Answer:
[tex]\sin(\theta) = \frac{2\sqrt{6}}{5}[/tex]
Step-by-step explanation:
[tex]\theta \in [\frac{\pi}{2}; \pi]\\cos(\theta) = \frac{-1}{5}[/tex]
By the Fundamental Theorem of Trigonometry => [tex]\sin(\theta) = \pm \sqrt{(1-\cos^2(\theta))}[/tex]
The positive version for [tex]\theta \in [0, \pi][/tex]. And the negative for [tex]\theta \in (\pi; 2\pi)[/tex]
Meaning sin is positive in the problem => [tex]\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1-\frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}[/tex]
