Finding angle between a given sky position and ecliptic pole Given a right ascension in degrees and declination in degrees (say, 88 degrees RA, -63 degrees DEC) I want to find the angle between that location and the ecliptic pole (90 degrees RA, -66 degrees DEC... are these the right values? What are the decimals following the 66? Most of my values that I'm working with are similar to the one I give, so which ecliptic pole, if any, does this correspond to?) How would I find this angle, I need a convenient formula (such as a dot product to find this angle). Thank you!
The angle between an ecliptic pole and another sky position is complementary to the ecliptic latitude of that position.
To find ecliptic (&lambda;, &beta;) coordinates, one method is to convert the equatorial (&alpha;, &delta;) to Cartesian form:
$$ \begin{align}
x_\mathrm{equ} &amp;= \cos \alpha \cos \delta \\
y_\mathrm{equ} &amp;= \sin \alpha \cos \delta \\
z_\mathrm{equ} &amp;= \sin \delta
\end{align}$$
then rotate by &epsilon; = 23.44&deg; on the x axis:
$$ \begin{align}
x_\mathrm{ecl} &amp;= x_\mathrm{equ} \\
y_\mathrm{ecl} &amp;= y_\mathrm{equ} \cos \varepsilon + z_\mathrm{equ} \sin \varepsilon \\
z_\mathrm{ecl} &amp;= z_\mathrm{equ} \cos \varepsilon - y_\mathrm{equ} \sin \varepsilon
\end{align} $$
and convert to spherical form:
$$ \begin{align}
\lambda &amp;= \mathtt{atan2}(y_\mathrm{ecl}, x_\mathrm{ecl}) \\
\beta &amp;= \arcsin z_\mathrm{ecl}
\end{align} $$

If you are not interested in the ecliptic longitude &lambda;, you can express the ecliptic latitude as
$$ \beta = \arcsin (\sin \delta \cos \varepsilon - \sin \alpha \cos \delta \sin \varepsilon) $$
Then the angle off the north ecliptic pole (&alpha;=18h0m, &delta;=+66.56&deg;) is 90&deg; - &beta;.

Working your example (&alpha;=5h52m, &delta;=-63&deg;),
I get (&lambda;=75.63&deg;, &beta;=-86.34&deg;),
and the angle off the south ecliptic pole (&alpha;=6h0m, &delta;=-66.56&deg;) is &beta; + 90&deg; = 3.66&deg;.