I am using
\documentclass[preprint,showpacs,preprintnumbers,amsmath,amssymb]{revtex4}
\usepackage{amsmath}
\newcommand\numberthis{\addtocounter{equation}{1}\tag{\theequation}}
I'd like to know how to incorporate either of the following two forms (TraditionalForm and StandardForm) produced by Mathematica into my document in some readable form (one in which material is not lost--like overflowing lines)
Mathematica provides me (using TradtionalForm) the code
\frac{1}{250} \left(50 (2 k+7)+\sqrt{50-10 \sqrt{5}} \left(-3 \sin \left(\frac{2}{5} \pi
(1-2 k)\right)+2 \sin \left(\frac{4 \pi k}{5}\right)-2 \sin \left(\frac{6 \pi
k}{5}\right)-3 \sin \left(\frac{1}{5} (\pi -6 \pi k)\right)+2 \sin \left(\frac{1}{5}
(\pi -4 \pi k)\right)-3 \left(\sin \left(\frac{2}{5} (3 \pi k+\pi )\right)+\sin
\left(\frac{1}{5} (4 \pi k+\pi )\right)\right)+2 \sin \left(\frac{1}{5} (6 \pi k+\pi
)\right)\right)+\sqrt{10 \left(5+\sqrt{5}\right)} \left(-2 \sin \left(\frac{2}{5} \pi
(1-4 k)\right)-2 \sin \left(\frac{2 \pi k}{5}\right)+2 \sin \left(\frac{8 \pi
k}{5}\right)-2 \sin \left(\frac{2}{5} \pi (k+1)\right)-3 \left(\sin \left(\frac{1}{5}
(\pi -2 \pi k)\right)-\sin \left(\frac{2}{5} (4 \pi k+\pi )\right)+\sin
\left(\frac{1}{5} (8 \pi k+\pi )\right)\right)+3 \cos \left(\frac{1}{10} (4 \pi
k+\pi )\right)\right)\right)
or (using StandardForm)
\frac{1}{250} \left(50 (7+2 k)+\sqrt{50-10 \sqrt{5}} \left(-3 \text{Sin}\left[\frac{2}{5}
(1-2 k) \pi \right]+2 \text{Sin}\left[\frac{4 k \pi }{5}\right]-2
\text{Sin}\left[\frac{6 k \pi }{5}\right]-3 \text{Sin}\left[\frac{1}{5} (\pi -6 k \pi
)\right]+2 \text{Sin}\left[\frac{1}{5} (\pi -4 k \pi )\right]-3
\left(\text{Sin}\left[\frac{2}{5} (\pi +3 k \pi )\right]+\text{Sin}\left[\frac{1}{5}
(\pi +4 k \pi )\right]\right)+2 \text{Sin}\left[\frac{1}{5} (\pi +6 k \pi
)\right]\right)+\sqrt{10 \left(5+\sqrt{5}\right)} \left(3 \text{Cos}\left[\frac{1}{10}
(\pi +4 k \pi )\right]-2 \text{Sin}\left[\frac{2}{5} (1-4 k) \pi \right]-2
\text{Sin}\left[\frac{2 k \pi }{5}\right]+2 \text{Sin}\left[\frac{8 k \pi
}{5}\right]-2 \text{Sin}\left[\frac{2}{5} (1+k) \pi \right]-3
\left(\text{Sin}\left[\frac{1}{5} (\pi -2 k \pi )\right]-\text{Sin}\left[\frac{2}{5}
(\pi +4 k \pi )\right]+\text{Sin}\left[\frac{1}{5} (\pi +8 k \pi
)\right]\right)\right)\right)