Module : Trigonométrie
Exercice
Sans calculatrice, calculez \(\sin\theta\), \(\cos\theta\) et \(tg\, \theta\) si
(a) \(\theta=\frac{5\pi}{6}\)
Réponse
\(\sin{\theta}=\frac{1}{2},\, \cos{\theta}=-\frac{\sqrt{3}}{2}\) et \(tg\,{\theta}=-\frac{\sqrt{3}}{3}\)
Aide
On a \(\sin{\frac{5\pi}{6}}=\sin{\frac{\pi}{6}}\), \(\cos{\frac{5\pi}{6}}=-\cos{\frac{\pi}{6}}\) et \(tg\,{\frac{5\pi}{6}}=\frac{\sin{\frac{5\pi}{6}}}{\cos{\frac{5\pi}{6}}}\).
Solution
On a \(\sin{\frac{5\pi}{6}}=\sin{\frac{\pi}{6}}=\frac{1}{2}\),
\(\cos{\frac{5\pi}{6}}=-\cos{\frac{\pi}{6}}=-\frac{\sqrt{3}}{2}\),
\(tg\,{\frac{5\pi}{6}}=\frac{\sin{\frac{5\pi}{6}}}{\sin{\frac{5\pi}{6}}}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{\sqrt{3}}{3}\).
Théorie
La théorie correspondant à cet exercice se trouve ici.
(b) \(\theta=315^{\circ}\)
Réponse
\(\sin{\theta}=-\frac{\sqrt{2}}{2},\, \cos{\theta}=\frac{\sqrt{2}}{2}\) et \(tg\,{\theta}=-1\)
Aide
On a \(\sin{315^{\circ}}=\sin{\frac{7\pi}{4}}\), \(\cos{315^{\circ}}=\cos{\frac{7\pi}{4}}\) et \(tg\,{315^{\circ}}=\tan{\frac{7\pi}{4}}\).
Solution
On a \(\sin{315^{\circ}}=\sin{\frac{7\pi}{4}}=-\sin{\frac{\pi}{4}}=-\frac{\sqrt{2}}{2}\),
\(\cos{315^{\circ}}=\cos{\frac{7\pi}{4}}=\cos{\frac{\pi}{4}}=\frac{\sqrt{2}}{2}\),
\(tg\,{315^{\circ}}=tg\,{\frac{7\pi}{4}}=-tg{\,\frac{\pi}{4}}=-1\).
Théorie
La théorie correspondant à cet exercice se trouve ici.