Finding cosine when negating its angle, given only the value of sine. If $\sin(\theta) = \frac{1}{\sqrt{5}}$, I must find $\cos(-\theta)$.
Using a right triangle (hypotenuse is $\sqrt{5}$ and opposite is $1$), it is easy to find that the adjacent side is $2$. Then since $\cos(-\theta) = \cos(\theta)$ I can simply find the answer as $\frac{2}{\sqrt{5}}$.
This is correct according to my homework. However, I was thinking, what if $\theta$ is in the second quadrant of the unit circle? Basically $\pi > \theta > \pi/2$.
Then, visually speaking if you look at the unit circle, there is no way that $\cos(\theta)$ can be positive. It has to be negative because the angle ends up in the third quadrant when you negate it as $-\theta$.
But my homework says that the only answer is the positive version, so I'm not sure what am I doing wrong.
 A: I think you have put more good thought into this than the author of your homework answers did, and I agree with your conclusion.
Representing the angle $\theta$ on the unit circle, we have one end of the angle at $(1,0)$ and the other end at any point where $y=\frac1{\sqrt5}$,
which is to say the other end can be at either
$\left(\frac2{\sqrt5},\frac1{\sqrt5}\right)$ or 
$\left(-\frac2{\sqrt5},\frac1{\sqrt5}\right)$.
There are infinitely many angles with sine $\frac1{\sqrt5}$, since you can go around the circle as many times as you like in either direction to reach one of those two points, but you must end at one of those two points.
Then the other end of the angle $-\theta$ is at
$\left(\frac2{\sqrt5},-\frac1{\sqrt5}\right)$ or 
$\left(-\frac2{\sqrt5},-\frac1{\sqrt5}\right)$.
In the end there are two possible values for $\cos(-\theta)$
found by taking the $x$ coordinates of the last two points:
$\frac2{\sqrt5}$ and $-\frac2{\sqrt5}.$
(Or you could realize that $\cos(-\theta)=\cos(\theta)$ and therefore when you found the endpoints of the angle $\theta$ you already had the answer.)
But if someone did not think about it as much as you did, they might not think about the answer $-\frac2{\sqrt5}.$ I guess that's what happened to the author of your homework answers.
A big reason why I was so explicit about using the unit circle in this answer is that it encourages thinking about all the quadrants, whereas the right-triangle definition of sine and cosine really only makes sense in the first quadrant.
