答案1
通过使用nccmath
和手动调整括号大小,您可以显著减少方程的大小:
\documentclass{article}
\usepackage{nccmath}
\begin{document}
\begin{align}\medmath{
\left\lbrace \begin{aligned}
\phi &= \arctan\left(\frac{j_z}{k_z}\right) \\
\theta &= \arctan\left(\frac{-i_z}{\sqrt{1-i_z^2}}\right) = \arcsin\left(-i_z\right) \\
\psi &= \arctan\left(\frac{i_y}{i_x}\right)
\end{aligned} \right .}
\end{align}
\end{document}
答案2
去掉分数周围不需要的括号可以节省大量(垂直)空间。使用内联分数表示法代替以下方法可以节省更多空间\frac
:
\documentclass{article}
\usepackage{amsmath} % for 'align' and 'aligned' environments
\begin{document}
\begin{align}
\text{with tall parens: }&
\left\lbrace \begin{aligned}
\phi &= \arctan\left(\frac{j_z}{k_z}\right) \\
\theta &= \arctan\left(\frac{-i_z}{\sqrt{1-i_z^2}}\right) = \arcsin\left(-i_z\right) \\
\psi &= \arctan\left(\frac{i_y}{i_x}\right)
\end{aligned} \right . \\
\text{without tall parens: } &
\left\lbrace \begin{aligned}
\phi &= \arctan\frac{j_z}{k_z} \\
\theta &= \arctan\frac{-i_z}{\sqrt{1-i_z^2}} = \arcsin(-i_z) \\
\psi &= \arctan\frac{i_y}{i_x}
\end{aligned} \right . \\
\text{inline fractions: } &
\left\lbrace \begin{aligned}
\phi &= \arctan(j_z/k_z) \\
\theta &= \arctan\bigl(-i_z/\sqrt{1-i_z^2}\,\bigr) = \arcsin(-i_z) \\
\psi &= \arctan(i_y/i_x)
\end{aligned} \right .
\end{align}
\end{document}