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I'm trying to understand in a practical/graphical view the derivative of $\sin(x)$ (that results in $\cos(x)$).

Is there any animation or illustration explaining that?

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marked as duplicate by Hans Lundmark, Asaf Karagila, TZakrevskiy, Aaron Maroja, Jack D'Aurizio Mar 7 at 20:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I don't know of any animations so you might have to do it yourself. Draw a nice sine curve, draw tangents to it at lots of places, approximate their gradients, and plot those values on another graph. It will look (very satisfyingly if you do it well) much like a cosine curve. –  Ragib Zaman Jul 18 '12 at 15:47
I know this isn't graphical, but if you're trying to understand why $\frac{d}{dx}\sin(x) = \cos(x)$ instead of just accepting this as "the rule", then why not expand sine in a Taylor series and differentiate term by term? Then see what you have left. –  Derek Allums Jul 18 '12 at 15:52
Please check this demonstration for graphical "proof" of $\sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$. This readily gives the derivative, putting $\alpha=x$ and $\beta = \Delta x$. –  Sasha Jul 18 '12 at 15:54

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up vote 1 down vote accepted

MIT OCW's single variable calculus course has a interactive mathlet explaining the derivates of sines and cosines (and few others) graphically. Please refer to the worked example at here.

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Wondeful! By the way, I think you mean "refer to the Mathlet". Thanks! –  Tom Brito Jul 18 '12 at 17:49

Think in polar coordinates.

Draw a circle centered at the origin. Pick an angle $\theta$ on the circle. Draw a ray between the origin and the circle with that angle.

Draw the tangent line on the circle there.

Now, draw a line through the origin parallel to that tangent line. Where does it intersect the circle? At exactly $\cos(\theta)$.

This works because the tangent line on the circle is normal to the ray leading to that point. That means that the line we draw through the origin also intersects that ray at right angles.

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