Deriving the rest of trigonometric identities from the formulas for $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, and $\cos (A-B)$ I am trying to study for a test and the teacher suggest we memorize $\sin(A+B)$,  $\sin(A-B)$, $\cos(A+B)$, $\cos (A-B)$, and then be able to derive the rest out of those. I have no idea how to get any of the other ones out of these, it seems almost impossible. I know the $\sin^2\theta + \cos^2\theta = 1$ stuff pretty well though. For example just knowing the above how do I express $\cot(2a)$ in terms of $\cot a$? That is one of my problems and I seem to get stuck half way through.

I'm not certain, but I think perhaps you need to revisit the basic trigonometric definitions.
$$\begin{array}{lll}
\sin\theta = \frac{opposite}{hypotenuse}&\csc\theta=\frac{hypotenuse}{opposite}\\
\cos\theta = \frac{adjacent}{hypotenuse}&\sec\theta=\frac{hypotenuse}{adjacent}\\
\tan\theta = \frac{opposite}{adjacent}&\cot\theta=\frac{adjacent}{opposite}
\end{array}$$
Having done that, we can now manipulate the definitions. For example
$$\cot\theta = \frac{adjacent}{opposite}\cdot\frac{\frac{1}{hypotemnuse}}{\frac{1}{hypotemnuse}}=\frac{\frac{adjacent}{hypotenuse}}{\frac{opposite}{hypotenuse}}=\frac{\cos\theta}{\sin\theta}$$
or alternatively
$$\cot\theta = \frac{adjacent}{opposite}=\frac{\frac{1}{opposite}}{\frac{1}{adjacent}}=\frac{\frac{1}{opposite}}{\frac{1}{adjacent}}\cdot\frac{hypotenuse}{hypotenuse}=\frac{\frac{hypotenuse}{opposite}}{\frac{hypotenuse}{adjacent}}=\frac{\csc\theta}{\sec\theta}$$

After that, move on to making 3 triangles that are similar to the adjacent-opposite-hypotenuse triangle. These triangles will rotate which side is equal to 1. This is where we get the familiar Pythagorean identities.

The next step is to derive $\cos (\alpha+\beta)$ and $\sin (\alpha+\beta)$ by setting the lengths (I would use the squares of the lengths though) of the green lines in the diagram equal to each other. It's easy to do, since the endpoints are well known.
Finally, lets derive $\cot(A+B)$
$$\begin{array}{lll}
\cot(A+B)&=&\frac{\sin(A+B)}{\cos(A+B)}\\
&=&\frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}\\
&=&\frac{\frac{\sin A\cos B + \cos A\sin B}{\sin A\sin B}}{\frac{\cos A\cos B - \sin A\sin B}{\sin A\sin B}}\\
&=&\frac{\frac{\cos B}{\sin B}+\frac{\cos A}{\sin A}}{\frac{\cos A\cos B}{\sin A\sin B}-1}\\
&=&\frac{\cot B+\cot A}{\cot A\cot B-1}
\end{array}$$
Note that $\cot 2A$ is just $\cot(A+B)$ where $A=B$, so
$$\cot 2A = \cot(A+A)=\frac{\cot A+\cot A}{\cot A\cot A-1}=\frac{2\cot A}{\cot^2 A-1}$$
Of course this list isn't comprehensive, for example we didn't derive the sine and cosines laws, nor did we consider the double angle and half angle formulas. But I hope that what little I did present here was helpful.
