<s>[INST] 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.
 [/INST] Three examples of algebraic derivations of trigonometric identities as an application of the addition and subtraction formulas. 
1. Example on how to deduce the logarithmic transformation formulas (sum to product formulas) from
$$\sin (a+b)=\sin a\cos b+\sin b\cos a,\qquad (1)$$
$$\sin (a-b)=\sin a\cos b-\sin b\cos a.\qquad (2)$$
If you write
$$\left\{ 
\begin{array}{c}
a=\frac{p+q}{2}, \\ 
b=\frac{p-q}{2},\end{array}\right. \Leftrightarrow \left\{ 
\begin{array}{c}
a+b=p, \\ 
a-b=q,\end{array}%
\right. $$
you get
$$\sin (a+b)+\sin (a-b)=2\sin a\cos b,$$
$$\sin (a+b)-\sin (a-b)=2\sin b\cos a,$$
and thus
$$\sin p+\sin q=2\sin \frac{p+q}{2}\cos \frac{p-q}{2},\qquad (3)$$
$$\sin p-\sin q=2\sin \frac{p-q}{2}\cos \frac{p+q}{2}.\qquad (4)$$
You can use $(3)$ to solve the equation
$$\sin (5x)+\sin x=\sin (3x)$$
that appeared in my exam in 1968. (See a comment of mine to this post ).
2. As for the example in your question we present the following derivation. From
$$\sin (a+b)=\sin a\cos b+\sin b\cos a,\qquad (5)$$
$$\cos (a+b)=\cos a\cos b-\sin a\sin b,\qquad (6)$$
we get
$$\cot (a+b)=\frac{\cos (a+b)}{\sin (a+b)}=\frac{\cos a\cos b-\sin a\sin b}{\sin a\cos b+\sin b\cos a},$$
or dividing the numerator and denominator by $\sin a\cos b$ 
$$\begin{eqnarray*}
\cot (a+b) &=&\dfrac{\dfrac{\cos a\cos b-\sin a\sin b}{\sin a\cos b}}{\dfrac{%
\sin a\cos b+\sin b\cos a}{\sin a\cos b}}=\dfrac{\dfrac{\cos a\cos b}{\sin
a\cos b}-\dfrac{\sin a\sin b}{\sin a\cos b}}{\dfrac{\sin a\cos b}{\sin a\cos b}%
+\dfrac{\sin b\cos a}{\sin a\cos b}} \\
&=&\dfrac{\dfrac{\cos a}{\sin a}-\dfrac{\sin b}{\cos b}}{1+\dfrac{\sin b\cos a}{%
\sin a\cos b}}=\dfrac{\cot a-\tan b}{1+\tan b\cot a}=\dfrac{\cot a-\dfrac{1}{%
\cot b}}{1+\dfrac{\cot a}{\cot b}} \\
&=&\dfrac{\dfrac{\cot a\cot b-1}{\cot b}}{\dfrac{\cot b+\cot a}{\cot b}}=\dfrac{%
\cot a\cot b-1}{\cot b+\cot a},\qquad (7)
\end{eqnarray*}$$
a particular case of which (for $a=b$) is given by 
$$\cot (2a)=\dfrac{\cot ^{2}a-1}{2\cot a}.\qquad (8)$$
3. The summation and subtraction formula for the tangent. From 
$$\sin (a\pm b)=\sin a\sin
b\pm \sin b\cos a$$
we get
$$\tan a\pm \tan b=\frac{\sin a}{\cos a}\pm \dfrac{\sin b}{\cos b}=\frac{\sin a\cos b\pm \sin b\cos a}{\cos a\cos b}=\frac{\sin (a\pm b)}{\cos a\cos b}.\qquad (9)$$
</s>