\begin{eqnarray}
\begin{split}
J_{cp} &=& \frac{1}{36} \bigg[\sqrt{3} \sin (2 \text{$\phi' $}) (\sin (\sigma ) \cos (\text{$\sigma' $}) \sin (2 \phi )-2 \sin (\text{$\sigma' $}) \cos (2 \phi ))\\
&\quad-3 \sin (\sigma ) \sin (2 \phi ) \cos (2 \text{$\phi' $})\bigg]\\
I_{1} &=& \frac{1}{288}\bigg[\left. 2 \sqrt{3} \sin (\phi) \sin (2 \phi') \left( 8 \sin (2 \sigma) \cos (\sigma') \sin (\phi) \cos^2 (\phi) \right. \right. \\
&\quad \left. \left. + 2 \sin (\sigma') \left( 8 \cos (\sigma) \cos^3 (\phi) - 4 \cos (2 \sigma) \sin (\phi) \cos^2 (\phi) + \sin (\phi) + 5 \sin (3 \phi) \right) \right. \right.\\
&\quad \left. \left. - 4 \sin (\sigma) \cos (\sigma') (\cos (\phi) + 3 \cos (3 \phi)) \right) \right. \\
&\quad \left. + 6 \cos (2 \phi') \left( \sin (4 \phi) (\sin (\sigma - 2 \sigma') - 3 \sin (\sigma)) \right. \right. \\
&\quad \left. \left. + \cos (\sigma') (-4 \sin (2 \phi) \sin (\sigma - \sigma') + 2 \sin (\sigma) \cos (4 \phi) \cos (\sigma - \sigma') \right. \right. \\
&\quad \left. \left. - \sin (2 \sigma - \sigma') + 3 \sin (\sigma')) - 2 \sin (2 \sigma') \cos (2 \phi) \right) \right. \\
&\quad \left. - 24 \sin (2 \sigma') \sin^3 (\phi) (2 \cos (\sigma) \cos (\phi) + \sin (\phi)) \right. \\
&\quad \left. - 12 \sin (2 \phi) \left( \sin (\sigma) (\cos (2 \sigma') + 3) \cos (2 \phi) + \sin (\sigma') (\sin (2 \phi) \cos (2 \sigma - \sigma')\right) \right) \\
&\quad \left. + 2 \sin (\sigma) \sin (\sigma') \right.\bigg]\\
I_{2} &=& \frac{1}{2304} \bigg[\left. 96 \cos (2 \phi') \left( \sin^3 (\phi) \left( 8 \sin (\sigma) \cos^2 (2 \sigma') \cos (\phi) \right. \right. \right.\\
&\quad \left. \left. + \sin (4 \sigma') (\sin (\phi) - 4 \cos (\sigma) \cos (\phi)) \right) \right. \\
&\quad \left. - 2 \cos (2 \sigma') \sin^2 (2 \phi) \sin (2 (\sigma - \sigma')) \right) \\
&\quad + 2 \sin (2 \sigma') \cos (4 \phi') \left( -24 \sin^3 (\phi) (\cos (2 \sigma') \sin (\phi) - 4 \cos (\phi) \cos (\sigma - 2 \sigma')) \right. \\
&\quad \left. + 8 \sin (2 \phi) (\sin (2 \phi) (\cos (2 \sigma) - 3 \cos (2 (\sigma - \sigma'))) + \cos (\sigma)) \right. \\
&\quad \left. - 36 \cos (\sigma) \sin (4 \phi) + 36 \cos (2 \phi) + 23 \cos (4 \phi) + 5 \right) \\
&\quad + 64 \sqrt{3} \cos (2 \sigma') \sin (\phi) \sin (2 \phi') (2 \cos (\sigma) \sin (2 \phi) - 3 \cos (2 \phi) - 1) \\
&\quad \left( 2 \sin (\sigma) \cos (\sigma') \cos (\phi) + \sin (\sigma') (\sin (\phi) - 2 \cos (\sigma) \cos (\phi)) \right) \\
&\quad + 32 \sqrt{3} \sin (2 \sigma') \sin (\phi) \sin (4 \phi') (-2 \cos (\sigma) \sin (2 \phi) + 3 \cos (2 \phi) + 1) \\
&\quad \left( \cos (\sigma') \sin (\phi) - 2 \cos (\phi) \cos (\sigma - \sigma') \right) \\
&\quad - 288 \sin^2 (\sigma') \cos (\sigma') (\sin (4 \phi) \sin (\sigma - \sigma') + \sin (\sigma') \cos (2 \phi)) \\
&\quad - 2 \sin (2 \sigma') (8 \sin (2 \phi) (\sin (2 \phi) (9 \cos (2 (\sigma - \sigma')) + \cos (2 \sigma)) \\
&\quad - 9 \cos (\sigma - 2 \sigma') + \cos (\sigma)) + 9 \cos (2 \sigma') (\cos (4 \phi) + 3) + 23 \cos (4 \phi) + 5) \bigg] \\
\end{split}
\end{eqnarray}
2 Answers
You never want to use eqnarray
.
Removing all unnecessary \left
and \right
instructions and fixing the number of &
, besides changing the incredibly wrong \text{$\phi'$}
into \phi'
, I get a decent result.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{split}
J_{cp} &= \frac{1}{36} \Bigl[\sqrt{3} \sin (2 \phi') (\sin (\sigma ) \cos (\sigma') \sin (2 \phi )-2 \sin (\sigma') \cos (2 \phi ))\\
&\quad-3 \sin (\sigma ) \sin (2 \phi ) \cos (2 \phi')\Bigr]\\
I_{1} &= \frac{1}{288}\Bigl[ 2 \sqrt{3} \sin (\phi) \sin (2 \phi') ( 8 \sin (2 \sigma) \cos (\sigma') \sin (\phi) \cos^2 (\phi) \\
&\quad + 2 \sin (\sigma') ( 8 \cos (\sigma) \cos^3 (\phi) - 4 \cos (2 \sigma) \sin (\phi) \cos^2 (\phi) + \sin (\phi) + 5 \sin (3 \phi) ) \\
&\quad - 4 \sin (\sigma) \cos (\sigma') (\cos (\phi) + 3 \cos (3 \phi)) ) \\
&\quad + 6 \cos (2 \phi') ( \sin (4 \phi) (\sin (\sigma - 2 \sigma') - 3 \sin (\sigma)) \\
&\quad + \cos (\sigma') (-4 \sin (2 \phi) \sin (\sigma - \sigma') + 2 \sin (\sigma) \cos (4 \phi) \cos (\sigma - \sigma') \\
&\quad - \sin (2 \sigma - \sigma') + 3 \sin (\sigma')) - 2 \sin (2 \sigma') \cos (2 \phi) ) \\
&\quad - 24 \sin (2 \sigma') \sin^3 (\phi) (2 \cos (\sigma) \cos (\phi) + \sin (\phi)) \\
&\quad - 12 \sin (2 \phi) ( \sin (\sigma) (\cos (2 \sigma') + 3) \cos (2 \phi) + \sin (\sigma') (\sin (2 \phi) \cos (2 \sigma - \sigma')) ) \\
&\quad + 2 \sin (\sigma) \sin (\sigma') \Bigr]\\
I_{2} &= \frac{1}{2304} \Bigl[ 96 \cos (2 \phi') ( \sin^3 (\phi) ( 8 \sin (\sigma) \cos^2 (2 \sigma') \cos (\phi) \\
&\quad + \sin (4 \sigma') (\sin (\phi) - 4 \cos (\sigma) \cos (\phi)) ) \\
&\quad - 2 \cos (2 \sigma') \sin^2 (2 \phi) \sin (2 (\sigma - \sigma')) ) \\
&\quad + 2 \sin (2 \sigma') \cos (4 \phi') ( -24 \sin^3 (\phi) (\cos (2 \sigma') \sin (\phi) - 4 \cos (\phi) \cos (\sigma - 2 \sigma')) \\
&\quad + 8 \sin (2 \phi) (\sin (2 \phi) (\cos (2 \sigma) - 3 \cos (2 (\sigma - \sigma'))) + \cos (\sigma)) \\
&\quad - 36 \cos (\sigma) \sin (4 \phi) + 36 \cos (2 \phi) + 23 \cos (4 \phi) + 5 ) \\
&\quad + 64 \sqrt{3} \cos (2 \sigma') \sin (\phi) \sin (2 \phi') (2 \cos (\sigma) \sin (2 \phi) - 3 \cos (2 \phi) - 1) \\
&\quad ( 2 \sin (\sigma) \cos (\sigma') \cos (\phi) + \sin (\sigma') (\sin (\phi) - 2 \cos (\sigma) \cos (\phi)) ) \\
&\quad + 32 \sqrt{3} \sin (2 \sigma') \sin (\phi) \sin (4 \phi') (-2 \cos (\sigma) \sin (2 \phi) + 3 \cos (2 \phi) + 1) \\
&\quad ( \cos (\sigma') \sin (\phi) - 2 \cos (\phi) \cos (\sigma - \sigma') ) \\
&\quad - 288 \sin^2 (\sigma') \cos (\sigma') (\sin (4 \phi) \sin (\sigma - \sigma') + \sin (\sigma') \cos (2 \phi)) \\
&\quad - 2 \sin (2 \sigma') (8 \sin (2 \phi) (\sin (2 \phi) (9 \cos (2 (\sigma - \sigma')) + \cos (2 \sigma)) \\
&\quad - 9 \cos (\sigma - 2 \sigma') + \cos (\sigma)) + 9 \cos (2 \sigma') (\cos (4 \phi) + 3) + 23 \cos (4 \phi) + 5) \Bigr] \\
\end{split}
\end{equation}
\end{document}
I suggest you switch to a single align*
environment and be more circumspect about where you place the line breaks. I'd also get rid of all (yes, all) \left
and \right
sizing directives and get rid of most parentheses to declutter the appearance of the equations. And, don't be shy to use not only round parentheses but also square brackets and curly braces to provide some hierarchical visual structure on the expressions.
\documentclass{article} % or some other suitable document class
\usepackage[T1]{fontenc}
\usepackage{amsmath} % for 'align*' environment
\begin{document}
\begin{align*}
J_{cp}
&= \frac{1}{36} \bigl\{\sqrt{3} \sin 2\phi' (\sin\sigma \cos\sigma' \sin 2\phi -2 \sin\sigma' \cos 2\phi ) \\
&\qquad-3 \sin\sigma \sin 2\phi \cos 2\phi' \bigr\} \\[\jot]
I_{1} &= \frac{1}{288}\bigl\{ 2 \sqrt{3} \sin\phi \sin 2\phi' \bigl[ 8 \sin 2\sigma \cos\sigma' \sin\phi \cos^2\phi\\
&\qquad\quad + 2 \sin\sigma' ( 8 \cos\sigma \cos^3\phi - 4 \cos 2\sigma \sin\phi \cos^2\phi + \sin\phi + 5 \sin 3\phi )\\
&\qquad\quad - 4 \sin\sigma \cos\sigma' (\cos\phi + 3 \cos 3\phi) \bigr]\\
&\qquad + 6 \cos 2\phi' \bigl[ \sin 4\phi (\sin(\sigma-2\sigma') - 3 \sin\sigma )
+ \cos\sigma' (-4 \sin 2\phi \sin (\sigma-\sigma')\\
&\qquad\quad + 2 \sin\sigma \cos 4\phi \cos (\sigma-\sigma')
- \sin (2 \sigma-\sigma') + 3 \sin\sigma') - 2 \sin 2\sigma' \cos 2\phi \bigr]\\
&\qquad - 24 \sin 2\sigma' \sin^3\phi (2 \cos\sigma \cos\phi + \sin\phi)\\
&\qquad - 12 \sin 2\phi \bigl[ \sin\sigma (\cos 2\sigma' + 3) \cos 2\phi \\
&\qquad\quad + \sin\sigma' (\sin 2\phi \cos (2 \sigma-\sigma')) \bigr]
+ 2 \sin\sigma \sin\sigma' \bigr\} \\[\jot]
I_{2} &= \frac{1}{2304} \bigl\{ 96 \cos 2\phi'
\bigl[ \sin^3\phi ( 8 \sin\sigma \cos^2 2\sigma' \cos\phi + \sin 4\sigma' (\sin\phi - 4 \cos\sigma \cos\phi) )\\
&\qquad\quad - 2 \cos 2\sigma' \sin^2 2\phi \sin (2 (\sigma-\sigma')) \bigr]\\
&\qquad + 2 \sin 2\sigma' \cos 4\phi'
\bigl[ -24 \sin^3\phi (\cos 2\sigma' \sin\phi - 4 \cos\phi \cos(\sigma-2\sigma'))\\
&\qquad\quad +8 \sin 2\phi \bigl( \sin 2\phi [\cos 2\sigma - 3\cos(2 (\sigma-\sigma'))] + \cos\sigma \bigr) \\
&\qquad\quad -36 \cos\sigma \sin 4\phi + 36 \cos 2\phi + 23 \cos 4\phi + 5 \bigr] \\
&\qquad + 64 \sqrt{3} \cos 2\sigma' \sin\phi \sin 2\phi' (2 \cos\sigma \sin 2\phi - 3\cos 2\phi - 1) \\
&\qquad\quad \times 4\sigma' \bigl( 2\sin\sigma \cos\sigma' \cos\phi + \sin\sigma' (\sin\phi-2\cos\sigma \cos\phi) \bigr) \\
&\qquad + 32 \sqrt{3} \sin 2\sigma' \sin\phi \sin 4\phi' (-2 \cos\sigma \sin 2\phi + 3 \cos 2\phi + 1) \\
&\qquad\quad \times 4\sigma' \bigl( \cos\sigma' \sin\phi - 2 \cos\phi \cos (\sigma-\sigma') \bigr) \\
&\qquad - 288 \sin^2\sigma' \cos\sigma' (\sin 4\phi \sin (\sigma-\sigma') + \sin\sigma' \cos 2\phi ) \\
&\qquad - 2 \sin 2\sigma' \bigl[8 \sin 2\phi (\sin 2\phi (9 \cos (2 (\sigma-\sigma')) + \cos 2\sigma) \\
&\qquad\quad - 9 \cos(\sigma-2\sigma') + \cos\sigma ) + 9 \cos 2\sigma' (\cos 4\phi + 3) + 23 \cos 4\phi + 5 \bigr] \bigr\}
\end{align*}
\end{document}
\begin{document}
etc. Also, please explain the problem more clearly. Do you get an error, or wrong output, how do you want the output to be? You say you want to use split but there is no split in the code, what exactly are you trying to do?eqnarray
but if you are usingamsmath
and have an error with itssplit
environment you should show a small document using that, and show the error that you get,