Working at proof derivative cos

[georg gill]

I have tried to formulate proof for derivative of cosine and this is what I ended up with:

Does anyone see any errors? And since I found this proof a bit long, does anyone know any shorter version which proves all the steps involved?

Step1

http://www.viewdocsonline.com/document/4bp3pb

Step2

http://www.viewdocsonline.com/document/8e1qd2

Step3

http://www.viewdocsonline.com/document/tx1cp3

Summary

http://www.viewdocsonline.com/document/5f54l8

 

[jgens]

Skimming through your proof, everything looks correct. If you utilize a geometric definition of sine and cosine, you pretty much have to go through all of the arguments above in order to get the limit relations, so there aren't really any shortcuts from that viewpoint (at least none that I can think of).

If you want a shorter proof of the derivatives, you can always utilize a more sophisticated definition of sine and cosine. For example, with the power series definition it is pretty much immediate (once you have some theorems on differentiating power series of course). Another way that makes things pretty quick is defining sine and cosine via differential equations.

 

[mathman]

You could use the identity cos(x) =(e^ix + e^-ix)/2, so the question reduces to the derivative of e^u.

 

[georg gill]

come to think of it one could also use L'hopital's on

$$\frac{sinh}{h}$$ as h goes to zero

Should not be any reason not for that?

 

[Mute]

come to think of it one could also use L'hopital's on

$$\frac{sinh}{h}$$ as h goes to zero

Should not be any reason not for that?

To use L'Hopital's rule you need to know that the derivative of sine is cosine, but to prove that the derivative of sine is cosine you have to be able to evaluate that limit, so it would be a circular argument.

 

[epenguin]

This is the second time this question has been posed here in a couple of weeks, see https://www.physicsforums.com/showpost.php?p=3617881&postcount=1 https://www.physicsforums.com/showthread.php?t=550860&highlight=Prove+the+differentiation+rule .

Yes I think there is a shorter version than this. The standard version given me and I think most people at school depends on the sine or cosine of sums of angles formulae which are not exactly self-evident and in fact an unnecessary slog. I propose this:

As the angle δx → 0 , the arc δx approaches ever closer to a very short straight line segment which I represent by ab in the Figure. Then our conclusion follows from the similarity of the little triangle oab to the large one OAB

http://img811.imageshack.us/img811/7476/dsincosproof042.jpg



$$\frac{d\, sin x}{dx} = \frac{\mid oa \mid}{|ab|} = \frac {OA}{AB} = cos\, \angle OAB = cos \,x$$

Quite similarly

$$\frac{d\, cos x}{dx} = \frac{-\mid ob \mid}{|bc|} = \frac {-\,OB}{AB} = -\, sin\, \angle OAB = -\,sin\,x$$

I think that is all that is necessary.