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May I ask what font is being used in this document? enter image description here

I am new to latex and I am used to the default font. I am making a lecture not and I wanna know also how to typeset my whole document using this font. Will the symbols get affected too? Any help is appreciated. thanks

james34
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    Welcome to TeX.SE. Did you take the screenshot you posted from a pdf file? If so, and if you use Acrobat or Acrobat Reader, what's the font information you get by clicking on File -> Properties -> Fonts? (If you use another pdf browser, it should have equivalent steps to obtain information about the font, or fonts, used in the document.) – Mico Apr 01 '21 at 06:33
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    Seems to be some Garamond – User Apr 01 '21 at 06:34
  • Hi..I have checked it as you instructed. I see Garamond, Helvetica, Courier...etc there are a lot of fonts used for the entire doc. – james34 Apr 01 '21 at 06:51
  • Forget what fonts are used in that document and focus on what fonts of the The LaText Font Catalogue you prefer. – Fran Apr 01 '21 at 06:57
  • Hi again..it is actually Adobe Garamond Pro. But how do I use this as my default font style for my latex doc? Will the symbols also change? – james34 Apr 01 '21 at 06:58
  • See https://tex.stackexchange.com/questions/534260/garamond-for-both-text-and-math for a related discussion. – Marijn Apr 01 '21 at 10:26
  • See also https://tex.stackexchange.com/q/112313/231952 – Ivan Apr 01 '21 at 12:03

1 Answers1

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Here I put a minimal working example where I use the garamond option with mathdesign: \usepackage[garamond]{mathdesign}. I use this site https://www.ctan.org/tex-archive/fonts/urw/garamond/ where you can download and install a clone of garamond: urw-garamond (GaramondNo8 Regular, Medium, Italic, and Medium Italic).

enter image description here

\documentclass[12pt]{article}
\usepackage[margin=2cm]{geometry}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage[garamond]{mathdesign}
\usepackage{parskip}
\begin{document}

$|2\sin^2 x-1|>\cos x$ $|\cos(2x)|>\cos x$  $\cos(2x)=\cos^2 x- \sin^2 x=2\cos^2 x-1=-(1-2\cos^2 x)$.
[|2\sin^2 x-1|>\cos x \iff |\cos(2x)|>\cos x ]
[\cos \alpha -\cos \beta =-2\sin \left( \frac{\alpha +\beta }{2} \right)\sin \left( \frac{\alpha -\beta }{2} \right)] ad $|\cos (2x)| - \cos x = 0$
$|a|=b \iff a=\pm b$; $\cos(2x)=\pm\cos(x)$.
\end{document}
Sebastiano
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