Calcule o cosseno de…
Trigonometria: Infinito 11 A - Parte 1 Pág. 97 Ex. 56
Enunciado
Sabendo que $\cos \frac{\pi }{8}=\frac{1}{2}\sqrt{2+\sqrt{2}}$, calcule o cosseno de:
- $-\frac{\pi }{8}$
- $\frac{3\pi }{8}$
- $\frac{5\pi }{8}$
- $\frac{9\pi }{8}$
- $-\frac{325\pi }{8}$
Resolução
- Ora,
\[\cos (-\frac{\pi }{8})=\cos \frac{\pi }{8}=\frac{1}{2}\sqrt{2+\sqrt{2}}\] - Ora,
\[\begin{array}{*{35}{l}}
\cos (\frac{3\pi }{8}) & = & \cos (\frac{\pi }{2}-\frac{\pi }{8}) \\
{} & = & sen\,\frac{\pi }{8} \\
{} & = & \sqrt{1-{{\left( \frac{1}{2}\sqrt{2+\sqrt{2}} \right)}^{2}}}\,\,\,\,\,\text{(Porqu }\!\!\hat{\mathrm{e}}\!\!\text{ ?)} \\
{} & = & \sqrt{1-\frac{1}{4}(2+\sqrt{2})} \\
{} & = & \sqrt{\frac{1}{4}(4-2-\sqrt{2})} \\
{} & = & \frac{1}{2}\sqrt{2-\sqrt{2}} \\
\end{array}\] - Ora,
\[\cos \frac{5\pi }{8}=\cos (\frac{\pi }{2}-(-\frac{\pi }{8}))=sen\,(-\frac{\pi }{8})=-sen\,\frac{\pi }{8}=-\frac{1}{2}\sqrt{2-\sqrt{2}}\] - Ora,
\[\cos \frac{9\pi }{8}=\cos (\pi +\frac{\pi }{8})=-\cos \frac{\pi }{8}=-\frac{1}{2}\sqrt{2+\sqrt{2}}\] - Ora,
\[\cos (-\frac{325\pi }{8})=\cos (-41\pi +\frac{3\pi }{8})=\cos (\frac{3\pi }{8})=\frac{1}{2}\sqrt{2-\sqrt{2}}\]
