\begin{exercice*}
  Vérifier que la fonction $f$ est $T$-périodique.
  \begin{enumerate}
  \item \begin{tabular}{@{}p{4cm}}$f:x\mapsto \sin \left(10\pi x\right)$\end{tabular} \quad  $T=0,2$
  \item \begin{tabular}{@{}p{4cm}}$f:x\mapsto \cos\left(4x+\dfrac{\pi}{3}\right)$\end{tabular} \quad $T=\dfrac{\pi }{2}$
  \item \begin{tabular}{@{}p{4cm}}$f:x\mapsto \sin\left(\dfrac{10x-1}{3}\right)$\end{tabular} \quad    $T=\dfrac{3 \pi }{5}$
  \item \begin{tabular}{@{}p{4cm}}$f:x\mapsto \dfrac{2}{5}\cos(3\pi x)$\end{tabular} \quad $T=\dfrac{2}{3}$
  \end{enumerate}
    \begin{corrige}
    \begin{enumerate}
  \item $f(x+0,2)$\par
  $=\sin (10\pi (x+0,2))$\par
  $=\sin(10\pi x+2\pi)=f(x)$
  \item $f(x+\frac{\pi}{2})$\par
  $= \cos(4(x+\frac{\pi}{2})+\frac{\pi}{3})$\par
  $=\cos(4x+2\pi+\frac{\pi}{3})=f(x)$
  \item $f(x+\frac{3\pi}{5})$\par
  $= \sin\left(\frac{10(x+\frac{3\pi}{5})-1}{3}\right)$\par
  $=\sin(\frac{10x+6\pi-1}{3})$\par
  $=\sin(\frac{10x-1}{3}+2\pi)=f(x)$
  \item $f(x+\frac{2}{3})$\par
  $=\frac{2}{5}\cos(3\pi (x+\frac{2}{3}))$\par
  $=\frac{2}{5}\cos(3\pi x+2\pi)=f(x)$
  \end{enumerate}
\end{corrige}
\end{exercice*}

