Source Latex
de la correction du devoir
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pdfauthor={Yoann Morel},
pdfsubject={Correction du devoir de mathématiques en 1S: trigonométrie et produit scalaire},
pdftitle={Corrigé du devoir de mathématiques: trigonométrie et produit scalaire},
pdfkeywords={Mathématiques, trigonométrie, cercle trigonométrique,
cosinus, sinus, valeurs remarquables, angles associés}
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\lfoot{Y. Morel - \href{https://xymaths.fr/Lycee/1S/Mathematiques-1S.php}{xymaths.fr - 1ère S}}
\rfoot{Correction du devoir de math\'ematiques - $1^{\text{ère}}S$ - \thepage/\pageref{LastPage}}
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\begin{document}
\ct{\bf\LARGE{Corrigé du devoir de math\'ematiques}}
\bgex
$\bgar[t]{ll}
A(x)
&=\sin\lp-x\rp-\cos\lp\dfrac{\pi}{2}-x\rp
+\cos\lp-x\rp
-\sin\lp\dfrac{\pi}{2}-x\rp \vspd\\
&=-\sin(x)-\sin(x)+\cos(x)-\cos(x) \vspd\\
&=-2\sin(x)
\enar$
\enex
\bgex
$\cos x =\dfrac{\sqrt{3}}{2}=\cos\dfrac{\pi}{6}
\iff
\la\bgar{ll}
x=\dfrac{\pi}{6}+k2\pi \\[0.3cm]
x=-\dfrac{\pi}{6}+k2\pi
\enar\right.
\quad,\quad k\in\Z
$.
Dans l'intervalle $]0;2\pi]$ les solutions sont donc,
$\mathcal{S}=\la \dfrac{\pi}{6};\dfrac{11\pi}{6}\ra$.
\enex
\bgex
\bgen
\item $\V{AB}\lp -14;26\rp$;\ $\V{AC}\lp 16;-5\rp$
\item $\|\V{AB}\|=AB=\sqrt{\lp-14\rp^2+26^2}\simeq 29,53$;\
$\|\V{AC}\|=AC=\sqrt{16^2+\lp-5\rp^2}=16,76$
\item $\V{AB}\cdot\V{AC}=-14\tm16+26\tm(-5)=-354$
\item On a aussi,
$\V{AB}\cdot\V{AC}
=AB\tm AC\tm\cos\lp\widehat{BAC}\rp
\simeq 19,53\tm16,76\tm\cos\lp\widehat{BAC}\rp
\simeq 494,92\cos\lp\widehat{BAC}\rp
$.
On en d\'eduit que
$\cos\lp\widehat{BAC}\rp\simeq \dfrac{-354}{494,92}$,
soit $\lp\widehat{BAC}\rp\simeq 135,7^\circ$
\enen
\enex
\bgex
\bgen[a.]
\item
$\V{MA}\cdot\V{MB}
=\lp\V{MI}+\V{IA}\rp\cdot\lp\V{MI}+\V{IB}\rp
=MI^2+\V{MI}\cdot\lp \V{IA}+\V{IB}\rp+\V{IA}\cdot\V{IB}
$
or, $\V{IA}+\V{IB}=\V{O}$ car $I$ est le milieu de $[AB]$,
et de m\^eme $\V{IA}\cdot\V{IB}=\V{IA}\cdot\lp -\V{IA}\rp=-IA^2$.
On a donc bien ainsi,
$\V{MA}\cdot\V{MB}=MI^2-IA^2$.
\item On a alors,
$M\in\mathcal{E}
\iff \V{MA}\cdot\V{MB}=7
\iff MI^2-IA^2=7
$
or, $IA=\dfrac{AB}{2}=3$, d'o\`u $IA^2=9$,
et donc,
$M\in\mathcal{E}\iff MI^2=7+IA^2=16$.
\item $\mathcal{E}$ est donc le cercle de centre $I$ et de rayon
$\sqrt{16}=4$.
\enen
\enex
\label{LastPage}
\end{document}
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