgo{A} - \lgo{B}; &\lgo{1} &=0 \\
\lgo{A^n} &= n\lgo{A}
\end{align*}
}
\SetVerticalCoffin\hrulex{\leftcolwidth+\rightcolwidth}{%
\color{blue}\rule{\leftcolwidth+\rightcolwidth}{0.5pt}
}
\SetHorizontalCoffin\vrulex{
\color{blue}\rule{0.5pt}{\CoffinTotalHeight\titulo+\CoffinTotalHeight\identidades +\CoffinTotalHeight\angdoble}
}
\SetHorizontalCoffin\vrulexx{
\color{blue}\rule{0.5pt}{\CoffinTotalHeight\identidades +\CoffinTotalHeight\angdoble}
}
% join
\JoinCoffins\Framex[l,t]\titulo[l,t]
\JoinCoffins\Framex[l,b]\identidades[l,t]
\JoinCoffins\Framex[\identidades-r,\identidades-t]\sumdif[l,t]
\JoinCoffins\Framex[\sumdif-l,\sumdif-b]\transprod[l,t]
\JoinCoffins\Framex[\identidades-l,\identidades-b]\angdoble[l,t]
\JoinCoffins\Framex[\transprod-l,\transprod-b]\transsum[l,t]
\JoinCoffins\Framex[\transsum-l,\transsum-b]\logaritmo[l,t]
% put the rules
\JoinCoffins\Framex[\titulo-l,\titulo-t]\hrulex[l,t]
\JoinCoffins\Framex[\angdoble-l,\angdoble-b]\hrulex[l,t]
\JoinCoffins\Framex[\titulo-l,\titulo-b]\hrulex[l,t]
\JoinCoffins\Framex[\titulo-l,\titulo-t]\vrulex[hc,t]
\JoinCoffins\Framex[\titulo-r,\titulo-t]\vrulex[hc,t]
\JoinCoffins\Framex[\identidades-r,\identidades-t]\vrulexx[hc,t]
% typeset the assembly
\noindent\TypesetCoffin\Framex
\end{document}
请参阅此非常有用的软件包的其他示例
答案2
这可以通过tcolorbox
包来完成,可以是raster
或poster
。poster
环境提供了更大的灵活性,但内容不能跨页面拆分,而 是rasters
可拆分的。为了获得 40:60 的列宽分割,我声明了 5 个海报列,内容跨越第一列材料的 1-2 列,跨越第二列材料的 3-5 列。代码中有一些注释来解释语法。
\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}
\usepackage{newtxtext,newtxmath} %Scalable fonts for text & math
\usepackage[most]{tcolorbox} % Load most of the tcolorbox libraries
\setlength{\parindent}{0cm} % don't indent paragraphs
% COMANDOS PERSONALES ----------------------------------------------
\newcommand{\sen}{\mathop{\rm sen}\nolimits}
\newcommand{\tg}{\mathop{\rm tg}\nolimits}
\newcommand{\ctg}{\mathop{\rm ctg}\nolimits}
\begin{document}
\begin{tcbposter}[
coverage = {spread},
fontsize=8pt, % base font size
poster = {
% showframe, %show frame to assist with bx placement, if necessary
columns=5, % page has 5 columns
colspacing=2mm, %space between columns
rowspacing=2mm %space between rows
},
boxes = {
size=tight, %no between the text content and the frame of the box
colback=white, %colour of box background
colframe=blue, %colour of box frame
title style={left color=black,right color=cyan}, %style for box title
center title, %box title alignment
fonttitle=\bfseries\normalfont\scshape, %font for title
fontupper=\scriptsize, %size of box content relative to fontsize=
}
]
\posterbox[
halign=center, %centre the subtitle
adjusted title = TITLE %box title
]{name= Box0,column=1,span=5,below=top} %box placement: col 1, span cols 1-5
{Trig cheatsheet} %subtitle
\posterbox[
adjusted title= Identidades trigonométricas,
]{name= Box1,column=1,span=2,below=Box0}{ %start in col 1, span cols 1-2, placed below Box0
\begin{align*}
\sen^2A+\cos^2A &= 1 \\
1 + \tg^2A &= \sec^2A \\
1 + \ctg^2A &= \csc^2A \\
\tg A &= \sen A/\cos A \\
\tg A &= \cos A/\sen A \\
\sec A &= 1/\cos A \\
\csc A &= 1/\sen A \\
\ctg A &= 1/\tg A
\end{align*}
}