<s>[INST] Tricky trig question from GRE 
Please evaluate 
  \begin{align}
  1-\sin^2\left(\arccos \frac{\pi}{12}\right)
\end{align}

What I've tried so far is to use Pythagorean identity and I got
\begin{align}
  \cos^2\left(\arccos \frac{\pi}{12}\right)
\end{align}
If \begin{align}
 \arccos \frac{\pi}{12}=y
\end{align}
then 
\begin{align}
 \cos y =\frac{\pi}{12}
\end{align}
and here I can't continue because the answers are in such form:


*

*(A) $\sqrt{\frac{1-\cos\frac\pi{24}}{2}}$

*(B) $\sqrt{\frac{1-\cos\frac\pi{6}}{2}}$

*(C) $\sqrt{\frac{1+\cos\frac\pi{24}}{2}}$

*(D) $\frac\pi6$


and one more is missing, but that's pdf's issue
 [/INST] Once you get to this stage:
$\cos^2 (\arccos \frac{\pi}{12})$,
Use the facts that $\cos(\arccos x) = x$ and $\cos^2 y = (\cos y)^2$ to get to the final answer.
In this case, that's simply $(\frac{\pi}{12})^2 = \frac{\pi^2}{144}$
</s>