TITLE: Trigonometry to the 24th power
QUESTION [30 upvotes]: 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?

REPLY [8 votes]: A note:   A general formula that is applicable is 
\begin{align}
\sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right].
\end{align}
When $n=6$ this reduces to
\begin{align}
\sin^{24}\left(\frac{\pi}{24}\right) + \cos^{24}\left( \frac{\pi}{24}\right) = \frac{1}{4^{11}} \left[ \sum_{r=0}^{5} \binom{24}{2r} \cos\left(1 - \frac{r}{6} \right) \pi \, + \frac{1}{2} \binom{24}{12} \right].
\end{align}
The remaining details have been given by Micah's solution.