What trig identities must one know to derive the others? My TA told me in problem section one day that every trig identity could be derived from just 2: the Pythagorean identity and the double-angle identity (or he might have said the half-angle identity).  I'm a bit dubious that every trig identity could be derived from just these two.  What would you say the minimum number of identities one must know to say derive every identity on the wikipedia page?
If it is possible to derive all of the identities from just 2, can anyone recommend a source that takes that approach?
 A: I think that Apostol's Calculus book might contain this idea in detail: 
If you are willing to believe that sine and cosine are continuous, and have proved that a continuous function on a dense subset of an interval has a unique continuous extension to the interval, then once you know
(1) $\sin(0) = 0$, $\cos(0) = 1$, $\sin(\pi/2) = 0$; $\cos(x) > 0$ for $x \in [0, \pi/2)$, 
(2) $\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$ for all $a, b$, and
(3) $\sin(-x) = -\sin(x)$ for all $x$.
you can derive $1 = \cos^2(a) + \sin^2(a)$ by setting $b = a$ in the second formula.
You then can find that cosine is even (set $a = 0$). 
Assumption 3 may not be necessary, but I confess, I forget how to show that the sine is odd without it.  
Then you can set $b = -a$ to get
$$cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1.$$
Applying this to $a = x/2$, you get
$$
\cos(x) = 2\cos^2(x/2) - 1 \\
\cos(x/2) = \sqrt{\frac{\cos(x) + 1}{2}}
$$
from this, you can determine cosine of all numbers of the form $\frac{\pi}{2} \frac{1}{2^n}$; using the addition formula, you can determine cosine at all points of the form  
$$\frac{\pi}{2} \frac{k}{2^n}$$
where $k$ is an integer. These form a dense subset of the interval $[0, \pi/2]$. 
You can then also show that for $x$ small, $\sin x$ is also small, so that (using the addition/subtraction formulas) cosine is continuous on this dense subset; it therefore has a unique continuous extension to $[0, \pi/2]$. The same goes for sine, and you're on your way.  
@Semiclassical suggested, in comments above, that an addition formula, together with $\sin^2 + \cos^2 = 1$, might suffice, but that a half-angle formula might not. The conjecture about half-angle formulas is correct, as the following shows:
Let
\begin{align}
Sin(x) = \begin{cases} 
\sin(x) & x = \frac{p}{2^k} \pi & \text{for $p, k, \in \mathbb  Z$} \\
0 & \text{otherwise}
\end{cases}\\
Cos(x) = \begin{cases} 
\cos(x) & x = \frac{p}{2^k} \pi & \text{for $p, k, \in \mathbb  Z$} \\
0 & \text{otherwise}
\end{cases}
\end{align}
Then $Sin$ and $Cos$ satisfy $Sin^2 + Cos^2 = 1$ and the half-angle formulas, but they are not the same functions as $\sin$ and $\cos$, and hence need not satisfy the other formulas. In particular, they fail to satisfy the addition formula. 
