Question

Given the following, use a Pythagorean identity to find sin() if is in quadrant II.

Answer 1

Answer:

$\sin(\theta) = \frac{2\sqrt{6}}{5}$

Step-by-step explanation:

$\theta \in [\frac{\pi}{2}; \pi]\\cos(\theta) = \frac{-1}{5}$

By the Fundamental Theorem of Trigonometry => $\sin(\theta) = \pm \sqrt{(1-\cos^2(\theta))}$

The positive version for $\theta \in [0, \pi]$. And the negative for $\theta \in (\pi; 2\pi)$

Meaning sin is positive in the problem => $\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1-\frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$