
\begin{enumerate}
\item $0<\dfrac{\pi}{5}<\dfrac{\pi}{2}$ donc $\sin\dfrac{\pi}{5}>0$.

De plus, d'après le cours, pour tout réel $x$, $\cos^2+\sin^2x=1$ donc, en prenant $x=\dfrac{\pi}{5}$, cela donne :
\begin{align*}
\left(\cos\dfrac{\pi}{5}\right)^2+\left(\sin\frac{\pi}{5}\right)^2=1 & \iff
\left(\sin\dfrac{\pi}{5}\right)^2=1-\left(\cos\frac{\pi}{5}\right)^2\\
& \iff \left(\sin\dfrac{\pi}{5}\right)^2=1-\left(\frac{1+\sqrt5}{4}\right)^2\\
& \iff \left(\sin\dfrac{\pi}{5}\right)^2=1-\dfrac{1+2\sqrt5+5}{16}\\
& \iff \left(\sin\dfrac{\pi}{5}\right)^2=\dfrac{16-6-2\sqrt5}{16}\\
& \iff \left(\sin\dfrac{\pi}{5}\right)^2=\dfrac{5-\sqrt5}{8}\\
& \iff \boxed{\sin\dfrac{\pi}{5}=\sqrt{\dfrac{5-\sqrt5}{8}}}
\end{align*}

\item $\dfrac{\pi}{2}\leqslant x \leqslant \pi$ donc $\cos x\leqslant0$.

De plus,
\begin{align*}
\cos^x + \sin^2x = 1 & \iff \cos^2x = 1-\sin^2x\\
& \iff \cos^2x = 1-\left(\dfrac{4}{5}\right)^2\\
& \iff \cos^2x=1-\dfrac{16}{25}\\
& \iff \cos^2x = \dfrac{9}{25}\\
& \iff \boxed{\cos x = -\dfrac{3}{5}}
\end{align*}

\item $\big(\sqrt{2+\sqrt3}\big)^2=2+\sqrt3$.

De plus, $\begin{aligned}[t]
\left(\dfrac{\sqrt2+\sqrt6}{2}\right)^2
& = \dfrac{2+2\sqrt{12}+6}{4}\\
& = \dfrac{8+2\sqrt{4\times3}}{4}\\
& = \dfrac{8+4\sqrt3}{4}\\
& = \dfrac{4(2+\sqrt3)}{4}\\
& =2+\sqrt3.
\end{aligned}$

Ainsi, $\big(\sqrt{2+\sqrt3}\big)^2= \left(\dfrac{\sqrt2+\sqrt6}{2}\right)^2$, donc $\sqrt{2+\sqrt3} = \dfrac{\sqrt2+\sqrt6}{2}$ (car les deux nombres sont positifs).

\medskip

$0<\dfrac{7\pi}{12}<\pi$ donc $\sin\dfrac{7\pi}{12}>0$. Donc,
\begin{align*}
\sin\dfrac{7\pi}{12} & =\sqrt{1-\cos^2\dfrac{7\pi}{12}}\\
& = \sqrt{1-\left(\dfrac{\sqrt2-\sqrt6}{4}\right)^2}\\
& = \sqrt{1-\dfrac{2-2\sqrt{12}+6}{16}}\\
& = \sqrt{\dfrac{8+2\sqrt{12}}{16}}\\
& = \sqrt{\dfrac{8+2\sqrt{4\times3}}{16}}\\
&  = \sqrt{\dfrac{8+4\sqrt{3}}{16}}\\
& = \sqrt{\dfrac{2+\sqrt3}{4}}\\
& = \dfrac{\sqrt{2+\sqrt3}}{2}\\
& = \dfrac{\sqrt2+\sqrt6}{4}\quad\text{d'après la question précédente}.
\end{align*}
\end{enumerate}
