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

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SUMMARY

The differentiation rule stating that the derivative of cosine, (cos x)', equals -sin x for any x is confirmed through various methods discussed in the forum. The primary approach involves using the definition of the derivative and applying trigonometric identities and limits. Key techniques include the limit properties of sin and cos as h approaches zero, and the application of the chain rule when necessary. The discussion emphasizes that the derivative of cos x is -sin x, not sin x, which is a common misconception.

PREREQUISITES
  • Understanding of the definition of a derivative
  • Familiarity with trigonometric identities, particularly for sine and cosine
  • Knowledge of limit properties, specifically lim (sin h)/h as h approaches 0
  • Basic understanding of the chain rule in calculus
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  • Study the formal definition of a derivative in calculus
  • Learn about trigonometric limits and their applications in differentiation
  • Explore the use of Taylor series for approximating sine and cosine functions
  • Investigate the implications of the existence and uniqueness theorem for differential equations
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Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of trigonometric differentiation and its applications in calculus.

  • #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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