Bonjour, je m'appelle Chloé je suis en 1ereS J'ai des exercices de mathématiques a faire pour la rentrée, ça parle de cercle trigonométrique et des mesures en r

Bonsoir,

Cercle trigonométrique en pièce jointe

[tex]1)\ M(cos(-\dfrac{3\pi}{4});sin(-\dfrac{3\pi}{4}))\\\\cos(-\dfrac{3\pi}{4})=-cos(\dfrac{\pi}{4})=-\dfrac{\sqrt{2}}{2}\\\\sin(-\dfrac{3\pi}{4})=-sin(\dfrac{3\pi}{4})=-sin(\dfrac{\pi}{4})=-\dfrac{\sqrt{2}}{2}\\\\M:(-\dfrac{\sqrt{2}}{2};-\dfrac{\sqrt{2}}{2})[/tex]

[tex]2)\ M(cos(\dfrac{7\pi}{6});sin(\dfrac{7\pi}{6}))\\\\cos(\dfrac{7\pi}{6})=-cos(\dfrac{\pi}{6})=-\dfrac{\sqrt{3}}{2}\\\\sin(\dfrac{7\pi}{6})=-sin(\dfrac{\pi}{6})=-\dfrac{1}{2}\\\\M:(-\dfrac{\sqrt{3}}{2};-\dfrac{1}{2})[/tex]

[tex]3)\ M(cos(\dfrac{23\pi}{4});sin(\dfrac{23\pi}{4})\\\\cos(\dfrac{23\pi}{4})=cos(\dfrac{24\pi}{4}-\dfrac{\pi}{4})=cos(6\pi-\dfrac{\pi}{4})=cos(-\dfrac{\pi}{4})=cos(\dfrac{\pi}{4})=\dfrac{\sqrt{2}}{2}[/tex]

[tex]sin(\dfrac{23\pi}{4})=sin(\dfrac{24\pi}{4}-\dfrac{\pi}{4})=sin(6\pi-\dfrac{\pi}{4})\\=sin(-\dfrac{\pi}{4})=-sin(\dfrac{\pi}{4})=-\dfrac{\sqrt{2}}{2}[/tex]

[tex]M:(\dfrac{\sqrt{2}}{2};-\dfrac{\sqrt{2}}{2})[/tex]

[tex]4)\ M(cos(\dfrac{17\pi}{3});sin(\dfrac{17\pi}{3}))\\\\cos(\dfrac{17\pi}{3})=cos(\dfrac{18\pi}{3}-\dfrac{\pi}{3})=cos(6\pi-\dfrac{\pi}{3})=cos(-\dfrac{\pi}{3})=cos(\dfrac{\pi}{3})=\dfrac{1}{2}[/tex]

[tex]sin(\dfrac{17\pi}{3})=sin(\dfrac{18\pi}{3}-\dfrac{\pi}{3})=sin(6\pi-\dfrac{\pi}{3})\\=sin(-\dfrac{\pi}{3})=-sin(\dfrac{\pi}{3})=-\dfrac{\sqrt{3}}{2}[/tex]

[tex]M:(\dfrac{1}{2};-\dfrac{\sqrt{3}}{2})[/tex]