How to break this long radical into multiple lines?

Question

I've got a very long radical in a display-math environment, so long that the equation spills in the right margin. How can I break that equation over multiple lines, so that it doesn't go into the right margin?

\documentclass[12pt,a4paper]{article}

\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{fourier}
\usepackage[left=3cm,right=3cm,top=2.5cm,bottom=2cm]{geometry}

\begin{document}
\[
    V_{DABC}=\dfrac{1}{6}\cdot DA \cdot DB \cdot DC \cdot \sqrt{1+2\cdot
    \cos \widehat{ADB}\cdot \cos \widehat{BDC}\cdot \cos \widehat{ADC}
    -\cos^2 \widehat{ADB} -\cos^2 \widehat{ADC} -\cos^2 \widehat{BDC}} 
    =\dfrac{2\sqrt{2}}{3}.
\]
\end{document}

Answer 1

I'd much prefer Jubobs's approach, but just to show another method, you can split into two lines the radicand:

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{fourier}
\usepackage[left=3cm,right=3cm,top=2.5cm,bottom=2cm]{geometry}
\begin{document}
\[
  V_{DABC}=\dfrac{1}{6}\cdot DA \cdot DB \cdot DC \cdot
  \sqrt{
    \begin{aligned}
     1+&2\cos\widehat{ADB}\cos\widehat{BDC}\cos\widehat{ADC} \\
     & -\cos^2 \widehat{ADB} -\cos^2 \widehat{ADC} -\cos^2 \widehat{BDC}
    \end{aligned}
  }
  =\dfrac{2\sqrt{2}}{3}.
\]
\end{document}

I removed the unnecessary centered dots.

Answer 2

As suggested by egreg in his comment, the angles take up a lot of horizontal space, which you can claim back by defining shorter variables for them (alpha, beta, gamma, in my code).

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb} % <---- for \triangleq
\usepackage{fourier}
\usepackage[left=3cm,right=3cm,top=2.5cm,bottom=2cm,showframe]{geometry}
\begin{document}
\begin{align*}
  V_{DABC} &=\dfrac{1}{6}\cdot DA \cdot DB \cdot DC \cdot
    \sqrt{
      1 + 2 \cos\alpha \cos\beta \cos\gamma
        -\cos^2 \alpha -\cos^2 \beta -\cos^2 \gamma
    } \\
           &= \dfrac{2\sqrt{2}}{3}\,,
\end{align*}
%
where \(\alpha \triangleq \widehat{BDC}\),
\(\beta \triangleq \widehat{ADC}\),
and \(\gamma \triangleq \widehat{ADB}\).
\end{document}

Answer 3

An approach that produces a result that's very similar to the one in @egreg's answer relies on the \splitfrac macro of the mathtools package.

In the example below, I also utilize a macro called \V that typesets its argument in text italics instead of math italics. The macro is applied to the strings "DABC", "ADB", "DA", etc to keep TeX from typesetting them as if they were separate variables named "A", "B", "C", etc.

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools} % for \splitfrac macro
\usepackage{fourier}
\usepackage[hmargin=3cm,top=2.5cm,bottom=2cm,showframe]{geometry} 
\newcommand{\V}[1]{\textit{#1}} % shorthand macro, typesets its argument in text italics
\begin{document}
\[
V_{\V{DABC}}
=\frac{1}{6}\, \V{DA} \cdot \V{DB} \cdot \V{DC} \cdot
\sqrt{\splitfrac{
1+2\cos \widehat{\V{ADB}} \cdot \cos \widehat{\V{BDC}} \cdot \cos \widehat{\V{ADC}}}{
-\cos^2 \widehat{\V{ADB}} -\cos^2 \widehat{\V{ADC}}  -\cos^2 \widehat{\V{BDC}}}\,} 
=\frac{2\sqrt{2}}{3}\,.
\]
\end{document}

Answer 4

I think this is one of those cases where a local definition may help; in the code below, I have introduced f(A,B,C,D) under your radical, and then defined it immediately below.

% arara: pdflatex
% !arara: indent: {overwrite: yes}
\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\usepackage[left=3cm,right=3cm,top=2.5cm,bottom=2cm]{geometry}
\begin{document}
\begin{align}
    V_{DABC} & =\dfrac{1}{6}\cdot DA \cdot DB \cdot DC \cdot \sqrt{f(A,B,C,D)} \\
             & =\dfrac{2\sqrt{2}}{3}.
\end{align}
where
\begin{align*}
    f(A,B,C,D)  = 1 & +2\cdot \cos \widehat{ADB}\cdot \cos \widehat{BDC}\cdot \cos \widehat{ADC} \\
                    & -\cos^2 \widehat{ADB}-\cos^2 \widehat{ADC}-\cos^2 \widehat{BDC}
\end{align*}
\end{document}