<s>[INST] Is there an error in this problem? $\sin^4(\frac{23\pi}{12})-\cos^4(\frac{13\pi}{12})$ 
$\sin^4(\frac{23\pi}{12})-\cos^4(\frac{13\pi}{12})$

If there were written $\frac{23\pi}{12}$ instead of $\frac{13\pi}{12}$ it could be solved as $$\begin{align}[\sin^2(\frac{23\pi}{12})+\cos^2(\frac{23\pi}{12})][\sin^2(\frac{23\pi}{12})-\cos^2(\frac{23\pi}{12})]=&\\ -[\cos^2(\frac{23\pi}{12})-\sin^2(\frac{23\pi}{12})]=& \\ -\cos(\frac{23\pi}{6})=&\\ -\cos(\frac{11\pi}{6})=&\\ -\frac{\sqrt{3}}{2}\end{align}$$ Otherwise, there seems to be no way towards solving it.
 [/INST] $\frac{23}{12} \pi = 2\pi - \frac{\pi}{12}$, and $\frac{13}{12} \pi = \pi + \frac{\pi}{12}$. So you have:
$$\sin^4 \frac{\pi}{12} - \cos^4 \frac{\pi}{12}$$
which surprisingly gives the same answer you reached.
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