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

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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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HallsofIvy said:
Indeed that integral requires integrating [itex]1/\sqrt{1- x^2}[/itex] 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.
 
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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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.