\begin{enumerate}
\item 
\begin{itemize}
\item Montrons d'abord que $f$ est $\pi$-périodique.
\begin{align*}
f(x+\pi) & = \big[\sin(x+\pi)\big]^3\cos\big[3(x+\pi)\big]\\
& = \big[-\sin x\big]^3\cos(3x+3\pi)\\
& = -\sin^3x\cos(3x+\pi)\\
& = -\sin^3x\big[-\cos(3x)\big]\\
& = \sin^3x\cos(3x)\\
\Aboxed{f(x+\pi) & = f(x)}
\end{align*}

La fonction $f$ est donc $\pi$-périodique ; on peut donc restreindre l'intervalle d'étude de $f$ à un intervalle d'amplitude $\pi$, par exemple \intervFF{-\dfrac{\pi}{2}}{\dfrac{\pi}{2}}.

\item Montrons que $f$ est impaire. Le domaine de définition de $f$ est centré en 0. De plus,
\begin{align*}
f(-x) & = \big[\sin(-x)\big]^3\cos(-3x)\\
& = \big[-\sin x\big]^3\cos(3x)\\
& = -\sin^3x\cos(3x)\\
\Aboxed{f(-x) & = -f(x)}
\end{align*}

La fonction $f$ est donc impaire ; on peut donc réduire l'intervalle d'étude précédent à sa moitié, donc à \intervFF{0}{\dfrac{\pi}{2}}.
\end{itemize}

\item Remarquons que $2\times\dfrac{\pi}{8}=\dfrac{\pi}{4}$ donc :
\begin{align*}
& \cos\left(2\times\dfrac{\pi}{8}\right) = 2\cos^2\dfrac{\pi}{8}-1\\
\iff & \dfrac{\sqrt{2}}{2}+1=2\cos^2\dfrac{\pi}{8}\\
\iff & \cos^2\dfrac{\pi}{8}=\dfrac{1}{2}\left(\dfrac{\sqrt{2}}{2}+1\right)\\
\iff & \boxed{\cos\dfrac{\pi}{8}=\dfrac{\sqrt{2+\sqrt{2}}}{2}}\quad\text{car }\cos\dfrac{\pi}{8}>0.
\end{align*}

De même,
\begin{align*}
& \cos\left(2\times\dfrac{\pi}{8}\right) = 1-2\sin^2\dfrac{\pi}{8}\\
\iff & \dfrac{\sqrt{2}}{2}=1-2\sin^2\dfrac{\pi}{8}\\
\iff & \sin^2\dfrac{\pi}{8}=\dfrac{1}{2}\left(1-\dfrac{\sqrt{2}}{2}\right)\\
\iff & \boxed{\sin\dfrac{\pi}{8}=\dfrac{\sqrt{2-\sqrt{2}}}{2}}\quad\text{car }\sin\dfrac{\pi}{8}>0.
\end{align*}

\item\label{ex09trigoq3} $\big(\sqrt{2}-1\big)^2=3-2\sqrt{2}$ donc $\sqrt{\big(\sqrt{2}-1\big)^2}=\sqrt{3-2\sqrt{2}}$, soit $\sqrt{3-2\sqrt{2}}=\sqrt{2}-1$ (car tous ces nombres sont positifs).

\item $f$ est de la forme $uv$ avec $u(x)=\sin^3x$ et $v(x)=\cos(3x)$.

Donc $f'=u'v+uv'$ avec $u'(x)=3\cos x\sin^2x$ et $v'(x)=-3\sin(3x)$.

Ainsi,
\begin{align*}
f'(x) & = 3\cos x\sin^2x\cos(3x)-3\sin(3x)\sin^3x\\
& = 3\sin^2x\big(\cos x\cos(3x)-\sin(3x)\sin x\big)\\
\Aboxed{f'(x) & = 3\sin^2x\cos(4x)}\quad\text{en utilisant la formule }\cos(a+b)=\cos a\cos b-\sin a \sin b
\end{align*}

$\forall\; x\in\intervFF{0}{\dfrac{\pi}{2}},\ 3\sin^2x\geqslant0$. De plus,
\begin{align*}
\cos(4x)>0 & \iff -\dfrac{\pi}{2}+2k\pi < 4x < \dfrac{\pi}{2}+2k\pi\\
& \iff -\dfrac{\pi}{8}+k\dfrac{\pi}{2} < x < \dfrac{\pi}{8}+k\dfrac{\pi}{2}
\end{align*}
\newpage
Donc sur \intervFF{0}{\dfrac{\pi}{2}}, $\cos(4x)\geqslant0 \iff x\in\intervFF{0}{\dfrac{\pi}{8}}\cup\intervFF{\dfrac{3\pi}{8}}{\dfrac{\pi}{2}}$ d'où le tableau suivant :
\begin{center}
\begin{tikzpicture}
\tkzTabInit[color,colorC=violet!20,colorL=violet!20,colorV=violet!20]{$x$/0.8,$f'(x)$/0.5,$f(x)$/1.5}{0,$\frac{\pi}{8}$,$\frac{3\pi}{8}$,$\frac{\pi}{2}$}
\tkzTabLine{,+,z,-,z,+}
\tkzTabVar{-/,+/,-/,+/}
\end{tikzpicture}
\end{center}
$\begin{aligned}[t]
f\left(\frac{\pi}{8}\right) & = \left(\sin\frac{\pi}{8}\right)^3\cos\frac{3\pi}{8}\\
& = \left[\dfrac{2-\sqrt{2}}{4}\times\dfrac{\sqrt{2-\sqrt{2}}}{2}\right]\times\cos\left(\dfrac{\pi}{4}+\dfrac{\pi}{8}\right)\\
& = \dfrac{\big(2-\sqrt{2}\big)\sqrt{2-\sqrt{2}}}{8}\left[\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{8}-\sin\dfrac{\pi}{4}\sin\dfrac{\pi}{8}\right]\\
& = \dfrac{\big(2-\sqrt{2}\big)\sqrt{2-\sqrt{2}}}{8}\times\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{2+\sqrt{2}}}{2}-\dfrac{\sqrt{2-\sqrt{2}}}{2}\right)\\
& = \dfrac{\big(2-\sqrt{2}\big)\sqrt{2-\sqrt{2}}}{8}\times\dfrac{\sqrt{4+2\sqrt{2}}-\sqrt{4-2\sqrt{2}}}{4}\\
& = \dfrac{\big(2-\sqrt{2}\big)\sqrt{\big(2-\sqrt{2}\big)\big(4+2\sqrt{2}\big)}-\big(2-\sqrt{2}\big)\sqrt{\big(2-\sqrt{2}\big)\big(4-2\sqrt{2}\big)}}{32}\\
& = \dfrac{2-\sqrt{2}}{32}\times\left(\sqrt{8+4\sqrt{2}-4\sqrt{2}-4}-\sqrt{8-4\sqrt{2}-4\sqrt{2}+4}\right)\\
& = \dfrac{2-\sqrt{2}}{32}\times\left(2-\sqrt{12-8\sqrt{2}}\right)\\
& = \dfrac{2-\sqrt{2}}{32}\times\left(2-2\sqrt{3-2\sqrt{2}}\right)\\
& = \dfrac{\big(2-\sqrt{2}\big)\big(1-\sqrt{3-2\sqrt{2}}\big)}{16}\\
& = \dfrac{\big(2-\sqrt{2}\big)\big(1-(\sqrt{2}-1)\big)}{16}\quad\text{d'après la question \ref{ex09trigoq3}}\\
& = \dfrac{\big(2-\sqrt{2}\big)\big(2-\sqrt{2}\big)}{16}\\
& = \dfrac{6-4\sqrt{2}}{16}\\
\Aboxed{f\left(\frac{\pi}{8}\right) & = \dfrac{3-2\sqrt{2}}{8}}
\end{aligned}$

\newpage
\item La courbe représentative de $f$ est donnée ci-dessous :
\begin{center}
\begin{tikzpicture}[>=latex,yscale=5,xscale=0.08]
\draw[thick,->] (-100,0) -- (100,0);
\draw[thick,->] (0,-0.8) -- (0,0.5);
\draw[violet] plot[domain=-90:90,samples=100] 
(\x,{cos(3*\x)*sin(\x)*sin(\x)*sin(\x)});
\draw[thick] ({180/8},0.01) -- ({180/8},-0.01) node[below,outer ysep=5pt] {$\frac{\pi}{8}$};
\draw[thick] ({-180/8},-0.01) -- ({-180/8},0.01) node[above] {$-\frac{\pi}{8}$};
\draw[thick] ({3*180/8},-0.03) -- ({3*180/8},0.03) node[above] {$\dfrac{3\pi}{8}$};
\draw[thick] ({-3*180/8},0.03) -- ({-3*180/8},-0.03) node[below] {$-\dfrac{3\pi}{8}$};
\node[below left] at (0,0) {0};
%\draw[dashed] (45,0) -- (45,-0.25) -- (0,-0.25) node[left] {$-\frac{1}{4}$};
%\draw[thick] (1.25,1) -- (-1.25,1) node[above left] {1};
%\draw[<->,thick,violet] (-10,1) -- (10,1);
%\draw[<->,thick,violet] (35,-0.25) -- (60,-0.25);
%\draw[<->,thick,violet] (-60,-0.25) -- (-35,-0.25);
%\draw[very thick,violet,->] (-90,0) -- (-65,0);
%\draw[very thick,violet,->] (90,0) -- (65,0);
%\draw[thick] (30,-0.03) -- (30,0.03) node[above] {$\frac{\pi}{6}$};
%\draw[thick] (-30,-0.03) -- (-30,0.03) node[above] {$-\frac{\pi}{6}$};
\end{tikzpicture}
\end{center}
\end{enumerate}