Question

Se sen x= -(12/13), com x no terceiro quadrante, determine cos x

Answer 1

$\displaystyle \text{sen x}=\frac{-1 2}{13} \ ; \ \text{cos x = ?} \ ; \ \text x \in 3^\circ \text Q \\\\\\ \underline{\text{Rela{\c c}{\~a}o fundamental da trigonometria}}:\\\\ \text{sen}^2 \text x+\text{cos}^2\text x=1 \\\\ \text{cos}^2\text x= 1-\text{sen}^2\text x \\\\ \text{cos x} = -\sqrt{1-\text{sen}^2\text x} \\ \text{(negativo pq cosseno no terceiro quadrante {\'e} negativo)}$

$\displaystyle \text{cos x}=-\sqrt{1-(-\frac{12}{13})^2} \\\\\\ \text{cos x}=-\sqrt{1-\frac{144}{169}} \\\\\\ \text{cos x}=-\sqrt{\frac{169-144}{169}}\\\\\\ \text{cos x}=-\sqrt{\frac{25}{169}} \\\\\\\ \huge\boxed{\ \text{cos x}=\frac{-5}{13}\ }\checkmark$