
\begin{enumerate}
\item On a: $\begin{aligned}[t]
\dfrac{1-\e^{\i\frac{\pi}{3}}}{1-\e^{\i\frac{\pi}{4}}} & = \dfrac{\e^{\i\frac{\pi}{6}}\left(\e^{-\i\frac{\pi}{6}} - \e^{\i\frac{\pi}{6}} \right)}{\e^{\i\frac{\pi}{8}}\left(\e^{-\i\frac{\pi}{8}} - \e^{\i\frac{\pi}{8}} \right)}\\
& = \dfrac{\e^{\i\frac{\pi}{6}}}{\e^{\i\frac{\pi}{8}}} \times \dfrac{ 2\i\times\frac{\e^{-\i\frac{\pi}{6}} - \e^{\i\frac{\pi}{6}}}{2\i}}{ 2\i\times\frac{\e^{-\i\frac{\pi}{8}} - \e^{\i\frac{\pi}{8}}}{2\i}}\\
 & = \e^{\i\frac{\pi}{24}}\times\frac{\sin\frac{\pi}{6}}{\sin\frac{\pi}{8}}\\
 & = \e^{\i\frac{\pi}{24}}\times\frac{\frac{1}{2}}{\frac{\sqrt{2-\sqrt2}}{2}}\\
\Aboxed{ \dfrac{1-\e^{\i\frac{\pi}{3}}}{1-\e^{\i\frac{\pi}{4}}} & = \dfrac{1}{\sqrt{2-\sqrt2}} \e^{\i\frac{\pi}{24}}}
\end{aligned}$

\item On a :
\begin{align*}
\dfrac{1-\e^{\i\frac{\pi}{3}}}{1-\e^{\i\frac{\pi}{4}}} & = \dfrac{1-\frac{1}{2}-\frac{\sqrt3}{2}\i}{1-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\i}\\
& = \dfrac{\left(\frac{1}{2}-\frac{\sqrt3}{2}\i\right) \left(\frac{2-\sqrt2}{2}+\frac{\sqrt2}{2}\i \right)}{\left( \frac{2-\sqrt2}{2} \right)^2 + \left(\frac{\sqrt2}{2}\right)^2}\\
& \text{(on a multiplié par le conjugué du dénominateur)}\\
& = \dfrac{\frac{2-\sqrt2}{4}+\frac{\sqrt6}{4} + \i\left(\frac{\sqrt2}{4} - \frac{2\sqrt3-\sqrt6}{4} \right)}{\frac{4-4\sqrt2+2+2}{4}}\\
& = \left( \frac{2-\sqrt2+\sqrt6}{4} + \frac{\sqrt2-2\sqrt3+\sqrt6}{4}\i \right) \times \frac{4}{8-4\sqrt2}\\
& = \dfrac{2-\sqrt2+\sqrt6}{8-4\sqrt2} + \dfrac{\sqrt2-2\sqrt3+\sqrt6}{8-4\sqrt2}\i\textcolor{red}{\times\dfrac{8+4\sqrt2}{8+4\sqrt2}}\\
\Aboxed{\dfrac{1-\e^{\i\frac{\pi}{3}}}{1-\e^{\i\frac{\pi}{4}}}& = \dfrac{1+\sqrt3+\sqrt6}{4} + \dfrac{1+\sqrt2-\sqrt3}{4}\i}
\end{align*}

\item Des questions 1 et 2, on déduit l'égalité suivante:
\[
\dfrac{1+\sqrt3+\sqrt6}{4} + \dfrac{1+\sqrt2-\sqrt3}{4}\i = \dfrac{1}{\sqrt{2-\sqrt2}} \e^{\i\frac{\pi}{24}}
\]
soit:
\[
 \e^{\i\frac{\pi}{24}} = \sqrt{2+\sqrt2}\left(\dfrac{1+\sqrt3+\sqrt6}{4} + \dfrac{1+\sqrt2-\sqrt3}{4}\i\right)
\]
D'où:
\begin{align*}
\cos\frac{\pi}{24} & =\sqrt{2-\sqrt2}\times\dfrac{1+\sqrt3+\sqrt6}{4}\\
& =\frac{1}{4}\sqrt{(2-\sqrt2)\big(1+\sqrt3+\sqrt6\big)^2}\\
& = \frac{1}{4}\sqrt{8+2\sqrt2+2\sqrt6}.
\end{align*}
et:
\begin{align*}
\sin\dfrac{\pi}{24} & = \sqrt{2-\sqrt2}\times\dfrac{1+\sqrt2+\sqrt3}{4}\\
& =\frac{1}{4}\sqrt{(2-\sqrt2)\big(1+\sqrt2+\sqrt3\big)^2}\\
& = \frac{1}{4}\sqrt{8-2\sqrt2-2\sqrt6}.
\end{align*}
\end{enumerate}
