
Remarquons que $\dfrac{7\pi}{12}=\dfrac{\pi}{3}+\dfrac{\pi}{4}$. Ainsi,
\begin{itemize}
\begin{multicols}{2}
\item $\begin{aligned}[t]
\cos\dfrac{7\pi}{12} & = \cos\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)\\
& = \cos\dfrac{\pi}{3}\cos\dfrac{\pi}{4}-\sin\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\\
& = \dfrac{1}{2}\times\dfrac{\sqrt2}{2} - \dfrac{\sqrt3}{2}\times\dfrac{\sqrt2}{2}\\
\Aboxed{\cos\dfrac{7\pi}{12} & = \dfrac{\sqrt2-\sqrt6}{4}}
\end{aligned}$

\item $\begin{aligned}[t]
\sin\dfrac{7\pi}{12} & = \sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)\\
& = \sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}+\cos\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\\
& = \dfrac{\sqrt3}{2}\times\dfrac{\sqrt2}{2} + \dfrac{1}{2}\times\dfrac{\sqrt2}{2}\\
\Aboxed{\sin\dfrac{7\pi}{12} & = \dfrac{\sqrt2+\sqrt6}{4}}
\end{aligned}$
\end{multicols}
\end{itemize}
