<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] You have that $\sin(\arccos(x))=\sqrt{1-x^2}$, therefore
$$\sin^2\left(\arccos\left(\frac{\pi}{12}\right)\right)=1-\frac{\pi^2}{144}\implies 1-\sin^2\left(\arccos\left(\frac{\pi}{12}\right)\right)=\frac{\pi^2}{144}$$
</s>