我可以很好地构建 tex，并且使用 minipage 可以让右侧的内容正确浮动在给定的“空白区域”上，我试图通过这样做来利用这一点，但左侧的内容仍然受到此影响。
我的目标是让“杂项”多列浮动在整个整体多列上，这样我就可以利用这个浪费的空间。这是有问题的部分。我对 latex 还不熟悉。
****编辑，我已按要求附上整个文档
\documentclass[8pt,letterpaper]{extarticle}
\usepackage[utf8x]{inputenc}
\usepackage{ucs}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{multirow}
\usepackage{array}
\usepackage{tikz}
\usepackage{graphicx}
\usepackage[hmargin={0in,0in},vmargin={0in,0in},portrait]{geometry}
\usepackage{setspace}
\usepackage{float}
\DeclareMathOperator{\sech}{sech}
\DeclareMathOperator{\csch}{csch}
\author{Jon}
\begin{document} % Beginning of Document
\begin{spacing}{1.3}
\begin{flushright}%unit circle
\begin{tabular}{c}
\multicolumn{1}{c}{Trig} \\
\scalebox{0.43}{
\begin{tabular}{c}
\begin{tikzpicture}
[scale=3.8,cap=round,>=latex]
% draw the coordinates
\draw[->] (-1.5cm,0cm) -- (1.5cm,0cm) node[right,fill=white] {$x$};
\draw[->] (0cm,-1.5cm) -- (0cm,1.5cm) node[above,fill=white] {$y$};
% draw the unit circle
\draw[thick] (0cm,0cm) circle(1cm);
\foreach \x in {0,30,...,360} {
% lines from center to point
\draw[gray] (0cm,0cm) -- (\x:1cm);
% dots at each point
\filldraw[black] (\x:1cm) circle(0.4pt);
% draw each angle in degrees
\draw (\x:0.6cm) node[fill=white] {$\x^\circ$};
}
% draw each angle in radians
\foreach \x/\xtext in {
30/\frac{\pi}{6},
45/\frac{\pi}{4},
60/\frac{\pi}{3},
90/\frac{\pi}{2},
120/\frac{2\pi}{3},
135/\frac{3\pi}{4},
150/\frac{5\pi}{6},
180/\pi,
210/\frac{7\pi}{6},
225/\frac{5\pi}{4},
240/\frac{4\pi}{3},
270/\frac{3\pi}{2},
300/\frac{5\pi}{3},
315/\frac{7\pi}{4},
330/\frac{11\pi}{6},
360/2\pi}
\draw (\x:0.85cm) node[fill=white] {$\xtext$};
\foreach \x/\xtext/\y in {
% the coordinates for the first quadrant
30/\frac{\sqrt{3}}{2}/\frac{1}{2},
45/\frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2},
60/\frac{1}{2}/\frac{\sqrt{3}}{2},
% the coordinates for the second quadrant
150/-\frac{\sqrt{3}}{2}/\frac{1}{2},
135/-\frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2},
120/-\frac{1}{2}/\frac{\sqrt{3}}{2},
% the coordinates for the third quadrant
210/-\frac{\sqrt{3}}{2}/-\frac{1}{2},
225/-\frac{\sqrt{2}}{2}/-\frac{\sqrt{2}}{2},
240/-\frac{1}{2}/-\frac{\sqrt{3}}{2},
% the coordinates for the fourth quadrant
330/\frac{\sqrt{3}}{2}/-\frac{1}{2},
315/\frac{\sqrt{2}}{2}/-\frac{\sqrt{2}}{2},
300/\frac{1}{2}/-\frac{\sqrt{3}}{2}}
\draw (\x:1.25cm) node[fill=white] {$\left(\xtext,\y\right)$};
% draw the horizontal and vertical coordinates
% the placement is better this way
\draw (-1.25cm,0cm) node[above=1pt] {$(-1,0)$}
(1.25cm,0cm) node[above=1pt] {$(1,0)$}
(0cm,-1.25cm) node[fill=white] {$(0,-1)$}
(0cm,1.25cm) node[fill=white] {$(0,1)$};
\end{tikzpicture}
\end{tabular}
} \\
\end{tabular}
\end{flushright}%End unit circle
\begin{tabular}{l l l l} %Derivates
\multicolumn{4}{c}{\textbf {\underline {Derivatives}}} \\
\textbf {\underline {Trig}} & \textbf {\underline {Inverse Trig}} & \textbf {\underline {Hyperbolic Trig}} & \textbf {\underline {Exponential/Log}} \\
$ \frac{d}{dx}\left(\sin x\right)=\cos x $
& $ \frac{d}{dx}\left(\sin^{-1} x\right)=\frac{1}{\sqrt{1-x^2}} $
& $ \frac{d}{dx}\left(\sinh x\right)=\cosh x $
& $ \frac{d}{dx}\left(a^x\right)=a^x \ln a $
\\ %end row1
$ \frac{d}{dx}\left(\cos x\right)=-\sin x $
& $ \frac{d}{dx}\left(\cos^{-1} x\right)=\frac{1}{\sqrt{1-x^2}} $
& $ \frac{d}{dx}\left(\cosh x\right)=\sinh x $
& $ \frac{d}{dx}\left(e^x\right)=e^x $
\\ % end row2
$ \frac{d}{dx}\left(\tan x\right)=\sec^2 x $
& $ \frac{d}{dx}\left(\tan^{-1} x\right)=\frac{1}{\sqrt{1+x^2}} $
& $ \frac{d}{dx}\left(\tanh x\right)=\sech^2 x $
& $ \frac{d}{dx}\left(\ln x\right)=\frac1 x , x > 0$
\\ %end row3
$ \frac{d}{dx}\left(\sec x\right)=\sec x \tan x $
& $\frac{d}{dx}\left(\sec^{-1} x\right)=\frac{1}{|x|\sqrt{x^2-1}}$
& $ \frac{d}{dx}\left(\sech x\right)=-\sech x \tanh x $
& $ \frac{d}{dx}\left(\ln |x|\right)=\frac1 x , x \neq 0$
\\%end row4
$ \frac{d}{dx}\left(\csc x\right)=-\csc x \cot x $
& $\frac{d}{dx}\left(\sec^{-1} x\right)=\frac{1}{|x|\sqrt{x^2-1}}$
& $ \frac{d}{dx}\left(\csch x\right)=-\csch x \coth x $
&
\\%end row5
$ \frac{d}{dx}\left(\cot x\right)=-\csc^2 x $
& $ \frac{d}{dx}\left(\cot^{-1} x\right)=\frac{1}{\sqrt{1+x^2}} $
& $ \frac{d}{dx}\left(\coth x\right)=-\csch^2 x $
&
%end row6
\end{tabular} %end of Derivates
%\begin{doublespace}
%emptyspace
%\end{doublespace}
\begin{tabular}[!htp]{l l} % Identities/Substitution
\multicolumn{2}{c}{\textbf {\underline {Identities/Substitution}}} \\
\textbf {\underline {Trig/Hyperbolic Trig}} & \textbf {\underline {Exp/log}} \\
$ \sqrt{a^2-b^2x^2} \Rightarrow x=\frac{a}{b} \sin \theta$
&$e^{- \infty} =0$
\\
$ \sqrt{a^2+b^2x^2} \Rightarrow x=\frac{a}{b} \tan \theta$
&$\pm \ln 0= \mp \infty$
\\
$ \sqrt{b^2x^2-a^2} \Rightarrow x=\frac{a}{b} \sec \theta$
& $\ln e = 1$
\\
$\sinh = \frac{1}{2} \left( e^x - e^{-x} \right)$
&$ \sum\limits_{n=0}^{\infty} = \frac{x^n}{n!} $
\\
$\cosh = \frac{1}{2} \left( e^x + e^{-x} \right)$
&
\\
$\sin^2 \theta + \cos^2 \theta = 1$
&
\\
$\tan^2 \theta + 1 = \sec^2 \theta$
&
\\
$\cot^2 \theta +1 = \csc^2 \theta$
&
\\
\end{tabular} %end identities/Substitution
\begin{tabular}{l l l l} %Integrals
\multicolumn{4}{c}{\textbf {\underline {Integrals}}} \\
\textbf {\underline {Trig}} & \textbf {\underline {Inverse Trig}} & \textbf {\underline {Hyperbolic Trig }}& \textbf {\underline {Exponential/Log}}\\
$ \int \cos u du = \sin u +c$
& $\int \frac{1}{\sqrt{a^2+u^2}}du = \frac1 a \sin^{-1} \left( \frac u a \right) +c$
&$\int \sinh u du=\cosh u+c$
& $\int \frac1 x dx \equiv \int x^{-1} =\ln|x|+c $
\\%end row1
$ \int \sin u du = -\cos u +c$
&$\int \frac{1}{a^2+u^2}du = \frac1 a \tan^{-1} \left( \frac u a \right) +c$
&$\int \cosh u du=\sinh u+c$
& $ \int \frac{1}{ax+b}dx=\frac1 a \ln |ax+b|+c$
\\%end row2
$ \int \tan u du = \ln |\sec u| +c$
&$\int \frac{1}{u \sqrt{u^2-a^2}}du = \frac1 a \sec^{-1} \left( \frac u a \right) +c$
&$\int \tanh u du=\ln (\cosh u)+c$
&$\int a^u du= \frac{a^u}{\ln a}+c$
\\%end row3
$ \int \sec u du = \ln |\sec u + \tan u|+c$
&
&$\int \sech u du=\tan^{-1}| \sinh u| +c$
&$\int e^u du =e^u +c$
\\%end row4
$ \int \csc u du = \ln |\csc u - \cot u|+c$
&
&$\int \sech u \tanh u du=-\sech u +c$
&$\int ue^u du =(u-1)e^u +c$
\\%end row5
$ \int cot u du = \ln |\sin u |+c$
&
&$\int \csch u \coth u du=-\csch u +c$
&$\int \ln u du = u \ln (u) - u+c$
\\%end row6
$ \int \sec u \tan u du = \sec u +c$
&
&$\int \sech^2 du=\tanh u +c$
&$\int \frac{1}{u \ln u}du = \ln| \ln u |+c$
\\%end row7
$ \int \csc u \cot u du =-\csc u +c$
& \begin{minipage}[t]{2in}
\begin{tabular}[t]{l}
\multicolumn{1}{l}{\textbf {\underline {Miscellaneous}} }\\
$\int \frac{1}{a^2-u^2}du = \frac{1}{2a}\ln | \frac{u+a}{u-a}|+c$
\\
$\int \frac{1}{u^2-a^2}du = \frac{1}{2a}\ln | \frac{u-a}{u+a}|+c$
\\
$\int \sqrt{a^2+u^2}du =\frac{u}{2} \sqrt{a^2+u^2}+ \frac{a^2}{2} \ln|u+ \sqrt{a^2+u^2}|+c $
\\
$\int \sqrt{u^2-a^2}du =\frac{u}{2} \sqrt{u^2-a^2}- \frac{a^2}{2} \ln|u+ \sqrt{u^2-a^2}|+c $
\\
$\int \sqrt{a^2-u^2}du =\frac{u}{2} \sqrt{a^2-u^2}+ \sin^{-1} \left( \frac{u}{a} \right)+c $
\end{tabular}
\end{minipage}
&$\int \csch^2 du=-\coth u +c$
&
\\%end row8
$ \int \sec^2 u du = \tan u +c$
&
&
&
\\%end row9
$ \int \csc^2 u du = -\cot u +c$
&
&
&
\\%end row10
$\int \cot^2u du = -x - \cot u +c$
&
&
&
\\%end row11
&
&
&
\\ %end row 12
&
&
&
\\%end row13
&
&
&
\\%end row14
\end{tabular} %end integrals
\end{spacing}
\end{document} % End of Document
答案1
有点难以理解您到底想做什么。我现在假设您想将稍宽的“杂项”部分挤压到“逆三角函数”下方，即“三角函数”最后几行的右侧。您遇到的问题是 太高，minipage增加了单元格的总高度，从而将所有其他行向下推，包括“三角函数”这一行。[t]和 的（顶部）选项minipage仅tabular控制基线的绘制位置，即总高度的哪一部分作为高度，哪一部分作为深度。在这种情况下，您只是有一个很大的深度，这对您没有多大帮助。
为了解决这个问题，您需要减小或删除 的垂直尺寸minipage。这可以通过将其放在 内部来实现\raisebox{0pt}[0pt][0pt]{<content>}。这会将其增加 0pt（强制参数），并将高度和深度设置为 0pt（第一个和第二个可选参数）：
\raisebox{0pt}[0pt][0pt]{%
\begin{minipage}[t]{2in}%
\begin{tabular}[t]{..}
columns ...
\end{tabular}%
\end{minipage}%
}
使用adjustbox带有一些正确设置的包/环境trim也可以达到同样的效果：
\begin{adjustbox}{trim=0pt {\depth} {\width-2in} 0pt}
\begin{tabular}[t]{..}
..
\end{tabular}
\end{adjustobx}
这会削减整个深度（替换\raisebox），并将宽度缩小到 2 英寸（替换{minipage}[t]{2in}）。但这只是逐字内容所必需的。无论如何，它可能会在此过程中避免一些“盒子过满”警告。