2

I need do align some (not all) of equations to the left. So also I want to avoid weird formatting of them

so that is what $\displaystyle ...$ does enter image description here

Interline space is tiny. To increase it, I can use \[ \] environment, but it centers equations: enter image description here

Though space between lines is ok. So I can try flalign*

{\setlength{\abovedisplayskip}{0pt}%
\begin{flalign*}
& f(\theta)=cos^a(\theta)&\\
& cos^{2\alpha}\theta^{\circ}_{\text{ск}}=1/2&\\
& \alpha=0.5\cdot\frac{log0.5}{log(cos(18^\circ))}=6.91&\\
& cos^{2\alpha}\theta^{\circ}_{\text{Д}}=t=0.013&\\
& \theta_\text{Д}=arccos(\sqrt[2\alpha]{t_\text{ед}})=43.24&\\
\end{flalign*}%
}

enter image description here

As you can see, interline space decreases again (but it is bigger than in $...$ variant), and there is also space before and after the block, though I tried to get rid of it as it is described here

I've triedalign* too

формулой для аппроксимации ДН элемента: \par\noindent
$\begin{aligned}
&f(\theta)=cos^a(\theta) &\\
&cos^{2\alpha}\theta^{\circ}_{\text{ск}}=1/2 &\\
&\alpha=0.5\cdot\frac{log0.5}{log(cos(18^\circ))}=6.91 &\\
&cos^{2\alpha}\theta^{\circ}_{\text{Д}}=t=0.013 &\\
&\theta_\text{Д}=arccos(\sqrt[2\alpha]{t_\text{ед}})=43.24 &\\
\end{aligned}$

enter image description here

It's OK about spaces before and after, but spacing between lines is decreased.

Also I can not use fleqn as I want to be left-aligned only part of the equations, not all of them.

SO what can I do?

P.S. Yes, I can add \vspace after each line, but maybe there is another,better way to do this. Thanks!

1
  • 3
    unrelated to alignment but use \sin and \cos never use math italic for multi-letter names, it is designed to look like a product of one-letter variables Dec 13, 2021 at 14:26

2 Answers 2

6

The simplest solution uses the fleqn environment, from nccmath:

\documentclass{article}
\usepackage{amsmath, nccmath}
\usepackage{showframe}
\renewcommand{\ShowFrameLinethickness}{0.4pt}

\begin{document}

This equation is left-aligned:
\begin{fleqn}
\begin{align*}
& f(\theta)=\cos^a(\theta) \\
& \cos^{2\alpha}\theta^{\circ}_{\text{ск}}=1/2 \\
& \alpha\=0.5\cdot\frac{\log0.5}{\log(\cos(18^\circ))}=6.91 \\
& \cos^{2\alpha}\theta^{\circ}_{\text{Д}}=t=0.013\\
& \theta_\text{Д}=\arccos(\sqrt[2\alpha]{t_\text{ед}})=43.24
\end{align*}%
\end{fleqn}

\end{document} 

enter image description here

3
  • As I said in the original post, I don't want to use fleqn because I want only some of equations flush left, not all. How can I centerothers&
    – lazba
    Dec 13, 2021 at 15:36
  • 6
    @lazba please look again at Bernard's answer. The package gives a fleqn environment specifically so you can apply it to just some equations Dec 13, 2021 at 15:42
  • Oh, really! my fault
    – lazba
    Dec 13, 2021 at 15:54
4

You may use the \IEEEeqnarray environment along with the * type glue to align the equations to the left. The related manual is available here. Appendix F of this manual describes the usage of the environment. Also, you can control the vertical spacing above and below the equation using the solution from the question you linked.

\documentclass{article}
\usepackage{amsmath}
\usepackage{IEEEtrantools}
\usepackage{lipsum}
\usepackage{showframe}

\begin{document}
\section*{Default Spacing}
\lipsum[1][1]
\begin{IEEEeqnarray*}{*l}
f(\theta)=cos^a(\theta) \\
\cos^{2\alpha}\theta^{\circ}_{\text{ск}}=1/2 \\
\alpha=0.5\cdot\frac{\log0.5}{\log(cos(18^\circ))}=6.91 \\
\cos^{2\alpha}\theta^{\circ}_{\text{Д}}=t=0.013 \\
\theta_\text{Д}=arccos(\sqrt[2\alpha]{t_\text{ед}})=43.24
\end{IEEEeqnarray*}
\lipsum[1][2]

\section*{User Defined Spacing}
\lipsum[1][1]
{\setlength{\abovedisplayskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
\begin{IEEEeqnarray*}{*l}
f(\theta)=cos^a(\theta) \\
\cos^{2\alpha}\theta^{\circ}_{\text{ск}}=1/2 \\
\alpha=0.5\cdot\frac{\log0.5}{\log(cos(18^\circ))}=6.91 \\
\cos^{2\alpha}\theta^{\circ}_{\text{Д}}=t=0.013 \\
\theta_\text{Д}=arccos(\sqrt[2\alpha]{t_\text{ед}})=43.24
\end{IEEEeqnarray*}}%
\lipsum[1][2]

\section*{Default Spacing}
\lipsum[1][1]
\begin{IEEEeqnarray*}{*l}
f(\theta)=cos^a(\theta) \\
\cos^{2\alpha}\theta^{\circ}_{\text{ск}}=1/2 \\
\alpha=0.5\cdot\frac{\log0.5}{\log(cos(18^\circ))}=6.91 \\
\cos^{2\alpha}\theta^{\circ}_{\text{Д}}=t=0.013 \\
\theta_\text{Д}=arccos(\sqrt[2\alpha]{t_\text{ед}})=43.24
\end{IEEEeqnarray*}
\lipsum[1][2]

\end{document} 

enter image description here

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