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?

Considering that the answer is apparently
$$ \frac{7 \left(489857+280140 \sqrt{3}\right)}{8388608}, $$
I doubt there's an easy way.
You know $\sin{\frac{\pi}{6}}=\frac{1}{2}$, $\cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}$, so using the half-angle formulae
$$ 2\sin^2{\frac{x}{2}} = 1-\cos{x},\\
2\cos^2{\frac{x}{2}} = 1+\cos{x} $$
will get us to $\sin{\frac{\pi}{24}}$ and $\cos{\frac{\pi}{24}}$, when applied twice.
In particular,
$$ 4\sin^4{\frac{x}{4}} = \left(1-\cos{\frac{x}{2}}\right)^2 = 1+\cos^2{\frac{x}{2}}-2\cos{\frac{x}{2}} \\
4\cos^4{\frac{x}{4}} = \left(1+\cos{\frac{x}{2}}\right)^2 = 1+\cos^2{\frac{x}{2}}+2\cos{\frac{x}{2}} $$
(At this point we see we have the nice identity
$$4(\sin^4{\theta}+\cos^4{\theta}) = 2(1+\cos^2{2\theta}) = 3+\cos{4\theta},$$
although sadly this is not of much help to us here.)
Therefore we need to raise everything to the power $6$, then divide by $4^6=2^{12}$. Set $a=\cos{(x/2)}$, then we have
$$ 4^6 ( \sin^{24}{\frac{x}{4}} + \cos^{24}{\frac{x}{4}} ) = (1-a)^{12}+(1+a)^{12} $$
At this point we break out the binomial theorem, and notice that quite a lot cancels: all the odd terms, in fact. Then
$$ \sin^{24}{\frac{x}{4}} + \cos^{24}{\frac{x}{4}} = \frac{1}{2^5} \left( 1 + a^6 + \binom{12}{2}(a^2 + a^{10}) + \binom{12}{4}(a^4+a^8) + \binom{12}{6}a^6 \right) = \frac{1}{2^{11}}(1+a^{12}+66(a^2+a^{10})+495(a^4+a^8)+924a^6) $$
Right, now the good news is that we don't need to know what $a$ is, just $a^2$. Indeed, the formula above gives
$$ \cos^2{\frac{\pi}{12}} = \frac{1}{2}\left(1+\cos{\frac{\pi}{6}}\right) = \frac{2+\sqrt{3}}{4}. $$
Therefore you just have to find $a^n$ for even $n$ up to $12$.
