Prove the differentiation rule: (cosx)' = sinx for any x.

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Homework Help Overview

The discussion revolves around proving the differentiation rule for the cosine function, specifically that the derivative of cos(x) is -sin(x). Participants explore various methods to approach this proof, including the limit definition of the derivative and trigonometric identities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants suggest starting with the definition of the derivative, while others question the validity of the original statement regarding the derivative of cosine. There are mentions of using trigonometric identities and the chain rule, with some expressing concerns about circular reasoning in the arguments presented.

Discussion Status

The discussion is active, with participants offering various perspectives on how to approach the proof. Some guidance has been provided regarding the use of limits and trigonometric identities, but there is no explicit consensus on the best method to use.

Contextual Notes

Participants note that the approach to proving the differentiation rule may depend on how cosine is defined, with references to definitions involving the unit circle and power series. There are also mentions of specific limits and identities that may be relevant to the proof.

  • #31
Halls, why would it require an integral ? The angle in the right triangle expressed in radians instead of degrees is defined as the length of the circular segment. The 'height' of the triangle is the number <sine x>, when x is the angle at center given in radians.

Why do you think there's a need for an integral ?
 
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  • #32
HallsofIvy said:
Indeed that integral requires integrating 1/\sqrt{1- x^2} which is undefined at that starting point, x= 1, y= 0. It is an improper integral and just proving the integral exists require some use of sine and cosine. So trying to define sine and cosine in that way is definitely circular!
I don't think that's accurate. See disregardthat's post #79 in the thread where we discussed these things in August: Link.
 
  • #33
More calculus without tears.

I don't know why I have never seen it done this way which makes it more obvious and expected. Hasn't anyone else?
2n28oz.jpg

Here is a unit circle.
The vertical height (BA) of A tends to sin x.

δx is the length of arc increment: as it gets small it gets as close as we please to a straight line segment.

So δ(sin x)/δx becomes the ratio of the vertical side to the hypotenuse of the little triangle Abc. Which is similar to the large triangle ABC. So it equals the corresponding ratio there (AB/AC) which is cos x (or tends to it as close as we please as we make δx as small as we please).

I did (sin x)' here; you can get d(cos x)/dx in the same way, in fact you could get it from the same figure, maybe I should put one more line in.

..
 
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  • #34
  • #35
This is pretty amazing!

We're now at the 35th post in this thread and OP has not replied even once !

Well, it has provided some interesting debate --- or whatever it might be called.
 

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