The angle $\theta$ is located in Quadrant III and $\cos(\theta) = -\frac{5}{8}$. What is the exact value of $\sin(\theta)$?
I'm at a loss. The fraction to me implies sides of a right triangle but when we get into quadrants that aren't Quadrant I, I don't understand how the angles and sides are related anymore since the angle is "larger" and wraps around the circle more instead of just being an angle in a triangle.
Edit: Is it $-\frac{\sqrt{39}}{8}$? (I did $\sqrt{8^2-5^2} = \sqrt{39}$ to get the length of the other "side" of the triangle and since it's Quadrant III I assume this to be negative, and so sin is just the ratio of that over the hypotenuse instead?)
Guide:
Information that can help you solve the problem:
$$\sin^2(\theta) +\cos^2(\theta)=1$$
and also in the third quadrant:
$$\sin(\theta) < 0$$
Reading task about trigonometry function, especially the part about unit circle definition.