Question

Ajutor va rog! =)))) ​

Answer 1

$\alpha \: \in \: (\pi, \frac{3\pi}{2} ) = >\alpha\: \in \: cadranul \: III$

$cos \alpha = - \frac{5}{13}$

$sin \alpha = ?$

${sin}^{2} \alpha + {cos}^{2} \alpha = 1$

${sin}^{2} \alpha + {( - \frac{5}{13} )}^{2} = 1$

${sin}^{2} \alpha + \frac{25}{169} = 1$

${sin}^{2} \alpha = 1 - \frac{25}{169}$

${sin}^{2} \alpha = \frac{144}{169}$

$sin \alpha = \pm \sqrt{ \frac{144}{169} }$

$sin \alpha = \pm \frac{ \sqrt{144} }{ \sqrt{169} }$

$sin \alpha = \pm \frac{12}{13}$

$a \: \in \: (\pi, \frac{3\pi}{2} ) = > sin \alpha < 0 = > sin \alpha = - \frac{12}{13}$