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Are there any easy ways or mnemonics to memorize the trigonometric identities like for example $$ \sin(3x) = 3\sin(x) - 4\sin^3(x) $$ I find them quite difficult to come up with, I almost always need to look them up.

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There are some ad hoc tricks but these are highly identity dependent. For instance, $\sin(3x)$ in terms of $\sin(x)$ cannot have $\sin^2(x)$ term since $\sin(3x)$ is odd. Hence, $ \sin(3x) = a \sin(x) + b \sin^3(x)$. $x = \pi/2 \implies a+b = -1$ and $x = \pi/2 \implies a + b/2 = 1$. This gives us $b=-4$ and $a=3$. In general, $\sin((2n+1)x)$ is a polynomial $2n+1$ degree polynomial in $\sin(x)$ with only odd powers. – user17762 Oct 18 '12 at 0:00
In addition to the other responses, it is fairly easy to derive them using the sum-difference trig formulas, so remembering those can be used to derive all other trig identities. – Amzoti Oct 18 '12 at 0:49

For that kind of trigonometric identity, you want to remember and use de Moivre's formula: $$ (\cos x + i \sin x)^n = \cos (nx) + i \sin (nx) $$ You need to be comfortable with complex numbers, though.

For the example you gave, you get $$ (\cos x + i \sin x)^3 = \cos^3 x + 3i \cos^2 x \sin x - 3 \cos x \sin^2 x - i\sin^3 x $$ and so, looking at the imaginary part, $$ \begin{eqnarray} \sin 3x &=& 3 \cos^2 x \sin x - \sin^3 x \\ &=& 3(1-\sin^2 x)\sin x- \sin^3 x \\ &=& 3\sin x -4\sin^3 x \end{eqnarray} $$

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In general, you can manipulate the following definitions of $\sin$ and $\cos$ in order to derive many identities:

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

$$\sin \theta := \frac{e^{i\theta}-e^{-i\theta}}{2 i}$$

$$\cos \theta := \frac{e^{i\theta}+e^{-i\theta}}{2}$$

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