在 aligned 和 aligned[t] 环境中对齐（2）

Question

以下代码几乎满足了我的要求。您会看到，在显示的方程式的最后五行中，大多数表达式都向右移动了。我希望这些行中的第一个“+”与第 2 行和第 3 行中的第一个“+”对齐。谢谢。

\documentclass[10pt]{amsart}
\usepackage{}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{newlfont}
\usepackage{mathtools}
\usepackage{tikz}

\begin{document}
\begin{alignat*}{3}
\vert x + y + z \vert^{2}
&= [\cos{a} + \cos{b} + \cos{c}]^{2} + [\sin{a} + \sin{b} + \sin{c}]^{2} \\
&= \begin{aligned}[t] 3 &+ [2\cos{a}\cos{b} + 2\cos{a}\cos{c} + 2\cos{b}\cos{c}] \\
&+ [2\sin{a}\sin{b} + 2\sin{a}\sin{c} + 2\sin{b}\sin{c}]
\end{aligned} \\
&= 3&& + \bigl[&&\cos(a + b) + \cos(a + c) + \cos(b + c) \\
  &&&&&+ \cos(a - b) + \cos(a - c) + \cos(b - c)\bigr] \\
  &&&+ \bigl[&&\cos(a - b) + \cos(a - c) + \cos(b - c) \\
  &&&&&- \bigl(\cos(a + b) + \cos(a + c) + \cos(b + c)\bigr) \bigr] \\
  &= 3&&\mathrlap{{}+ 2\bigl(\cos(a - b) + \cos(a - c) + \cos(b - c)\bigr).}
\end{alignat*}

\end{document}

Answer 1

参考你的上一个问题：

\mathrlap您也可以在前三行使用。

\documentclass[10pt]{amsart}
\usepackage{mathtools}% loads also amsmath

\begin{document}

\begin{alignat*}{3}
  \vert x + y + z \vert^{2}
    &= \mathrlap{[\cos{a} + \cos{b} + \cos{c}]^{2} + [\sin{a} + \sin{b} + \sin{c}]^{2}} \\
  &= 3 &&\mathrlap{{}+ [2\cos{a}\cos{b} + 2\cos{a}\cos{c} + 2\cos{b}\cos{c}]} \\
  &&&\mathrlap{{}+ [2\sin{a}\sin{b} + 2\sin{a}\sin{c} + 2\sin{b}\sin{c}]}\\
  &= 3&& + \bigl[&&\cos(a + b) + \cos(a + c) + \cos(b + c) \\
  &&&&&+ \cos(a - b) + \cos(a - c) + \cos(b - c)\bigr] \\
  &&&+ \bigl[&&\cos(a - b) + \cos(a - c) + \cos(b - c) \\
  &&&&&- \bigl(\cos(a + b) + \cos(a + c) + \cos(b + c)\bigr) \bigr] \\
  &= 3&&\mathrlap{{}+ 2\bigl(\cos(a - b) + \cos(a - c) + \cos(b - c)\bigr).}
\end{alignat*}

\end{document}

结果：

\mathrlap或者仅在第一行使用，并aligned在其他行使用。

\documentclass[10pt]{amsart}
\usepackage{mathtools}% loads also amsmath

\begin{document}

\begin{alignat*}{2}
  \vert x + y + z \vert^{2}
    &= \mathrlap{[\cos{a} + \cos{b} + \cos{c}]^{2} + [\sin{a} + \sin{b} + \sin{c}]^{2}} \\
  &= 3 &&
    \!\begin{aligned}[t]
      & + [2\cos{a}\cos{b} + 2\cos{a}\cos{c} + 2\cos{b}\cos{c}] \\
      &+ [2\sin{a}\sin{b} + 2\sin{a}\sin{c} + 2\sin{b}\sin{c}]
    \end{aligned} \\
  &= 3&&
    \!\begin{aligned}[t]
      {}+ \bigl[&\cos(a + b) + \cos(a + c) + \cos(b + c) \\
      &+ \cos(a - b) + \cos(a - c) + \cos(b - c)\bigr]
    \end{aligned} \\
  &&&
    \!\begin{aligned}[t]
      {}+ \bigl[&\cos(a - b) + \cos(a - c) + \cos(b - c) \\
      &- \bigl(\cos(a + b) + \cos(a + c) + \cos(b + c)\bigr) \bigr]
    \end{aligned}\\
  &= 3&&+ 2\bigl(\cos(a - b) + \cos(a - c) + \cos(b - c)\bigr).
\end{alignat*}

\end{document}

结果：

前面\!的\begin{aligned}[t]是必需的，因为aligned插入一个\,

Answer 2

据我了解，您想要获得以下内容（哦！）：

\documentclass[10pt]{amsart}
\usepackage{}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{newlfont}
\usepackage{mathtools}
\usepackage{tikz}

\begin{document}
\begin{alignat*}{3}
\vert x + y + z \vert^{2}
&= [\cos{a} + \cos{b} + \cos{c}]^{2} + [\sin{a} + \sin{b} + \sin{c}]^{2} \\
&= \begin{aligned}[t] 3 &+ [2\cos{a}\cos{b} + 2\cos{a}\cos{c} + 2\cos{b}\cos{c}] \\
&+ [2\sin{a}\sin{b} + 2\sin{a}\sin{c} + 2\sin{b}\sin{c}]
\end{aligned}\\ 
&= \begin{aligned}[t]3 &+ \bigl[\cos(a + b) + \cos(a + c) + \cos(b + c) \\
 & + \cos(a - b) + \cos(a - c) + \cos(b - c)\bigr] \\
  &+ \bigl[\cos(a - b) + \cos(a - c) + \cos(b - c) \\
  &- \bigl(\cos(a + b) + \cos(a + c) + \cos(b + c)\bigr) \bigr] \\
\end{aligned} \\
  &= 3\mathrlap{{}+ 2\bigl(\cos(a - b) + \cos(a - c) + \cos(b - c)\bigr).}
\end{alignat*}

\end{document}

我不确定最后一行是否应该看起来有点不同，正如您的示例所建议的那样，但我保留了它。然而，我建议使用以下版本：

\documentclass[10pt]{amsart}
\usepackage{}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{newlfont}
\usepackage{mathtools}
\usepackage{tikz}

\begin{document}
\begin{alignat*}{3}
\vert x + y + z \vert^{2}
&= [\cos{a} + \cos{b} + \cos{c}]^{2} + [\sin{a} + \sin{b} + \sin{c}]^{2} \\
&= \!\begin{aligned}[t] 3 &+ [2\cos{a}\cos{b} + 2\cos{a}\cos{c} + 2\cos{b}\cos{c}] \\
&+ [2\sin{a}\sin{b} + 2\sin{a}\sin{c} + 2\sin{b}\sin{c}]
\end{aligned}\\ 
&=\! \begin{aligned}[t]3 &+ \bigl[\cos(a + b) + \cos(a + c) + \cos(b + c) \\
 & + \cos(a - b) + \cos(a - c) + \cos(b - c)\bigr] \\
  &+ \bigl[\cos(a - b) + \cos(a - c) + \cos(b - c) \\
  &- \bigl(\cos(a + b) + \cos(a + c) + \cos(b + c)\bigr) \bigr] \\
\end{aligned} \\
%  &= 3\mathrlap{{}+ 2\bigl(\cos(a - b) + \cos(a - c) + \cos(b - c)\bigr).}
 &= 3{{}+ 2\bigl(\cos(a - b) + \cos(a - c) + \cos(b - c)\bigr).}
\end{alignat*}

\end{document}