Finding $\cos(s+t)$ I am attempting to find $\cos(s+t)$ and $\cos(s-t)$
I am given that they are in quadrant II and that $\cos s = -1/5$ and $\sin t = 3/5$. I have no idea what the relation between these numbers $s$ or what I am suppose to do exactly. What is $s$ and what is $t$? I know that $\cos$ is $x/r$ so $x=-1$ and $r=5$, $\sin$ is $y/r$ so $y=3$ but I don't know what to do from here really.
 A: $s$ and $t$ are the "names" of two angles. You should use three things:


*

*The addition formulas for cosine:
$$\begin{align*}
\cos(\alpha+\beta) &= \cos\alpha\cos\beta - \sin\alpha\sin\beta\\
\cos(\alpha-\beta) &= \cos\alpha\cos\beta + \sin\alpha\sin\beta.
\end{align*}$$
But in order to use these formulas, you need to know $\cos s$, $\cos t$, $\sin s$, and $\sin t$. You only know $\cos s$ and $\sin t$, so you need to figure out the other two.

*To figure out $\sin s$, you should use the fact that
$$\cos^2 s + \sin^2 s = 1.$$
That will tell you that $\sin^2 s = 1-\cos^2 s$; finally, use the fact that $s$ is in quadrant II to decide if you should take the positive or negative square root.

*Likewise, to figure out $\cos t$, you should use the fact that
$$\cos^2 t + \sin^2 t = 1,$$
and again use the fact that $t$ is in the second quadrant to determine if you use the positive or negative square roots.

The fact that cosine of an angle can be defined to be the length of the adjacent side divided by the length of the hypotenuse (what you describe as $x/r$), and sine as the length of the opposite side divided byt he length of the hypotenuse (what you call $y/r$) does not need to play a role here. 
You can use it (it's essentially steps 2 and 3), but you should notice that you would have two different triangles (one with angle $t$ and one with angle $s$, so the $x,y,r$ you use for $\cos s$ would be different from the $x,y,r$ you would use for the $\sin t$).
From $\cos s = -1/5$ you imagine a right triangle with $x=-1$ and $r=5$. Since $x^2+y^2 = r^2$, you get $y^2 = 25-(-1)^2 = 24$; then figure out if you get $y=\sqrt{24}$ or $y=-\sqrt{24}$, using the fact that you are in quadrant II. This gives you $\sin s = y/r$.
You could then use the same idea to find $\cos t$, from $\sin t = 3/5$, $y=3$, $r=5$ (new $y$, because it's a different angle; same $r$ only by chance). 
A: More than any formula of any kind: Did you draw a picture? Do you have any clue what you are really looking for?
A: Do you know that $\cos (s+t)=\cos s\cdot \cos t - \sin s \cdot \sin t$?  You can just apply that.
