\begin{enumerate}
\item $\begin{aligned}[t]
\cos\dfrac{7\pi}{12} & = \cos\left(\dfrac{\pi}{4}+\dfrac{\pi}{3}\right)\\
& = \cos\dfrac{\pi}{4}\cos\dfrac{\pi}{3}-\sin\dfrac{\pi}{4}\sin\dfrac{\pi}{3}\\
& = \dfrac{\sqrt{2}}{2}\times\dfrac{1}{2}-\dfrac{\sqrt{2}}{2}\times\dfrac{\sqrt{3}}{2}\\
& = \dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{6}}{4}\\
\Aboxed{\cos\dfrac{7\pi}{12} & =\dfrac{\sqrt{2}-\sqrt{6}}{4}}
\end{aligned}$

\item\label{q2extrigo} $\begin{aligned}[t]
\sin\dfrac{\pi}{12} & = \sin\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)\\
& = \sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}-\sin\dfrac{\pi}{4}\cos\dfrac{\pi}{3}\\
& = \dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}\times\dfrac{1}{2}\\
\Aboxed{\sin\dfrac{\pi}{12} & = \dfrac{\sqrt{6}-\sqrt{2}}{4}}
\end{aligned}$

\item $\begin{aligned}[t]
\cos\dfrac{\pi}{4} & = \cos\left(2\times\dfrac{\pi}{8}\right)\\
\dfrac{\sqrt{2}}{2} & = 1-2\sin^2\dfrac{\pi}{8}\\
2\sin^2\dfrac{\pi}{8} & = 1-\dfrac{\sqrt{2}}{2}\\
\sin^2\dfrac{\pi}{8} & = \dfrac{2-\sqrt{2}}{4}\\
\Aboxed{\sin\dfrac{\pi}{8} & = \dfrac{\sqrt{2-\sqrt{2}}}{2}}\quad\text{car }\dfrac{\pi}{8}\in\intervFF{0}{\pi}\text{ donc }\sin\dfrac{\pi}{8}>0.
\end{aligned}$

On en déduit alors :

\medskip

$\begin{aligned}[t]
\cos^2\dfrac{\pi}{8} & = 1-\sin^2\dfrac{\pi}{8}\\
\cos^2\dfrac{\pi}{8} & = 1-\dfrac{2-\sqrt{2}}{4}\\
\cos^2\dfrac{\pi}{8} & = \dfrac{2+\sqrt{2}}{4}\\
\Aboxed{\cos\dfrac{\pi}{8} & = \dfrac{\sqrt{2+\sqrt{2}}}{2}}\quad\text{car }\cos\dfrac{\pi}{8}>0.
\end{aligned}$


\item Nous allons ici utiliser le fait que :
\[
\cos^2\dfrac{\pi}{12}+\sin^2\dfrac{\pi}{12}=1\qquad\text{donc}\qquad\cos\dfrac{\pi}{12}=\sqrt{1-\sin^2\dfrac{\pi}{12}}
\]
et que :
\[
\cos\left(2\times\dfrac{\pi}{24}\right)=1-2\sin^2\dfrac{\pi}{24}.
\]
On a :
\begin{align*}
\cos\dfrac{\pi}{12} & = \cos\left(2\times\dfrac{\pi}{24}\right)\\[8pt]
\sqrt{1-\sin^2\dfrac{\pi}{12}} & = 1-2\sin^2\dfrac{\pi}{24}\\[8pt]
\sqrt{1-\left(\dfrac{\sqrt{6}-\sqrt{2}}{4}\right)^2} & = 1-2\sin^2\dfrac{\pi}{24}\qquad\text{d'après la question \ref{q2extrigo}}\\[8pt]
\sqrt{1-\dfrac{8-4\sqrt{3}}{16}} & = 1-2\sin^2\dfrac{\pi}{24}\\
\sqrt{\dfrac{8+4\sqrt{3}}{16}} & = 1-2\sin^2\dfrac{\pi}{24}\\
\dfrac{\sqrt{2+\sqrt{3}}}{2} & = 1-2\sin^2\dfrac{\pi}{24}\\
\sin^2\dfrac{\pi}{24} & = \dfrac{1}{2}\left(1-\dfrac{\sqrt{2+\sqrt{3}}}{2}\right)\\
\sin^2\dfrac{\pi}{24} & = \dfrac{2-\sqrt{2+\sqrt{3}}}{4}\\
\Aboxed{\sin\dfrac{\pi}{24} & = \dfrac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}}\qquad\text{car }\sin\dfrac{\pi}{24}>0.
\end{align*}
\end{enumerate}