
\begin{itemize}
\item $\begin{aligned}[t]
\cos\left(2\times\dfrac{\pi}{8}\right)=2\cos^2\left(\dfrac{\pi}{8}\right)-1 & \iff \cos\dfrac{\pi}{4} = 2\cos^2\left(\dfrac{\pi}{8}\right)-1\\
& \iff \dfrac{\sqrt2}{2}=2\cos^2\left(\dfrac{\pi}{8}\right)-1\\
& \iff \cos^2\dfrac{\pi}{8}=\dfrac{\sqrt2+2}{4}\\
& \iff \boxed{\cos\dfrac{\pi}{8}=\dfrac{\sqrt{\sqrt2+2}}{2}}
\end{aligned}$

\item $\begin{aligned}[t]
\cos\left(2\times\dfrac{\pi}{8}\right)=1-2\sin^2\left(\dfrac{\pi}{8}\right) & \iff \dfrac{\sqrt2}{2} = 1-2\sin^2\left(\dfrac{\pi}{8}\right)\\
& \iff \sin^2\dfrac{\pi}{8} = \dfrac{2-\sqrt2}{4}\\
& \iff \boxed{\sin\dfrac{\pi}{8} = \dfrac{\sqrt{2-\sqrt2}}{2}}
\end{aligned}$
\end{itemize}

\begin{rem}
On prend ici la valeur positive du sinus et du cosinus car $0 < \dfrac{\pi}{8} < \dfrac{\pi}{2}$.
\end{rem}
