Question

Dado o valor de tg α=√3/3.calcule cos  α e sen α 

Answer 1

$\text{tg}~\alpha=\dfrac{\text{sen}~\alpha}{\text{cos}~\alpha}$.

$\text{sen}^2~\alpha+\text{cos}^2~\alpha=1~~(i)$.

Pelo enunciado, $\dfrac{\text{sen}~\alpha}{\text{cos}~\alpha}=\dfrac{\sqrt{3}}{3}$. 

Elevando os dois lados ao quadrado, $\dfrac{\text{sen}^2~\alpha}{\text{cos}^2~\alpha}=\dfrac{3}{9}=\dfrac{1}{3}$.

Daí, tiramos que, $\text{cos}^2~\alpha=3\text{sen}^2~\alpha$. Substituindo em $(i)$:

$\text{sen}^2~\alpha+3\text{sen}^2~\alpha=1$

$4\text{sen}^2~\alpha=1~~\Rightarrow~~\text{sen}^2~\alpha=\dfrac{1}{4}~~\Rightarrow~~\text{sen}~\alpha=\sqrt{\dfrac{1}{4}}~~\Rightarrow~~\text{sen}~\alpha=\dfrac{1}{2}$

$\text{cos}^2~\alpha=3\cdot\dfrac{1}{4}~~\Rightarrow~~\text{cos}^2~\alpha=\dfrac{3}{4}~~\Rightarrow~~\text{cos}~\alpha=\sqrt{\dfrac{3}{4}}~~\Rightarrow~~\text{cos}~\alpha=\dfrac{\sqrt{3}}{4}$.