\begin{questions}
\item
\begin{questions}
\item $\cos\frac{5\pi}{3} = \cos\left( \frac{6\pi}{3}-\frac{\pi}{3}\right) = \cos\left(2\pi - \frac{\pi}{3}\right) = \cos\frac{\pi}{3}=\frac{1}{2}$.

\item $\sin\frac{7\pi}{6} = \sin\left( \frac{6\pi}{6} + \frac{\pi}{6} \right) = \sin\left( \pi + \frac{\pi}{6} \right) = -\sin\frac{\pi}{6} = -\frac{1}{2}$.

\item $\begin{aligned}[t]
\sin\frac{113\pi}{3} & = \sin\left(\frac{114\pi}{3}-\frac{\pi}{3} \right) = \sin\left( 38\pi-\frac{\pi}{3} \right)\\
& = \sin\left(19 \times 2\pi - \frac{\pi}{3}\right) = -\sin\frac{\pi}{3} = -\frac{\sqrt3}{2}.
\end{aligned}$

\item $\begin{aligned}[t]
\cos\left(-\frac{37\pi}{4}\right) & = \cos\left(-\frac{40\pi}{4}+\frac{3\pi}{4} \right) = \cos\left( -10\pi + \frac{3\pi}{4}\right)\\
& = \cos\left( \frac{3\pi}{4}\right) = -\frac{\sqrt2}{2}.
\end{aligned}$
\end{questions}

\item $\begin{aligned}[t]
S & = {\color{DR}\sin\frac{\pi}{8} - \sin\frac{7\pi}{8}} + {\color{DB}\sin\frac{5\pi}{8} - \sin\frac{3\pi}{8}}\\
& = {\color{DR}\left[ \sin\frac{\pi}{8} - \sin\left(\frac{8\pi}{8}-\frac{\pi}{8}\right) \right]} + {\color{DB}\left[ \sin\left(\frac{8\pi}{8}-\frac{3\pi}{8}\right) - \sin\frac{3\pi}{8} \right]}\\
& = \left[ \sin\frac{\pi}{8} - \sin\left(\pi-\frac{\pi}{8}\right) \right] + \left[ \sin\left(\pi-\frac{3\pi}{8}\right) - \sin\frac{3\pi}{8} \right]\\
& = \left[ \sin\frac{\pi}{8} - \sin\frac{\pi}{8} \right] + \left[ \sin\frac{3\pi}{8} - \sin\frac{3\pi}{8} \right]\\
& = 0.
\end{aligned}$

\begin{Remarque}nous avons utilisé ici la propriété suivante:
\[
\forall x\in\R,\ \sin(\pi-x) = \sin(x).
\]
\end{Remarque}

\item Sachant que $\pi < x < \frac{3\pi}{2}$ et que $\sin x = -\frac{\sqrt2}{2}$, déterminons la valeur de $\cos x$ à l'aide de l'égalité fondamentale de trigonométrie:
\begin{align*}
\cos^2x + \sin^2x = 1 & \iff \cos^2x + \left(-\frac{\sqrt2}{2}\right)^2 = 1\\
& \iff \cos^2x = 1-\frac{1}{2}=\frac{1}{2}
\intertext{Or $\pi < x < \frac{3\pi}{2}$ donc $\cos x < 0$.}
& \iff \cos x = -\sqrt{\frac{1}{2}}=-\frac{\sqrt2}{2}.
\end{align*}
\end{questions}
