Evaluating $\cos(A+B)$, given $\cos A$ and $\sin(B)$ Find the exact value:

Find $\cos(A+B)$ given that $\cos A=1/3$, with $A$ in the first quadrant, and $\sin B = -1/4$, with $B$ in the fourth quadrant.

 A: HINT:
Recall the following formula:
\begin{align}
\cos(A+B) & = \cos(A) \cos(B) - \sin(A) \sin(B) \tag{$\spadesuit$}\\
\cos^2(C) + \sin^2(C) & = 1 \tag{$\clubsuit$}
\end{align}
Plug in $C=A$ and $C=B$ separately in $(\clubsuit)$, to obtain $\sin(A)$ and $\cos(B)$. 
Also remember that since $A$ is in the first quadrant, we have $\sin(A)$, $\cos(A)$ to be positive. Since $B$ is in the fourth quadrant, $\sin(B)$ is negative and $\cos(B)$ is positive.
A: Remember that cosine represents adjacent over hypotenuse, and sine represents opposite over hypotenuse.
$\sin (A) = -\dfrac {1}{4} $ is therefore tells us that we have a triangle with a side length of one, and a hypotenuse of length four. Using Pythagorean theorem, we can find the third side, which is equal to $\sqrt {1^2 + 4^2} $, or  $\sqrt{15}$; this lets us find the cosine of that same angle.
By definition, $\cos (B)$ then equals $\dfrac{\sqrt{15}}{4}$.
We then do the same with $\cos (A) = \dfrac {2}{3} $ to get that $\sin A = \dfrac{\sqrt{8}}{3}$.
We can then use the identity $\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ to solve for $\cos(A + B)$ exactly, as we know $\sin(A), \cos(A), \sin(B)$, and $\cos(B)$.
This then leaves us with $\cos(A + B) = \dfrac {\sqrt{15} + \sqrt{8}}{12}$.
