Mnemonic for derivative/integral of $\sin x$ and $\cos x$ I'd love to know if anyone has a good mnemonic for answers of the following:
$$\frac{\mathrm{d}}{\mathrm{d}x} \, \sin x$$
$$\frac{\mathrm{d}}{\mathrm{d}x} \, \cos x$$
$$\int \sin x \,\mathrm{d}x$$
$$\int \cos x \,\mathrm{d}x$$
I know the first one by heart, and derive the others from it. This sometimes takes me up to five seconds, longer if I'm not really thinking clearly! Does anyone have a good mnemonic for the answers to these common occurrences?
 A: In the trigonometric unit circle. Differentiation is clockwise and integration is anticlockwise. For example if you were to search for the derivative on $\sin(x)$ then clockwise next is the derivative which is $\cos(x)$. Similarly, the antiderivative of $-\sin(x)$ is found by going anticlockwise on the unit circle, and we get $\cos(x)$.

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A: Vulgar, but how I remember it:
$$ \begin{matrix}
  D & I \\
  C & S
 \end{matrix}~~~~~~~~\text{ are negative }
$$
Derivative/integral of cosine and sine are negative. 
A: $$\large  \sin(x) \xrightarrow{\huge \frac{\text{d}}{\text{d}x}} \cos(x) \xrightarrow{\huge \frac{\text{d}}{\text{d}x}} -\sin(x) \xrightarrow{\huge \frac{\text{d}}{\text{d}x}} -\cos(x) \xrightarrow{\huge \frac{\text{d}}{\text{d}x}} \sin(x)$$
$$\\\\$$
$$\large \sin(x) \xrightarrow{\huge \int} - \cos(x) \xrightarrow{\huge \int} -\sin(x) \xrightarrow{\huge \int} \cos(x) \xrightarrow{\huge \int} \sin(x)$$
$$\\\\$$
(leaving apart the constants of integration).
A: I have three approaches:


*

*If I am in a period in life when I’m using them a lot, I just remember the cos is a bad guy and bad guy gives minus.

*But when I for some reason forget derivatives I derive them with simple logic in my head. I know by heart that $\sin$ and $\cos$ can be written in terms of exponents, and that $\cos$ is symmetric and $\sin$ is anti-symmetric. So I can immediately remember $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.$$Derivative of exponent is very simple and I already know the result, just prefactor is in question. I then in head just see how when I put imaginary number in denominator I get minus sign.
$$\frac{d}{dx}\frac{e^{ix}+e^{-ix}}{2}=i\frac{e^{ix}-e^{-ix}}{2}=-\frac{e^{ix}-e^{-ix}}{2i} $$
So when you get the logic, all that remains, is to remember that when doing derivative of $\cos$ you get $i$ in the numerator that you have to put in the denominator.

*I came up with this one just now, when I was thinking about the answer, and it seems great to me. All you have to remember is that if you write the matrix
$$\begin{pmatrix}
 f_{1}(x)=&f_{1}'(x)\\
 f_{2}(x)=&f_{2}'(x)\\
 \end{pmatrix}=\begin{pmatrix}
 \cos(x)&-\sin(x)\\
 \sin(x)&\cos(x)\\
 \end{pmatrix},$$
you get $\cos(x)$ on the diagonal. Because all you have to remember is a sign in front of the functions by this information you already know everything. Even more, identities for integrating are also already there. And even more, now you also have a very nice trick for memorizing rotational matrix, because
$$R(x)=\begin{pmatrix}
 \cos(x)&-\sin(x)\\
 \sin(x)&\cos(x)\\
 \end{pmatrix}$$
is a matrix that describes rotation of a vector for angle $x$.
$$\begin{pmatrix}
 \cos(\phi)&-\sin(\phi)\\
 \sin(\phi)&\cos(\phi)\\
 \end{pmatrix}\begin{pmatrix}
 x\\
 y\\
 \end{pmatrix}=\begin{pmatrix}
 x\cos(\phi)-y\sin(\phi)\\
 x\sin(\phi)+y\cos(\phi)\\
 \end{pmatrix}.$$
It works also in the opposite direction. If you know rotational matrix by heart, you just have to remember that the fist block describes functions and the second one derivatives of them.
