Question

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Answer 1

Se $\alpha \in \left]\dfrac{\pi}{2},\,\pi \right[$, então $\alpha$ é um arco do 2º quadrante.

$\mathrm{sen\,}\alpha=\dfrac{1}{3}$


Identidades utilizadas:

$\begin{array}{rclc} \mathrm{sen}^{2\,}\alpha+\cos^{2}\alpha &=& 1&\;\;\;\text{(i)}\\ \\ \mathrm{tg\,}\alpha&=&\dfrac{\mathrm{sen\,}\alpha}{\cos \alpha}&\;\;\;\text{(ii)} \end{array}$


$\mathrm{sen}^{2\,}\alpha+\cos^{2}\alpha=1\\ \\ \cos^{2}\alpha=1-\mathrm{sen}^{2\,}\alpha\\ \\ \cos^{2}\alpha=1-\left(\dfrac{1}{3} \right )^{2}\\ \\ \cos^{2}\alpha=1-\dfrac{1}{9}\\ \\ \cos^{2}\alpha=\dfrac{9-1}{9}\\ \\ \cos^{2}\alpha=\dfrac{8}{9}\\ \\ \cos \alpha=\pm \sqrt{\dfrac{8}{9}}\\ \\ \cos \alpha=\pm \dfrac{\sqrt{8}}{\sqrt{9}}\\ \\ \cos \alpha=\pm \dfrac{\sqrt{2^{2}\cdot 2}}{3}\\ \\ \cos \alpha=\pm \dfrac{\sqrt{2^{2}}\cdot \sqrt{2}}{3}\\ \\ \cos \alpha=\pm \dfrac{2\sqrt{2}}{3}$


Como $\alpha$ é do 2º quadrante, então o cosseno é negativo. Logo,

$\cos \alpha=-\dfrac{2\sqrt{2}}{3}$


Então,

$\mathrm{tg\,}\alpha=\dfrac{\mathrm{sen\,}\alpha}{\cos \alpha}\\ \\ \mathrm{tg\,}\alpha=\dfrac{\,^{1}\!\!\!\diagup\!\!_{3}}{\,^{-2\sqrt{2}}\!\!\!\diagup\!\!_{3}}\\ \\ \mathrm{tg\,}\alpha=\dfrac{1}{\diagup\!\!\!\! 3}\cdot \dfrac{\diagup\!\!\!\! 3}{-2\sqrt{2}}\\ \\ \mathrm{tg\,}\alpha=-\dfrac{1}{2\sqrt{2}}\\ \\ \mathrm{tg\,}\alpha=-\dfrac{1\cdot \sqrt{2}}{2\sqrt{2}\cdot \sqrt{2}}\\ \\ \mathrm{tg\,}\alpha=-\dfrac{\sqrt{2}}{2\cdot \left(\sqrt{2} \right )^{2}}\\ \\ \mathrm{tg\,}\alpha=-\dfrac{\sqrt{2}}{2\cdot 2}\\ \\ \mathrm{tg\,}\alpha=-\dfrac{\sqrt{2}}{4}$