Aligning of equations within derivations

Question

Okay, so I have the following derivation so far in my latex document:

\documentclass[a4paper,oneside,11pt]{report}

\usepackage[english]{babel} %francais, polish, spanish, ...
\usepackage[T1]{fontenc}
\usepackage[ansinew]{inputenc}
\usepackage[cm]{fullpage}
\usepackage{float}
\usepackage{lmodern} 
\usepackage{graphicx}
\usepackage{caption}
\usepackage{subcaption}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{a4wide}
\setlength{\marginparwidth}{3cm}
\setlength{\topmargin}{0cm}
\setlength{\voffset}{0cm}
\setlength{\headsep}{0cm}

\usepackage{etoolbox}
\makeatletter
\patchcmd{\chapter}{\if@openright\cleardoublepage\else\clearpage\fi}{}{}{}
\makeatother

\makeatletter
\def\@makechapterhead#1{%
  \vspace*{10\p@}%
  {\parindent \z@ \raggedright \normalfont
    %\ifnum \c@secnumdepth >\m@ne
    %    \huge\bfseries \@chapapp\space \thechapter
    %    \par\nobreak
    %    \vskip 20\p@
    %\fi
    \interlinepenalty\@M
    \Huge \thechapter \space \space \space \bfseries #1\par\nobreak
    \vskip 10\p@
  }}
  \makeatother

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% DOCUMENT
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

\begin{align}
    F_{x_t}(h-h_3) + (F_{z_t} + mgcos\phi - 2F_{z_1})a_2 - 2h(F_{x_1} + F_{x_2}) - 2F_{z_1}a_1 - F_{z_t}(a_2 + b_1) = 0 \nonumber 
\end{align} 
\begin{align}
    2F_{z_1}(a_1 + a_2) &= F_{x_t}(h - h_3) + (F_{z_t} + mgcos\phi)a_2 -2h(F_{x_1} + F_{x_2}) - F_{z_t}(a_2 + b_1) \nonumber \\
    &= A_2sin\phi(h - h_3) + mga_2cos\phi - 2hA_1sin\phi - b_1(A_3sin\phi + B_3cos\phi) \nonumber \\
    &= sin\phi\big(A_2(h - h_3) -2hA_1 -b_1A_3 \big) + cos\phi \big(mga_2 - b_1B_3 \big) \nonumber
\end{align}
\begin{align}
\therefore F_{z_1} = A_4sin\phi + B_4cos\phi
\end{align}
\end{document}

As you can see, there are 5 lines in the derivation (i.e between \begin{align} and \end{align} statements. However, the middle 3 need to be aligned w.r.t to the equals sign, and the top and bottom line should be centred. I've sort of managed to achieve it this way, but I want a way of doing it without 3 align blocks, is this possible?

Answer 1

You could use ordinary equation and/or equation* environments for the first and last equation. Also, if every single line in an align environment gets a \notag statement to suppress an equation number, it's much better to use an align* environment and drop all those \notag statements.

\documentclass[a4paper,oneside,11pt]{report}
\usepackage[cm]{fullpage}
\usepackage{lmodern,amsmath,amssymb}
\usepackage{a4wide}
\setlength{\marginparwidth}{3cm}
\setlength{\topmargin}{0cm}
\setlength{\voffset}{0cm}
\setlength{\headsep}{0cm}

\begin{document}

\begin{equation*}
F_{x_t}(h-h_3) + (F_{z_t} + mg\cos\phi - 2F_{z_1})a_2 - 2h(F_{x_1} + F_{x_2}) - 2F_{z_1}a_1 - F_{z_t}(a_2 + b_1) = 0
\end{equation*}    
\begin{align*}
2F_{z_1}(a_1 + a_2) &= F_{x_t}(h - h_3) + (F_{z_t} + mg\cos\phi)a_2 -2h(F_{x_1} + F_{x_2}) - F_{z_t}(a_2 + b_1) \\
&= A_2\sin\phi(h - h_3) + mga_2\cos\phi - 2hA_1\sin\phi - b_1(A_3\sin\phi + B_3\cos\phi) \\
&= \sin\phi\big(A_2(h - h_3) -2hA_1 -b_1A_3 \big) + \cos\phi \big(mga_2 - b_1B_3 \big)
\end{align*}         
\begin{equation}
\therefore F_{z_1} = A_4\sin\phi + B_4\cos\phi
\end{equation}
\end{document}

---

Addendum If you want to eliminate the extra whitespace between the groups of equations, you could proceed by using a split environment inside a gather environment (with \notag directives for all but the final expression):

\documentclass[a4paper,oneside,11pt]{report}
\usepackage[cm]{fullpage}
\usepackage{lmodern,amsmath,amssymb}
\usepackage{a4wide}
\setlength{\marginparwidth}{3cm}
\setlength{\topmargin}{0cm}
\setlength{\voffset}{0cm}
\setlength{\headsep}{0cm}
\begin{document}
\begin{gather}
F_{x_t}(h-h_3) + (F_{z_t} + mg\cos\phi - 2F_{z_1})a_2 - 2h(F_{x_1} + F_{x_2}) - 2F_{z_1}a_1 - F_{z_t}(a_2 + b_1) = 0\notag\\
\begin{split}
2F_{z_1}(a_1 + a_2) &= F_{x_t}(h - h_3) + (F_{z_t} + mg\cos\phi)a_2 -2h(F_{x_1} + F_{x_2}) - F_{z_t}(a_2 + b_1) \notag\\
&= A_2\sin\phi(h - h_3) + mga_2\cos\phi - 2hA_1\sin\phi - b_1(A_3\sin\phi + B_3\cos\phi) \notag\\
&= \sin\phi\big(A_2(h - h_3) -2hA_1 -b_1A_3 \big) + \cos\phi \big(mga_2 - b_1B_3 \big) \end{split} \notag\\ 
\therefore F_{z_1} = A_4\sin\phi + B_4\cos\phi
\end{gather}
\end{document}

Answer 2

Perhaps

Note you should always use \cos not cos and similar multi-letter identifiers The default math italic font is not designed for multi-letter words.

\documentclass[a4paper,oneside,11pt]{report}

\usepackage[english]{babel} %francais, polish, spanish, ...
\usepackage[T1]{fontenc}
\usepackage[ansinew]{inputenc}
\usepackage[cm]{fullpage}
\usepackage{float}
\usepackage{lmodern} 
\usepackage{graphicx}
\usepackage{caption}
\usepackage{subcaption}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{a4wide}
\setlength{\marginparwidth}{3cm}
\setlength{\topmargin}{0cm}
\setlength{\voffset}{0cm}
\setlength{\headsep}{0cm}

\usepackage{etoolbox}
\makeatletter
\patchcmd{\chapter}{\if@openright\cleardoublepage\else\clearpage\fi}{}{}{}
\makeatother

\makeatletter
\def\@makechapterhead#1{%
  \vspace*{10\p@}%
  {\parindent \z@ \raggedright \normalfont
    %\ifnum \c@secnumdepth >\m@ne
    %    \huge\bfseries \@chapapp\space \thechapter
    %    \par\nobreak
    %    \vskip 20\p@
    %\fi
    \interlinepenalty\@M
    \Huge \thechapter \space \space \space \bfseries #1\par\nobreak
    \vskip 10\p@
  }}
  \makeatother

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% DOCUMENT
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

\begin{gather}
\begin{aligned}
                F_{x_t}(h-h_3) + (F_{z_t} + mg\cos\phi - 2F_{z_1})a_2 - 2h(F_{x_1} + F_{x_2}) - 2F_{z_1}a_1 - F_{z_t}(a_2 + b_1) = 0\\[\jot]
\begin{aligned}
                2F_{z_1}(a_1 + a_2) &= F_{x_t}(h - h_3) + (F_{z_t} + mg\cos\phi)a_2 -2h(F_{x_1} + F_{x_2}) - F_{z_t}(a_2 + b_1)  \\
                &= A_2\sin\phi(h - h_3) + mga_2\cos\phi - 2hA_1\sin\phi - b_1(A_3\sin\phi + B_3\cos\phi)  \\
                &= \sin\phi\big(A_2(h - h_3) -2hA_1 -b_1A_3 \big) + \cos\phi \big(mga_2 - b_1B_3 \big)
\end{aligned}\\[\jot]
                \therefore F_{z_1} = A_4\sin \phi + B_4\cos\phi 
\end{aligned}
\end{gather}
\end{document}

Answer 3

i suggest wrapping the whole thing in a single gather, use aligned to group the middle three lines, and use \nonumber on the first line and just before the (explicit) line break after \end{aligned} so that only one number appears for the whole structure.

\begin{gather}
  F_{x_t}(h-h_3) + (F_{z_t} + mg\cos\phi - 2F_{z_1})a_2 - 2h(F_{x_1} + F_{x_2}) - 2F_{z_1}a_1 - F_{z_t}(a_2 + b_1) = 0 \nonumber \\
  \begin{aligned}
    2F_{z_1}(a_1 + a_2) &= F_{x_t}(h - h_3) + (F_{z_t} + mg\cos\phi)a_2 -2h(F_{x_1} + F_{x_2}) - F_{z_t}(a_2 + b_1) \\
       &= A_2\sin\phi(h - h_3) + mga_2\cos\phi - 2hA_1\sin\phi - b_1(A_3\sin\phi + B_3\cos\phi) \\
       &= \sin\phi\big(A_2(h - h_3) -2hA_1 -b_1A_3 \big) + \cos\phi \big(mga_2 - b_1B_3 \big)
  \end{aligned} \nonumber \\
  \therefore F_{z_1} = A_4\sin\phi + B_4\cos\phi
\end{gather}

any "line" within the scope of gather will be centered; the aligned block will be treated as a single "line" for this purpose.

as Mico suggests in a comment, "sin" and "cos" are more appropriately set as operators with the commands \sin and \cos.

Answer 4

You can do it without align environment, instead using stackengine

\documentclass[a4paper,oneside,11pt]{report}
\usepackage[english]{babel} %francais, polish, spanish, ...
\usepackage[T1]{fontenc}
\usepackage{lmodern} 
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{a4wide}
\setlength{\marginparwidth}{3cm}
\setlength{\topmargin}{0cm}
\setlength{\voffset}{0cm}
\setlength{\headsep}{0cm}
\usepackage[usestackEOL]{stackengine}[2013-09-11]
\begin{document}
\stackMath\setstackgap{S}{1.5em}\setstackgap{L}{1.2\baselineskip}
\begin{equation}
\savestack{\temp}{\Longunderstack[l]{%
  = F_{x_t}(h - h_3) + (F_{z_t} + mg\cos\phi)a_2 -2h(F_{x_1} + F_{x_2}) - F_{z_t}(a_2 + b_1)\\
 = A_2\sin\phi(h - h_3) + mga_2\cos\phi - 2hA_1\sin\phi - b_1(A_3\sin\phi + B_3\cos\phi)\\
  = \sin\phi\big(A_2(h - h_3) -2hA_1 -b_1A_3 \big) + \cos\phi \big(mga_2 - b_1B_3 \big)}%
}
\Shortstack{%
F_{x_t}(h-h_3) + (F_{z_t} + mg\cos\phi - 2F_{z_1})a_2 - 2h(F_{x_1} + F_{x_2}) - 2F_{z_1}a_1 - F_{z_t}(a_2 + b_1) = 0\\
2F_{z_1}(a_2 + b_1)\mathrel{\temp}\\%
\therefore F_{z_1} = A_4\sin\phi + B_4\cos\phi%
}
\end{equation}
\end{document}