Trigonometry to the 24th power How can I find the value of
$$\sin^{24}\frac{\pi}{24} + \cos^{24}\frac{\pi}{24}$$
Specifically, is there some easy method that I am overlooking?

I would use complex exponentials. We have:
\begin{align}
\sin^{24}\frac{\pi}{24}+\cos^{24}\frac{\pi}{24}&=
  \left(\frac{e^{i\frac{\pi}{24}}-e^{-i\frac{\pi}{24}}}{2i}\right)^{24}
  +\left(\frac{e^{i\frac{\pi}{24}}+e^{-i\frac{\pi}{24}}}{2}\right)^{24}\\
 &=\frac{1}{2^{24}}\left(\sum_{k=0}^{24}\binom{24}{k} (-1)^k e^{i \frac{24-2k}{24}\pi}\right)
  +\frac{1}{2^{24}}\left(\sum_{k=0}^{24}\binom{24}{k} e^{i \frac{24-2k}{24}\pi}\right)\\
 &= \frac{1}{2^{23}}\sum_{l=0}^{12} \binom{24}{2l} e^{i \frac{12 - 2l}{12}\pi}
\end{align}
after canceling every other term in the two sums. We can rewrite this last expression in terms
of trig functions again, as
$$
\frac{1}{2^{22}} \left(\sum_{m=0}^{5} \binom{24}{2m} \cos \left(\pi-\frac{m\pi}{6}\right) +
 \frac{1}{2} \binom{24}{12}\right)
$$
(here we're folding the sum in half and taking advantage of the fact that
$\binom{n}{k}=\binom{n}{n-k}$, which is why the middle term is anomalous).
Now, as $\cos x=-\cos(\pi - x)$, this collapses to a reasonable number of terms:
$$
\frac{1}{2^{22}}\left[\frac{1}{2}\binom{24}{12} - 1 
+ \left(\binom{24}{10} - \binom{24}{2}\right) \cos \frac{\pi}{6}
+ \left(\binom{24}{8} - \binom{24}{4}\right) \cos \frac{\pi}{3}
\right] 
$$
which simplifies into
$$
\frac{1}{2^{22}} \left(\frac{3428999}{2} + 1960980\frac{\sqrt{3}}{2}\right)
=\frac{3428999 + 1960980 \sqrt{3}}{2^{23}}
$$
