Bonjour Sheita
Exercice 1
[tex]a)\ a=\dfrac{4\pi}{3}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\b=\dfrac{\pi}{4}+2k\pi\ \ (k\in\ \mathbb{Z})[/tex]
[tex]b)\ (\overrightarrow{OI};\overrightarrow{OA})=(\overrightarrow{OI};\overrightarrow{OB})+(\overrightarrow{OB};\overrightarrow{OA})\\\\(\overrightarrow{OB};\overrightarrow{OA})=(\overrightarrow{OI};\overrightarrow{OA})-(\overrightarrow{OI};\overrightarrow{OB})[/tex]
[tex](\overrightarrow{OB};\overrightarrow{OA})=\dfrac{4\pi}{3}-\dfrac{\pi}{4}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\(\overrightarrow{OB};\overrightarrow{OA})=\dfrac{16\pi}{12}-\dfrac{3\pi}{12}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\(\overrightarrow{OB};\overrightarrow{OA})=\dfrac{13\pi}{12}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\\boxed{(\overrightarrow{OA};\overrightarrow{OB})=-\dfrac{13\pi}{12}+2k\pi\ \ (k\in\ \mathbb{Z})}[/tex]
Exercice 2
[tex]a)\ \dfrac{5\pi}{4}-2\pi=\dfrac{5\pi}{4}-\dfrac{8\pi}{4}=-\dfrac{3\pi}{4}\in\ ]-\pi;\pi][/tex]
D'où la détermination principale de 5π/4 est [tex]\boxed{-\dfrac{3\pi}{4}}\in\ ]-\pi;\pi][/tex]
[tex]b)\ -\dfrac{7\pi}{2}+2\times2\pi=-\dfrac{7\pi}{2}+4\pi=-\dfrac{7\pi}{2}+\dfrac{8\pi}{2}=\dfrac{\pi}{2}\in\ ]-\pi;\pi][/tex]
D'où la détermination principale de -7π/2 est [tex]\boxed{\dfrac{\pi}{2}}\in\ ]-\pi;\pi][/tex]
c) La détermination principale de 5π/6 est [tex]\boxed{\dfrac{5\pi}{6}\in\ ]-\pi;\pi]}[/tex]
[tex]d)\ -\dfrac{11\pi}{4}+2\pi=-\dfrac{11\pi}{4}+\dfrac{8\pi}{4}=-\dfrac{3\pi}{4}\in\ ]-\pi;\pi][/tex]
La détermination principale de -11π/4 est [tex]\boxed{-\dfrac{3\pi}{4}\in\ ]-\pi;\pi]}[/tex]
Exercice 3
[tex]a)\ \cos(-\dfrac{\pi}{3})=\cos(\dfrac{\pi}{3})=\dfrac{1}{2}\\\\b)\ \sin(\dfrac{5\pi}{6})=\sin(\pi-\dfrac{5\pi}{6})=\sin(\dfrac{6\pi}{6}-\dfrac{5\pi}{6})=\sin(\dfrac{\pi}{6})=\dfrac{1}{2}\\\\c)\ \cos(\dfrac{5\pi}{4})=\cos(\pi+\dfrac{\pi}{4})=-\cos(\dfrac{\pi}{4})=-\dfrac{\sqrt{2}}{2}\\\\d)\ \sin(\dfrac{7\pi}{4})=\sin(\dfrac{7\pi}{4}-2\pi)=\sin(\dfrac{7\pi}{4}-\dfrac{8\pi}{4})=\sin(-\dfrac{\pi}{4})\\\\=-\sin(\dfrac{\pi}{4})=-\dfrac{\sqrt{2}}{2}[/tex]
Exercice 4
[tex]1)\ Si\ a=\dfrac{2\pi}{3},\ alors\ \cos(a)=-\dfrac{1}{2}\\\\2)\ \cos(2x-\dfrac{\pi}{6})=-\dfrac{1}{2}\\\\\cos(2x-\dfrac{\pi}{6})=\cos(\dfrac{2\pi}{3})\\\\2x-\dfrac{\pi}{6}=\dfrac{2\pi}{3}+2k\pi\ \ ou\ \ 2x-\dfrac{\pi}{6}=-\dfrac{2\pi}{3}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\2x=\dfrac{2\pi}{3}+\dfrac{\pi}{6}+2k\pi\ \ ou\ \ 2x=-\dfrac{2\pi}{3}+\dfrac{\pi}{6}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\2x=\dfrac{4\pi}{6}+\dfrac{\pi}{6}+2k\pi\ \ ou\ \ 2x=-\dfrac{4\pi}{6}+\dfrac{\pi}{6}+2k\pi\ \ (k\in\ \mathbb{Z})[/tex][tex]\\\\2x=\dfrac{5\pi}{6}+2k\pi\ \ ou\ \ 2x=-\dfrac{3\pi}{6}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\2x=\dfrac{5\pi}{6}+2k\pi\ \ ou\ \ 2x=-\dfrac{\pi}{2}+2k\pi\ \ (k\in\ \mathbb{Z})\\\\\boxed{x=\dfrac{5\pi}{12}+k\pi\ \ ou\ \ x=-\dfrac{\pi}{4}+k\pi\ \ (k\in\ \mathbb{Z})}[/tex]
