Source Latex
de la correction du devoir
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pdfauthor={Yoann Morel},
pdfsubject={Correction du devoir de mathématiques: produit scalaire et trigonométrie},
pdftitle={Corrigé du devoir de mathématiques: produit scalaire et trigonométrie},
pdfkeywords={trigoométrie, produit scalaire, géométrie, vecteurs, STI2D,
STI, première, Mathématiques}
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\lfoot{Y. Morel - \url{https://xymaths.fr/Lycee/1STI/}}
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\rfoot{Corrigé du devoir de mathématiques - 1STI2D - \thepage/\pageref{LastPage}}
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\begin{document}
\ct{\bf\LARGE{Corrig\'e du devoir de math\'ematiques}}
\bgex
On a $\V{AB}\lp -14;26\rp$;\ $\V{AC}\lp 16;-5\rp$,
$\|\V{AB}\|=AB=\sqrt{\lp-14\rp^2+26^2}\simeq 29,53$,
et
$\|\V{AC}\|=AC=\sqrt{16^2+\lp-5\rp^2}=16,76$
et donc, $\V{AB}\cdot\V{AC}=-14\tm16+26\tm(-5)=-354$
\medskip
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 $\lp\widehat{BAC}\rp\simeq 135,7^\circ$
\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
\bgmp{9cm}
Soit la fonction $f$ d\'efinie sur $\R$,
p\'eriodique de p\'eriode $T=2$, d\'efinie par
\[f(t)=\la\bgar{ll}
-t+1 & \text{ si } 0\leqslant t < 1 \\
t-1 & \text{ si } 1\leqslant t \leqslant 2
\enar\right.\]
\enmp
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\enex
\bgex
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\bgen
\item Graphiquement, on trouve $T=\dfrac{\pi}{2}$.
\item $\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{\frac{\pi}{2}}
=2\pi\tm\dfrac{2}{\pi}=4$.
\item
Par exemple, en $t=\dfrac{\pi}{8}$, on a graphiquement
$f(t)=f\lp\dfrac{\pi}{8}\rp
=2$.
Or $f(t)=a\sin(\omega t)=a\sin(4t)$, et donc
$f\lp\dfrac{\pi}{8}\rp=a\sin\lp 4\dfrac{\pi}{8}\rp=
a\sin\lp \dfrac{\pi}{2}\rp=a$.
On en d\'eduit donc que $a=2$.
\vspd
On peut aussi d\'eterminer le coefficient $a$ graphiquement en
observant que le maximum de $f$ est~$2$.
Or, $f(t)=a\sin(\omega t)$, et le maximum d'un sinus est $1$.
Ainsi, le maximum de $f$ est $a\tm1=2$, d'o\`u $a=2$.
\vspt
L'expression compl\`ete de la fonction est donc
$f(t)=2\sin\lp 4t\rp$.
\enen
\enex
\label{LastPage}
\end{document}
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