Is there a smart way to get a consistent accuracy with fp package?

Question

Is there a smart way to get a consistent accuracy with fp package? I don't want to use round for sure!

\documentclass[preview,border=12pt]{standalone}

\usepackage[nomessages]{fp}
\FPeval\Sixty{trunc(cos(pi/3):12)}
\FPeval\ThreeHundred{trunc(cos(5*pi/3):12)}
\usepackage{amsmath}

\begin{document}
$\!
\begin{aligned}
    \cos 60^\circ &= \Sixty\\
    \cos 300^\circ &= \ThreeHundred
\end{aligned}
$
\end{document}

Answer 1

In your case the calculation of cos(5*pi/3) is done with higher precision, but the choice of trunc is worse than round:

\documentclass[preview,border=12pt]{standalone}

\usepackage[nomessages]{fp}
\FPeval\Sixty{trunc(cos(pi/3):12)}
\FPeval\ThreeHundredTrunc{trunc(cos(5*pi/3):12)}
\FPeval\ThreeHundredRound{round(cos(5*pi/3):12)}
\FPeval\Tmp{cos(5*pi/3)}
\FPeval\TruncDelta{\Tmp-\ThreeHundredTrunc}
\FPeval\RoundDelta{\Tmp-\ThreeHundredRound}
\usepackage{amsmath}

\begin{document}
$\!
\begin{aligned}
    \cos 60^\circ &= \Sixty\\
    \cos 300^\circ &= \ThreeHundredTrunc~\text{(trunc)}
    & \Delta &= +\TruncDelta\\
    \cos 300^\circ &= \ThreeHundredRound~\text{(round)}
    & \Delta &= \RoundDelta
\end{aligned}
$
\end{document}

In this case the final error of the calculation using trunc is 999999 times greater than the variant using round.

Float arithmetic with decimal (or binary) numbers has its limitations, because already rational numbers cannot be expressed with a finite number of digits in the decimal (or binary) system. The typical workaround is using a higher precision in the internal representation and rounding the result. Using truncating makes it worse.

Answer 2

Maybe LaTeX3 is an option. Related to the documentation of l3fp:

\documentclass[11pt]{article}
\usepackage{expl3,l3fp}
\usepackage{xparse, siunitx}
\ExplSyntaxOn
\NewDocumentCommand { \calcnum } { m }
  { \num [  round-mode=places , round-precision=2 ] { \fp_to_decimal:n {#1} } }
\ExplSyntaxOff


\begin{document}
\[\cos(60)=\calcnum { cos( pi/3 ) }\]

\[\cos(300)=\calcnum { cos( 5*pi/3 ) }\]
\end{document}