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将环境与多个分离的值对齐

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将环境与多个分离的值对齐

我试图显示一个分为 3 列的方程式，如下图所示。

在此处输入图片描述

我设法使用不间断空格（〜）创建了上述图像，但由于它们没有正确对齐，因此在编辑器和文档中看起来显然都是垃圾。

\begin{align*}
\mathrm{Area~of~\Delta ABD}~~~~~~~~~~~~ &= ~~~~~~~~~~\mathrm{area~of~\Delta ABC}  ~~      &+ ~~~~~~~ \mathrm{area~of~\Delta ACD} \\
\frac{x*z*\sin(\alpha+\beta)}{2}~~~~~~~~~~~~ &= ~~~~~~~~~~~~ \frac{x*y*\sin\alpha}{2} &+ ~~~~~~~~~~~ \frac{z*y*\sin\beta}{2}
\end{align*}

Both sides are multiplied by 2:
\begin{align*}
    x*z*\sin(\alpha+\beta)~~~~~~~~~~ &= ~~~~~~~~~~~~~~~ x*y*\sin\alpha &+ ~~~~~~~~~~ z*y*\sin\beta
\end{align*}

Both sides are divided by $x*z$:
\begin{align*}
    \sin(\alpha+\beta)~~~~~~~~~~~~~~~~ &= ~~~~~~~~~~~~~~~~~~ \frac{y*\sin\alpha}{z} &+ ~~~~~~~~~~~ \frac{y*\sin\beta}{x}
\end{align*}

As $\frac{y}{z}=\cos(\beta)$ and $\frac{y}{x}=\cos(\alpha)$ we get:
\begin{align*}
\sin(\alpha+\beta) ~~~~~~~~~~~~~~~~~~~ &= ~~~~~~~~~~~~~~~ \cos\beta*\sin\alpha &+ ~~~~~~~ \cos\alpha*\sin(\beta)
\end{align*}

我想知道怎样才能用一种不那么可怕的方式来做这件事？

答案1

如果列内容居中以及符号对齐=对+您来说很重要，我建议您采用array如下四列环境：

在此处输入图片描述

\sin除了删除 和参数周围不需要的（且适得其反的）花括号之外\cos，我还删除了所有乘法符号，并用内联数学分数等价物*替换了一些表达式。\frac

\documentclass{article} % or some other suitable document class
\usepackage{array} % for "\newcolumntype" macro
\newcolumntype{L}{>{$}l<{$}} % automatic text mode, left-aligned

\begin{document}

\[
\renewcommand\arraystretch{1.667}
\begin{array}{@{} l @{\qquad\qquad} c @{{}={}} c @{{}+{}} c @{}}
& \textnormal{Area of $\Delta\mathit{ABD}$} 
& \textnormal{Area of $\Delta\mathit{ABC}$}
& \textnormal{Area of $\Delta\mathit{ACD}$} \\[0.5ex]
& \frac{1}{2}x z\sin(\alpha+\beta) 
& \frac{1}{2}x y\sin\alpha 
& \frac{1}{2}z y\sin\beta \\
\multicolumn{4}{@{}L}{Multiply both sides by 2:}\\
& xz\sin(\alpha+\beta) 
& xy\sin\alpha 
& zy\sin\beta\\
\multicolumn{4}{@{}L}{Divide both sides by $x z$:}\\
& \sin(\alpha+\beta) 
& (y/z)\sin\alpha 
& (y/x)\sin\beta \\
\multicolumn{4}{@{}L}{As $y/z=\cos\beta$ and $y/x=\cos\alpha$ we get}\\
& \sin(\alpha+\beta)  
& \cos\beta\sin\alpha 
& \cos\alpha\sin\beta\,.
\end{array}
\]

\end{document}

答案2

您可以像这样对齐，尽管您的第一个“文本”行相当宽，因此对齐的 + 看起来有点间隔。

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\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{alignat*}{2}
\text{Area of $\Delta ABD$} &= \text{area of $\Delta ABC$}&&+  \text{area of $\Delta ACD$} \\
\frac{x*z*\sin(\alpha+\beta)}{2} &=  \frac{x*y*\sin\alpha}{2} &&+  \frac{z*y*\sin\beta}{2}\\
\intertext{Both sides are multiplied by $2$:}
    x*z*\sin(\alpha+\beta) &=  x*y*\sin\alpha &&+  z*y*\sin\beta
\intertext{Both sides are divided by $x*z$:}
    \sin(\alpha+\beta) &=  \frac{y*\sin\alpha}{z} &&+  \frac{y*\sin\beta}{x}
\intertext{As $\frac{y}{z}=\cos(\beta)$ and $\frac{y}{x}=\cos(\alpha)$ we get:}
\sin(\alpha+\beta)  &=  \cos\beta*\sin\alpha &&+  \cos\alpha*\sin(\beta)
\end{alignat*}
\end{document}

答案3

我建议用三种方法来解决这个问题。第一种方法与你的方法类似；对于第二种方法，我使用中间列居中，以尽量减少空白。

在第三种情况下，不寻求任何对齐：每个公式都有三个部分，读者将能够在各个步骤中匹配它们。

请避免使用 ∗ 进行乘法。

\documentclass{article} % or some other suitable document class
\usepackage{amsmath,mathtools}
\usepackage{IEEEtrantools}

\newcommand{\IEEEintertext}[1]{\noalign{#1\vspace{1ex}}}

\begin{document}

\section{Your approach}

\begin{alignat*}{2}
\text{Area of $\Delta\mathit{ABD}$} &= \text{Area of $\Delta\mathit{ABC}$}
  &&+ \text{Area of $\Delta\mathit{ACD}$} \\
\frac{x z\sin(\alpha+\beta)}{2} &= \frac{x y\sin\alpha}{2} &&+ \frac{z y\sin\beta}{2} \\
\shortintertext{Multiply both sides by 2:}
xz\sin(\alpha+\beta) &= xy\sin\alpha &&+ zy\sin\beta\\
\shortintertext{Divide both sides by $x z$:}
\sin(\alpha+\beta) &= \frac{y\sin\alpha}{z} &&+ \frac{y\sin\beta}{x} \\
\shortintertext{As $y/z=\cos\beta$ and $y/x=\cos\alpha$ we get}
\sin(\alpha+\beta) &= \cos\beta\sin\alpha &&+ \cos\alpha\sin\beta\,.
\end{alignat*}

\section{With slightly different alignment}

\begin{IEEEeqnarray*}{rCcCl}
\text{Area of $\Delta\mathit{ABD}$} &=& \text{Area of $\Delta\mathit{ABC}$}
  &+& \text{Area of $\Delta\mathit{ACD}$} \\[1ex]
\frac{x z\sin(\alpha+\beta)}{2} &=& \frac{x y\sin\alpha}{2} &+& \frac{z y\sin\beta}{2} \\[1ex]
\IEEEintertext{Multiply both sides by 2:}
xz\sin(\alpha+\beta) &=& xy\sin\alpha &+& zy\sin\beta\\[1ex]
\IEEEintertext{Divide both sides by $x z$:}
\sin(\alpha+\beta) &=& \frac{y\sin\alpha}{z} &+& \frac{y\sin\beta}{x} \\[1ex]
\IEEEintertext{As $y/z=\cos\beta$ and $y/x=\cos\alpha$ we get}
\sin(\alpha+\beta) &=& \cos\beta\sin\alpha &+& \cos\alpha\sin\beta\,.
\end{IEEEeqnarray*}

\section{An altogether different method}

We start from the fact that
\[
\text{Area of $\Delta\mathit{ABD}$} = \text{Area of $\Delta\mathit{ABC}$}
  + \text{Area of $\Delta\mathit{ACD}$}
\]
Writing down the formulas and removing the common denominator $2$ gives
\[
xz\sin(\alpha+\beta) = xy\sin\alpha + zy\sin\beta
\]
Now we can divide both sides by $xz$ to obtain
\[
\sin(\alpha+\beta)=\frac{y}{z}\sin\alpha+\frac{y}{x}\sin\beta
\]
However, by definition, $y/z=\cos\beta$ and $y/x=\cos\alpha$, so we finally get
\[
\sin(\alpha+\beta)=\cos\beta\sin\alpha+\cos\alpha\sin\beta
\]

\end{document}

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