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.

Since $\displaystyle\cot(2a) = \frac{\cos(2a)}{\sin(2a)}$, you would have (assuming you know the addition formulas for sines and cosines):
$$\begin{align*}
\cos(2a) &= \cos(a+a) = \cos(a)\cos(a) - \sin(a)\sin(a)\\
         &= \cos^2(a) - \sin^2(a);\\
\sin(2a) &= \sin(a+a) = \sin(a)\cos(a) + \cos(a)\sin(a)\\
&= 2\sin(a)\cos(a),
\end{align*}$$
and therefore
$$\begin{align*}
\cot(2a) &= \frac{\cos(2a)}{\sin(2a)} = \frac{\cos^2(a) - \sin^2(a)}{2\sin(a)\cos(a)}\\
&= \frac{1}{2}\left(\frac{\cos^2(a)}{\sin(a)\cos(a)}\right) - \frac{1}{2}\left(\frac{\sin^2(a)}{\sin(a)\cos(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cos(a)}{\sin(a)} - \frac{\sin(a)}{\cos(a)}\right)\\
&= \frac{1}{2}\left(\cot(a) - \tan(a)\right)\\
&= \frac{1}{2}\left(\cot(a) - \frac{1}{\cot(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cot^2(a)}{\cot(a)} - \frac{1}{\cot(a)}\right)\\
&= \frac{1}{2}\left(\frac{\cot^2(a) - 1}{\cot (a)}\right).
\end{align*}$$
P.S. Now, as it happens, I don't know the formulas for double angles, nor most identities involving tangents, cotangents, etc. I never bothered to memorize them.  What I know are:


*

*The definitions of tangent, cotangent, secant, and cosecant in terms of sine and cosine;

*That sine is odd ($\sin(-x) = -\sin(x)$) and cosine is even ($\cos(-x)=\cos(x)$);

*The addition formulas for sine and cosine;

*The values of sine and cosine at $0^{\circ}$, $30^{\circ}$, $45^{\circ}$, $60^{\circ}$, and $90^{\circ}$.


(I can derive $\sin^2\theta + \cos^2\theta = 1$ from the above, but in all honesty that one comes up so often that I do know it as well). I do not know the addition or double angle formulas for tangents nor cotangents, so the above derivation was done precisely "on the fly", as I was typing. I briefly thought that I might need to $\cos(2a)$ with one of the following equivalent formulas:
$$\cos^2(a)-\sin^2(a) = \cos^2(a) + \sin^2(a) - 2\sin^2(a) = 1 - 2\sin^2(a)$$
or
$$\cos^2(a) - \sin^2(a) = 2\cos^2(a) - \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1,$$
if the first attempt had not immediately led to a formula for $\cot(2a)$ that involved only $\cot(a)$ and $\tan(a) = \frac{1}{\cot(a)}$. 
