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Working at proof derivative cos

  1. Dec 11, 2011 #1

    georg gill

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  2. jcsd
  3. Dec 11, 2011 #2

    jgens

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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.
     
  4. Dec 11, 2011 #3

    mathman

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    You could use the identity cos(x) =(eix + e-ix)/2, so the question reduces to the derivative of eu.
     
  5. Dec 12, 2011 #4

    georg gill

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

    [tex] \frac{sinh}{h}[/tex] as h goes to zero

    Should not be any reason not for that?
     
  6. Dec 12, 2011 #5

    Mute

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    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.
     
  7. Dec 15, 2011 #6

    epenguin

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    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 [Broken]



    [tex]\frac{d\, sin x}{dx} = \frac{\mid oa \mid}{|ab|} = \frac {OA}{AB} = cos\, \angle OAB = cos \,x[/tex]

    Quite similarly

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

    I think that is all that is necessary.
     
    Last edited by a moderator: May 5, 2017
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