For what it's worth:
I just memorize one Pythagorean identity and one of the sum identities. Many of the others (besides the obvious ones: the reciprocal, periodicity, and Pythagorean) can be derived
starting with one of the sum formulas.
So, you could just memorize how to derive them. Of course, in a test scenario, this may waste precious time...
Reciprocal identities
The reciprocal identities follow from the definitions of the trigonometric functions.
$$\eqalign
{ \sec\theta&= {1\over \cos\theta} \qquad \tan\theta= {\sin\theta\over \cos\theta} \cr
\csc\theta&= {1\over \sin\theta} \qquad \cot\theta= {1\over \tan\theta} \cr
}
$$
Periodicity relations
The Periodicity relations follow easily by considering the
involved angles on the unit circle.
$$\def\ts{}\eqalign
{ \sin(\theta)&= \sin(\theta \pm2k\pi) \qquad \csc(\theta)= \csc(\theta \pm2k\pi) \cr
\cos(\theta)&= \cos(\theta \pm2k\pi) \qquad \sec(\theta)= \sec(\theta \pm2k\pi)\cr
\tan(\theta)&= \tan(\theta \pm k\pi)\phantom{2} \qquad \cot(\theta)= \cot(\theta \pm k \pi) \cr
}
$$
$$\eqalign
{ \sin(\theta)&= - \sin(\theta -\pi) \qquad \csc(\theta)= - \csc(\theta -\pi) \cr
\cos(\theta)&= - \cos(\theta -\pi) \qquad \sec(\theta)= - \sec(\theta -\pi) \cr
\tan(\theta)&= - \tan(\theta -\ts{\pi\over2}) \qquad \kern-3pt \cot(\theta)= - \cot(\theta -\ts{\pi\over2}) \cr
}
$$
Pythagorean Identities
The first Pythagorean Identity follows from the Pythagorean Theorem (look at the unit circle). The other two
Pythagorean Identities follow from the first by dividing both sides by the appropriate expression (divide through by $\sin$ or by $\cos$ to obtain the other two).
$$\eqalign
{ \sin^2\theta +\cos^2\theta&=1\cr
1+ \cot^2\theta& =\csc^2\theta\cr
\tan^2\theta + 1& = \sec^2\theta}
$$
Sum and difference formulas
Memorize the first sum and difference formula. The second one can be derived from the first using the fact that $\sin$ is an odd function.
One can then derive the last two sum identities by using the first two and the fact that $\cos(\theta-\pi/2)=\sin\theta$.
$$\eqalign{
\cos(x+y)&=\cos x\cos y-\sin x\sin y\cr
\cos(x-y)&=\cos x\cos y+\sin x\sin y\cr
\sin(x+y)&=\sin x\cos y+\sin y\cos x\cr
\sin(x-y)&=\sin x\cos y-\sin y\cos x\cr
}
$$
Double angle formulas
The Double Angle formulas for $\sin$ and $\cos$ are derived by using the Sum and Difference formulas by writing, for example $\cos(2\theta)=\cos(\theta+\theta)$ and using the Pythagorean Identities for the $\cos$ formula (I suppose the formula for $\tan$ should be memorized).
$$\eqalign{
\sin(2\theta)&=2\sin\theta\cos\theta \cr
\tan(2\theta)&= {2\tan \theta\over 1-\tan^2\theta } \cr
\cos(2\theta)&= \cos^2\theta-\sin^2\theta \cr
&=2\cos^2\theta -1\cr
&=1-2\sin^2\theta\cr
}
$$
Half angle formulas
The Half-Angle formulas for $\sin$ and $\cos$ are then obtained from the Double Angle formula for $\cos$ by writing, for example, $\cos\theta=\cos(2\cdot{\theta\over2})$
The $\tan$ formula here can easily be obtained from the other two.
(Note the forms for the $\cos$ and $\sin$ formulas. These aren't to hard to memorize)
$$\eqalign{
\cos{\theta\over2}&= \pm\sqrt{1+\cos\theta\over2}\cr
\sin{\theta\over2}&= \pm\sqrt{1-\cos\theta\over2}\cr
\tan{\theta\over2}&=\pm\sqrt{1-\cos\theta\over1+\cos\theta}
}$$
(Image from Wikipedia Commons)
Check this picture out! Using the Pythagorean Theorem, you can find the length of the dotted line yourself, so finding $\sin(60)$ and $\sin(30)$ will never be more than a triangle away.
By drawing a 45-45-90 triangle, you can easily get $\sin(45)$ yourself too.
Don't worry about the memorization part: after using these enough, trust me, you won't have to look them up!
(Oh, and if the need arises, you can also get $\sin(15)$ and $\sin(75)$ yourself with the addition formulas. But I've never really used these or even bothered to try to memorize them.)
Best Answer
I always found recalling $e^{ix}=\cos x+i\sin x$ useful for quickly deriving the sum of angles formulae, e.g.
$$e^{i(x+y)}=\cos(x+y)+i\sin(x+y).\tag{1}$$
But
$$e^{ix+iy}=e^{ix}e^{iy}=(\cos x+i\sin x)(\cos y+i\sin y).\tag{2}$$
Expanding (2), equating with (1) and separating real and imaginary parts gives you the formulae.
You can then get the double angle formulae easily.
Wait, there's more!
We have $(e^{ix})^n=(\cos x+i\sin x)^n$.
But we also have $$(e^{ix})^n=e^{inx}=\cos(nx)+i\sin(nx),$$ so we get
$$(\cos x+i\sin x)^n = \cos(nx)+i\sin(nx).$$
For example, consider $n=2$, then expanding gives:
$$\cos^2 x-\sin^2 x = \cos(2x)$$ and $$2\sin x\cos x=\sin(2x).$$
This is another way to get the double angle formulae, but you can get more trig identities by letting $n=3, 4, \ldots$. In general, for positive integer $n$ we have
$$\cos(nx) = \Re\left((\cos x+i\sin x)^n\right) =\Re\sum_{k=0}^n{n\choose k}i^k\cos^{n-k}(x)\sin ^k(x)$$ and $$\sin(nx) = \Im\left((\cos x+i\sin x)^n\right)=\Im\sum_{k=0}^n{n\choose k}i^k\cos^{n-k}(x)\sin^k(x).$$
Expanding and simplifying will give you nice trig identities. This is called De Moivre's Theorem.