Calcule o valor numérico
Trigonometria: Infinito 11 A - Parte 1 Pág. 96 Ex. 55
Enunciado
- Se $\cos (\alpha -\frac{\pi }{2})=\frac{4}{5}$, calcule o valor numérico de $\begin{matrix}
\cos (-\alpha )-\cos (\pi +\alpha ) & \wedge & \alpha \in 1.{}^\text{o}Q \\
\end{matrix}$. - Determine o valor exato de $tg\,(-\alpha )+\cos (\alpha -\frac{\pi }{2})$, sabendo que $\begin{matrix}
sen\,(-\pi -\alpha )=\frac{3}{7} & \wedge & \alpha \in 2.{}^\text{o}Q \\
\end{matrix}$.
Resolução
- Ora, $\cos (-\alpha )-\cos (\pi +\alpha )=\cos \alpha +\cos \alpha =2\cos \alpha $.
Por outro lado, $\cos (\alpha -\frac{\pi }{2})=\frac{4}{5}\Leftrightarrow \cos (\frac{\pi }{2}-\alpha )=\frac{4}{5}\Leftrightarrow sen\,\alpha =\frac{4}{5}$.
Como $\alpha \in 1.{}^\text{o}Q$, então $\cos \alpha =+\sqrt{1-{{(\frac{4}{5})}^{2}}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$.
Logo, $\cos (-\alpha )-\cos (\pi +\alpha )=2\times \frac{3}{5}=\frac{6}{5}$.
- Ora, $tg\,(-\alpha )+\cos (\alpha -\frac{\pi }{2})=-tg\,\alpha +\cos (\frac{\pi }{2}-\alpha )=-tg\,\alpha +sen\,\alpha $.
Por outro lado, $sen\,(-\pi -\alpha )=\frac{3}{7}\Leftrightarrow sen\,(\pi +\alpha )=-\frac{3}{7}\Leftrightarrow sen\,\alpha =\frac{3}{7}$.
Como $\alpha \in 2.{}^\text{o}Q$, então $\cos \alpha =-\sqrt{1-{{(\frac{3}{7})}^{2}}}=\sqrt{1-\frac{9}{49}}=\frac{2\sqrt{10}}{7}$.
Logo, \[\begin{matrix}
tg\,(-\alpha )+\cos (\alpha -\frac{\pi }{2}) & = & -\frac{\frac{3}{7}}{\frac{2\sqrt{10}}{7}}+\frac{3}{7} \\
{} & = & \frac{3}{2\sqrt{10}}+\frac{3}{7} \\
{} & = & \frac{3\sqrt{10}}{20}+\frac{3}{7} \\
{} & = & \frac{21\sqrt{10}+60}{140} \\
\end{matrix}\]
