How to remember the trigonometric identities

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)

$$\eqalign{ \cos{\theta\over2}&= \pm\sqrt{1+\cos\theta\over2}\cr \sin{\theta\over2}&= \pm\sqrt{1-\cos\theta\over2}\cr \tan{\theta\over2}&=\pm\sqrt{1-\cos\theta\over1+\cos\theta} }$$

My favourite trick: I don't remember any of them. :-) The only thing I have in mind is that this matrix

$$ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} $$

rotates vectors in the plane by an angle $\theta$ and matrix multiplication is the same as composition. Hence, you have identities like

$$ \begin{pmatrix} \cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} $$

from which it follows

$$ \cos(2\theta) = \cos^2\theta - \sin^2\theta $$

and

$$ \sin(2\theta) = 2\sin\theta\cos\theta \ . $$

Alternatively, as yoyo says, you could use Euler's identity,

$$ e^{i\theta} = \cos\theta + i \sin\theta $$

to find, for instance, that

$$ \cos(\theta + \phi) + i\sin(\theta + \phi) = e^{i(\theta + \phi)} = e^{i\theta}e^{i\phi} = (\cos\theta + i\sin\theta) (\cos\phi + i\sin\phi) \ . $$

Hence,

$$ \cos(\theta + \phi) = \cos\theta\cos\phi - \sin\theta\sin\phi $$

and

$$ \sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi \ . $$