Question

Se tgx = - √5, então quanto vale senx, sabendo que x está no quarto quadrante?

Answer 1

$\displaystyle \text{Tg(x)} = -\sqrt5$

abrindo a tangente :

$\displaystyle \frac{\text{sen(x)}}{\text{cos(x)}}=-\sqrt5$

$\text{sen(x)}=-\sqrt5.\text{cos(x)}$

elevando ao quadrado :

$\text{sen}^2(\text x) = 5.\text{cos}^2(\text x)$

Sabemos pela relação fundamental da trigonometria que :

$\text{sen}^2(\text x) +\text{cos}^2(\text x) = 1\\\\ \text{cos}^2(\text x) = 1-\text{sen}^2(\text x)$

Substituindo o cos²(x) :

$\text{sen}^2(\text x) = 5.\text{cos}^2(\text x)$

$\text{sen}^2(\text x) = 5.(1-\text{sen}^2(\text x) )$

$\text{sen}^2(\text x) = 5-5.\text{sen}^2(\text x)$

$6.\text{sen}^2(\text x) = 5$

$\displaystyle\text{sen}^2(\text x) =\frac{5}{6}$

$\displaystyle\text{sen}(\text x) =\pm\frac{\sqrt{5}}{\sqrt{6}}$

x está no 4º quadrante e sabemos que no 4º quadrante o seno é negativo, portanto :

$\displaystyle\text{sen}(\text x) =-\frac{\sqrt{5}}{\sqrt{6}}$

racionalizando :

$\displaystyle\text{sen}(\text x) =-\frac{\sqrt{5}}{\sqrt{6}}.\frac{\sqrt{6}}{\sqrt{6}}$

Portanto :

$\huge\boxed{\displaystyle\text{sen}(\text x) =\frac{-\sqrt{30}}{6}}\checkmark$