How to remember trig identities?

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.

I find that four suffice. $$\cos^2 (x) + \sin^2(x) = 1 \tag{1}$$ $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) \tag{2}$$ $$\sin(a+b)=\sin(a)cos(b)+\sin(b)\cos(a) \tag{3}$$ $$\text{trig}(x) = \text{cotrig}(\frac{\pi}{2}-x) \tag{4}$$

The Pythagorean identity $(1)$ is easy to manipulate. Divide through by $cos^2(x)$ alternatively by $sin^2(x)$ to find the other forms

$$1 + \tan^2(x) = sec^2(x) \tag{5}$$ $$\cot^2(x) +1 = csc^2(x) \tag{6}$$

For the angle addition formulas $(2)$ and $(3)$, we can apply odd and even identities to quickly derive the angle subtraction identities:

$$\cos(a+\color{red}{(-b)})=\cos(a)\cos(\color{red}{(-b)})-\sin(a)\sin(\color{red}{(-b)})$$ $$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) \tag{7}$$

and

$$\sin(a+\color{red}{(-b)})=\sin(a)cos(\color{red}{(-b)})+\sin(\color{red}{(-b)})\cos(a)$$ $$\sin(a-b)=\sin(a)cos(b)-\sin(b)\cos(a) \tag{8}$$

We can also let $a=b$ and substitute in $(2)$ and $(3)$ to find double angle forms

$$\cos(a+a)=\cos(a)\cos(a)-\sin(a)\sin(a)$$ $$\cos(2a)=\cos^2(a)-\sin^2(a) \tag{9}$$

$$\sin(a+a)=\sin(a)cos(a)+\sin(a)\cos(a)$$ $$\sin(2a)=2\sin(a)cos(a) \tag{10}$$

Combining $(9)$ with the Pythagorean identity $(1)$ gives two more.

$$\cos(2a)=1-2\sin^2(a) \tag{11}$$ $$\cos(2a)=2\cos^2(a)-1 \tag{12}$$

If you want a half angle formula, you may as well let $u=2a$ in the previous four identities. Just mind your squares and roots. What happening to tangent? Divide any related $sin$ by $cos$ to get what you need.

Let's not forget product to sum identities! If we take $(2)$ and $(7)$ and add the equation, we find

$$\cos(a+b) + \cos(a-b)=2\cos(a)\cos(b)$$ $$\cos(a)\cos(b) = \frac12 (\cos(a+b) + \cos(a-b)) \tag{13}$$

Similar combinations will give the remaining product to sum identities.

As for $(4)$, $trig(x) = cotrig(\frac{\pi}{2}-x)$, I'm referring to cofunction identities, which all have the same form. For example, $\sin(x) = \cos(\frac{\pi}{2}-x).$ That's essentially six more identities.

We have over twenty identities at our disposal now, including the few that I've mentioned but neglected to outright derive.