Sine and Cosine Derivatives The derivatives of:

$$\frac{d}{dx}\sin(x)=\cos(x)$$
  $$\frac{d}{dx}\cos(x)=-\sin(x)$$

I currently trying to teach a friend of mine calculus, because he does not know it yet.  He keeps forgetting how to take the derivatives of $\sin(x)$ and $\cos(x)$.  Is there a simple way, or trick to remember the derivatives of $\sin(x)$ and $\cos(x)$?
 A: I prefer to use unit circle (for almost everything in trigonometry)
A simple way to memorize these formula is concentration on increasing & decreasing in their plots.
I highly recommend using a good applet. Visualizing  is  very helpful
Try these to find a suitable applet. I suggest this one for example.
A: If you've memorized the graphs of sine and cosine, and know that the derivatives are also some form of sine and cosine, then use the fact that the derivative at any point is the slope of the graph at that point. 
For example, the slope of the sine function at $x=0$ is $1$, and it falls to zero at $x=\pi/2$. This should remind you of the cosine function. Similarly the slope of the cosine function at $x=0$ is zero, and it decreases to $-1$ at $x=\pi/2$, leading to the negative of the sine function. This geometric interpretation should get you the signs right as well as the form of the function.
A: One way you can remember them that is fairly easy, so long as you remember the shapes of their graphs, is to recognize that the slope of $\sin x$ at $0$ is positive (it looks like it's $1$), then tends to $0$ at $\pi/2$, then becomes negative (it looks like it's $-1$) at $\pi$, then becomes $0$ again at $3\pi/2$ and finally ends up positive at $2\pi$. If you "connect the dots", you will see that the graph is that of $\cos x$. A similar procedure applies for remembering the derivative of $\cos x$.
A: First we should remember that these two functions are derivative one for other, except the sign. Now we want to determine the sign.
The derivative in a point is "how fast does the function rise". $\sin(x)$ is rising at $x=0$ so $(\sin(x))'$ must be positive at $x=0$. OK, $\cos(x)$ is suitable. The same way for $\cos(x)$: it doesn't change near $0$, then decreases. So $(\cos(x))'$ must be null at $x=0$, then negative. $-\sin(x)$ is suitable.
Of course it is not a way to find the derivative, just to remember.
A: Using complex numbers:
$$\cos x+i\sin x=e^{ix}$$
then
$$\cos'x+i\sin'x=(e^{ix})'=ie^{ix}=-\sin x+i\cos x.$$
By this method you can obtain the derivatives for higher orders
$$\cos''x+i\sin''x=i^2e^{ix}=-\cos x-i\sin x,$$
$$\cos'''x+i\sin'''x=i^3e^{ix}=\sin x-i\cos x,$$
 and so on.

Alternatively, if you give the angle $x$ a small increment, the corresponding point on the trigonometric circle will move in the direction of the tangent to the circle, which is orthogonal to the position vector.
$$(\cos x,\sin x)\to(-\sin x, \cos x).$$

A: Not that I know of.  Practice, practice, and then more practice.  That's how I've remembered them since 2001.  (Teaching also helps.)
Only slightly related to your question, here's a nice little factoid:

\begin{alignat}{3}
  \frac d{dx} \sin x &= \cos x & \qquad\qquad \frac d{dx} \cos x &= -\sin x\\[0.3cm]
  \frac d{dx} \sec x &= \sec x \tan x & \frac d{dx} \csc x &= -\csc x \cot x\\[0.3cm]
  \frac d{dx} \tan x &= \sec^2 x & \frac d{dx} \cot x &= -\csc^2 x\\[0.3cm]
\end{alignat}

The derivatives of all the functions beginning with "co" have minus signs on them.  The others do not.
