Any Errors in My Proof for Derivative of Cosine?

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The discussion centers on the proof for the derivative of cosine, with participants reviewing the original proof and suggesting alternatives. The initial proof is deemed correct, but it is noted that using geometric definitions of sine and cosine requires thorough steps without shortcuts. Suggestions for shorter proofs include utilizing power series, differential equations, or L'Hôpital's rule, though the latter may lead to circular reasoning. One participant proposes a geometric approach based on the similarity of triangles, which simplifies the proof significantly. Overall, the conversation highlights various methods to derive the derivatives of sine and cosine efficiently.
georg gill
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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.
 
You could use the identity cos(x) =(eix + e-ix)/2, so the question reduces to the derivative of eu.
 
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?
 
georg gill said:
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.
 
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.
 
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