Question

sabendo-se que cosx=1/5 e 3pi/2 ≤x ≤2pi, o valor de senx é igual a

Answer 1

$\displaystyle\underline{\text{rela{\c c}{\~a}o fundamental da trigonometria}}: \\\\\ \text{sen}^2(\text x)+\text{cos}^2(\text x) = 1 \\\\ \underline{\text{temos}}: \\\\ \text{cos x}=\frac{1}{5} \ \ ; \ \ \frac{3\pi}{2} \leq \ \text x \ \leq 2\pi \\\\\\ \text{sen}^2(\text x) = 1-\text{cos}^2(\text x) \\\\\\ \text{sen(x)}=-\sqrt{1-\text{cos}^2(\text x)} \to \text{sinal negativo porque x est{\'a} no }4^\circ \ \text{quadrante} \\\\ \text{sen(x)}=-\sqrt{1-\frac{1^2}{5^2}}$

$\displaystyle \text{sen(x)}=-\sqrt{\frac{25-1}{25}} \\\\\\ \text{sen(x)} = -\frac{\sqrt{24}}{5} \\\\\\\ \huge\boxed{\ \text{sen(x)}=\frac{-2\sqrt{6}}{5}\ }\checkmark$