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why is $$\frac{d}{dx}\cos(x)=-\sin(x)$$ I am studying for a differential equation test and I seem to always forget \this, and i am just wondering if there is some intuition i'm missing, or is it just one of those things to memorize? and i know this is not very differential equation related, just one of those things that I've never really understood.

and alliteratively why is $$ \int\sin(x)dx=-\cos(x)$$ any good explanations would be greatly appreciated.

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What I do is remember that one is the derivative of the other one and vice versa, modulo a negation somewhere, and then I draw (or imagine) a quick sketch of the graphs whenever I need to remember where the minus sign goes. –  Henning Makholm Nov 3 '11 at 1:26
Related question: math.stackexchange.com/questions/392/… –  leonbloy Nov 3 '11 at 1:36
Write sin(x) in it power series form and then differentiate it. –  simplicity Nov 3 '11 at 1:45
Recall that $\cos x=\sin(\pi/2-x)$ and $\sin x=\cos(\pi/2-x)$. Differentiate both sides of $\cos x=\sin(\pi/2-x)$. We get $(\cos x)'=-\cos(\pi/2-x)=-\sin x$. This is not an answer, since I doubt it is a helpful memory aid. But it is an excuse to mention the useful fact that $\cos x=\sin(\pi/2-x)$. –  André Nicolas Nov 3 '11 at 1:54

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I am with Henning Makholm on this and produce a sketch

enter image description here

The slope of the blue line is the red line, and the slope of the red line is the green line. I know the blue line is sine and the red line is cosine; the green line can be seen to be the negative of sine. Similarly the partial area under the red line is the blue line, though this sort of approach leads to $\int \sin(x)dx=1-\cos(x)$.

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Here is a geometric interpretation that is easy to remember: the unit circle is parametrized by $(\cos t, \sin t)$ and hence its tangent vector is orthogonal to the position vector. Rotating the position vector by 90 degrees gives you $(-\sin t, \cos t)$ and so $\cos'=-\sin$ and $\sin'=\cos$.

This argument has a simple interpretation in terms of complex numbers. $\exp(it) = \cos t + i\,\sin t$ and so $i \exp(it) = \exp(it)'= \cos' t + i\,\sin' t$. But $i \exp(it) = -\sin t +i\,\cos t$.

These two arguments are the same, really.

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Well if you find trouble remembering them maybe you can use the formulas $$\cos{x} = \frac{e^{ix} + e^{-ix}}{2} \quad \sin{x} = \frac{e^{ix} - e^{-ix}}{2i}$$

which you can get from Euler's formula and then if you forget what the derivative of $\cos{x}$ or of $\sin{x}$ is then you can just differentiate those exponentials, which is rather easy.

And I guess that you will not have problems remembering what the derivative of the exponential is.

Also using the power series representations for the sine and the cosine you can differentiate them term by term and verify easily that $(\cos{x})' = -\sin{x}$ and $(\sin{x})' = \cos{x}$.

But in any case, depending on how you define the trigonometric functions, there may be different ways to prove that each derivative is what it is.

I hope this helps a little.

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Pause before you answer the question. The derivative of sine is (....) cosine. The integral of sine is (....) minus the cosine. The derivative of cosine is (....) minus the sign. The integral of cosine is (...) sine. Never say the answer without the pause. Think of your best reason for the s-i-g-n (several given above, or the cosine curve begins by decreasing), and put that thought into the pause. Thereby train yourself to remember.

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