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?
 A: 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.
A: 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 $12$.
A: 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
$$
\frac{1}{2^{22}} \left(\frac{3428999}{2} + 1960980\frac{\sqrt{3}}{2}\right)
=\frac{3428999 + 1960980 \sqrt{3}}{2^{23}}
$$
A: Let:
$$ a_n = \sin^{2n}\frac{\pi}{24}+\cos^{2n}\frac{\pi}{24}.\tag{1}$$
Then trivially $a_0=2,a_1=1$ and:
$$ a_n - a_1 a_{n-1} = -\left(\sin^2\frac{\pi}{24}\cos^2\frac{\pi}{24}\right) a_{n-2}\tag{2} $$
so that $a_{12}$ can be computed in a few steps through the recurrence:
$$ a_n = a_{n-1}-\frac{2-\sqrt{3}}{16} a_{n-2}.\tag{3}$$
