Senos, co-senos e tangentes de alguns ângulos
Trigonometria: Infinito 11 A - Parte 1 Pág. 95 Ex. 47
Enunciado
Desenhe no círculo trigonométrico os ângulos seguintes, indicando os valores dos seus senos, cossenos e tangentes.
- $\frac{7\pi }{3}$
- $\frac{4\pi }{3}$
- $\frac{71\pi }{4}$
- $-\frac{5\pi }{4}$
- $\frac{23\pi }{6}$
- $-\frac{425\pi }{6}$
- $\frac{712\pi }{6}$
- $-\frac{135\pi }{4}$
Resolução
- Ora,
\[sen\,\,\frac{7\pi }{3}=sen\,(2\pi +\frac{\pi }{3})=sen\,\frac{\pi }{3}=\frac{\sqrt{3}}{2}\]
\[\cos \,\frac{7\pi }{3}=\cos (2\pi +\frac{\pi }{3})=\cos \frac{\pi }{3}=\frac{1}{2}\]
\[tg\,\,\frac{7\pi }{3}=tg\,(2\pi +\frac{\pi }{3})=tg\,\frac{\pi }{3}=\sqrt{3}\] - Ora,
\[sen\,\,\frac{4\pi }{3}=sen\,(\pi +\frac{\pi }{3})=-sen\,\frac{\pi }{3}=-\frac{\sqrt{3}}{2}\]
\[\cos \,\frac{4\pi }{3}=\cos (\pi +\frac{\pi }{3})=-\cos \frac{\pi }{3}=-\frac{1}{2}\]
\[tg\,\,\frac{4\pi }{3}=tg\,(\pi +\frac{\pi }{3})=tg\,\frac{\pi }{3}=\sqrt{3}\] - Ora,
\[sen\,\,\frac{71\pi }{4}=sen\,(18\pi -\frac{\pi }{4})=sen\,(-\frac{\pi }{4})=-sen\,\frac{\pi }{4}=-\frac{\sqrt{2}}{2}\]
\[\cos \,\frac{71\pi }{4}=\cos (18\pi -\frac{\pi }{4})=\cos (-\frac{\pi }{4})=\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}\]
\[tg\,\,\frac{71\pi }{4}=tg\,(18\pi -\frac{\pi }{4})=tg\,(-\frac{\pi }{4})=-tg\,\frac{\pi }{4}=-1\] - Ora,
\[sen\,\,(-\frac{5\pi }{4})=sen\,(2\pi -\frac{5\pi }{4})=sen\,(\frac{3\pi }{4})=sen\,\frac{\pi }{4}=\frac{\sqrt{2}}{2}\]
\[\cos \,(-\frac{5\pi }{4})=\cos (2\pi -\frac{5\pi }{4})=\cos (\frac{3\pi }{4})=-\cos \frac{\pi }{4}=-\frac{\sqrt{2}}{2}\]
\[tg\,\,(-\frac{5\pi }{4})=tg\,(2\pi -\frac{5\pi }{4})=tg\,(\frac{3\pi }{4})=-tg\,\frac{\pi }{4}=-1\] - Ora,
\[sen\,\,\frac{23\pi }{6}=sen\,(4\pi -\frac{\pi }{6})=sen\,(-\frac{\pi }{6})=-sen\,\frac{\pi }{6}=-\frac{1}{2}\]
\[\cos \,\frac{23\pi }{6}=\cos (4\pi -\frac{\pi }{6})=\cos (-\frac{\pi }{6})=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\]
\[tg\,\frac{23\pi }{6}=tg\,(4\pi -\frac{\pi }{6})=tg\,(-\frac{\pi }{6})=-tg\,\frac{\pi }{6}=-\frac{\sqrt{3}}{3}\] - Ora,
\[sen\,\,(-\frac{425\pi }{6})=sen\,(72\pi -\frac{\pi }{6})=sen\,(\frac{7\pi }{6})=sen\,(\pi +\frac{\pi }{6})=-sen\,\frac{\pi }{6}=-\frac{1}{2}\]
\[\cos \,(-\frac{425\pi }{6})=\cos (72\pi -\frac{\pi }{6})=\cos (\frac{7\pi }{6})=\cos (\pi +\frac{\pi }{6})=-\cos \frac{\pi }{6}=-\frac{\sqrt{3}}{2}\]
\[tg\,(-\frac{425\pi }{6})=tg\,(71\pi -\frac{\pi }{6})=tg\,\frac{\pi }{6}=\frac{\sqrt{3}}{3}\] - Ora,
\[sen\,\,\frac{712\pi }{6}=sen\,(118\pi +\frac{4\pi }{6})=sen\,\frac{2\pi }{3}=sen\,(\pi -\frac{\pi }{3})=sen\,\frac{\pi }{3}=\frac{\sqrt{3}}{2}\]
\[\cos \,\frac{712\pi }{6}=\cos (118\pi +\frac{4\pi }{6})=\cos (\frac{2\pi }{3})=\cos (\pi -\frac{\pi }{3})=-\cos \frac{\pi }{3}=-\frac{1}{2}\]
\[tg\,\frac{712\pi }{6}=tg\,(118\pi +\frac{4\pi }{6})=tg\,\frac{2\pi }{3}=tg\,(\pi -\frac{\pi }{3})=-tg\,\frac{\pi }{3}=-\sqrt{3}\] - Ora,
\[sen\,\,(-\frac{135\pi }{4})=sen\,(34\pi -\frac{135\pi }{4})=sen\,\frac{\pi }{4}=\frac{\sqrt{2}}{2}\]
\[\cos \,(-\frac{135\pi }{4})=\cos (34\pi -\frac{135\pi }{4})=\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}\]
\[tg\,(-\frac{135\pi }{4})=tg\,(34\pi -\frac{135\pi }{4})=tg\,\frac{\pi }{4}=1\]