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Proof limit derivatave of sin(x)

  1. Aug 9, 2015 #1
    Dear PF Forum,
    Continuing our debate discussion in differential in slice of X.
    I read this particulare website. About proofing the derivative of sine(x).
    In there, the web writes

    Code (Text):

    arc AC < |AB| + |BC|

           < |AB| + |BD|
           = |AD|

    I don't know what the "equal sign" means.
    Is this?
    Code (Text):

    arc AC < |AB| + |BC|
    arc AC < |AB| + |BD|
    arc AC = |AD|

    or this?
    Code (Text):

    arc AC < |AB| + |BC|
    arc AC < |AB| + |BD|, where |AB| + |BD| = |AD|

  2. jcsd
  3. Aug 9, 2015 #2
    Your first interpretation is correct. Get used to that kind of terse writing, it's pretty common.
  4. Aug 9, 2015 #3
    Which one?
    arc AC < |AB| + |BC|
    arc AC < |AB| + |BD|
    arc AC = |AD|,

    but arc AC can't be |AD|. This is my instinct. I just one to know exactly what it means, because in math, a little misinterpretation can lead to big mistake, more with "proof"
  5. Aug 9, 2015 #4

    Moderator note: The answer above is wrong.
    Last edited by a moderator: Aug 9, 2015
  6. Aug 9, 2015 #5
    Ok, I understand now.
    I read subsequent paragraphs, it's just the way this website formats its answer.
    It is.
    arc AC < |AB| + |BC|
    arc AC < |AB| + |BD|, where |AB| + |BD| = |AD|
  7. Aug 9, 2015 #6
    I remember vaguely learning about this. I think they consider that equality for extremely small values of theta. In this case, the limit as theta goes to 0.
  8. Aug 9, 2015 #7
    No, Mirero, it's just the proof. But
    is tens paragraphs below. It's just the way the web produces its answer. And yes, as theta goes to zero, then arc sin AC = |AD|. But it's not funny to put your answer in the first paragraph.
  9. Aug 9, 2015 #8
    Sorry, I just looked at the diagram again and your interpretation seems to be right. That kind of writing seems weird to me though, and is certainly not conventional.
  10. Aug 9, 2015 #9
    You should see this:
    This should be arranged to

    ##\theta = \text{arc AC}##
    ##\tan\theta = \frac{\sin \theta}{\cos\theta}##
    ##\text{arc AC} < \tan \theta## or
    ##\text{arc AC} < \frac{\sin \theta}{\cos \theta}## or
    ##\theta < \tan \theta## or
    ##\theta < \frac{\sin \theta}{\cos \theta}##
    And the web simply writes
    ##\theta = \text {arc AC} < \tan \theta = \frac{\sin \theta}{\cos \theta}##
  11. Aug 9, 2015 #10
    Actually, I would prefer what was written on the website in that case :P

    Seems much more succint in my opinion.
  12. Aug 9, 2015 #11


    Staff: Mentor

    NO!!! The second version is the correct one, not the first.

    Version B says this:
    arc AC < |AB| + |BC| < |AB| + |BD| = |AD|

    In short, arc AC < |AD|
  13. Aug 9, 2015 #12
    Yea, I sorta realized that a little too late. I'm much more used to reading flows of equations of the first form, and I guess I do that automatically now. Guess I should be more careful instead of haphazardly going off on my instincts next time. :H
  14. Aug 9, 2015 #13
    Yeah, that's just the way the website format its output. You of all people know Mark44 :smile: that a little incorrect symbol in math will translate a very different thing. Especially for the one who tries to learn mark. But I've solved this problem, it's just how it formats its output.
  15. Aug 9, 2015 #14


    Staff: Mentor

    As already discussed, no, this is not the same as what's at the top here. Notice that at the top, "arc AC" appears only once, not three times, as you have it just above.

    Mixing inequality symbols with equal signs is very common in math texts. For a chain of expressions like this one --
    Code (Text):
    arc AC < |AB| + |BC|
           < |AB| + |BD|
           = |AD|
    -- the < sign on the 2nd line refers back to the right side of the previous line. The = sign on the 3rd line refers to the previous (2nd) line.
  16. Aug 9, 2015 #15
    the proof is fairly easy, when you think in terms of geometry. We construct the diagram using a unit circle. Therefore the radius is 1.

    We have 3 triangles in the diagram. Think of area 1< area 2< area 3.

    I can go further.. However i'll give you a hint. what area formulas do you know? Then it is simple algebraic manipulation. Very trivial. Hint. Area of sector is needed.

    Another key part in the proof. I won't give the detail. But it is along the same lines.

    x^2> x.

    however (1/x)> (1/(x^2))
    Last edited: Aug 9, 2015
  17. Aug 9, 2015 #16
    This post will not be helpful for the OP, so I suggest that he skips it. This is mainly intended for MidgetDwarf.

    This proof is one of those proofs which are very misleading. It looks very easy, and which has very simply ideas. But when you start thinking deeper, all hell breaks loose. The main problem I always had with this proof is that it used geometry. The problem with geometry is that it is never fleshed out in calculus or analysis textbooks, so strictly speaking, you shouldn't be using it until you do flesh it out. But what the specific problem is, is that you need to have a theory of "areas". Now areas appear very simple. But it is very tricky to define it. In fact, I conjecture that areas such as the area of a circle appear so simple because we have been told this over and over again in elementary school until it was second nature to us. But there is nothing simple about the area of a circle.

    Let's start with the question how we would define the area of a circle (or a more general shape). Well, calculus actually gives a very nice answer: the Riemann integral. Yeah, we can just define the area of (half of) a circle to be ##\int_{-1}^1 \sqrt{1-x^2}dx##. OK, but how do we evaluate this? Ah, easy, we use substitution ##x = \sin(u)##. But for that, don't we first need to know that the derivative of the sine equals the cosine? Hmm, yes, we do. But we can easily prove that! Indeed, most calculus books show that by first showing that ##\lim_{h\rightarrow 0} \frac{\sin(h)}{h} = 1##. And how do we prove that? Well, the proof is in the OP, we just compare the area of a circle with... Hmm, so we enter in a circular argument this way...

    So yes, I hope I convinced you this way that the proof of ##\lim_{h\rightarrow 0} \frac{\sin(h)}{h}= 1## is not as straightforward as it appears. Now there are essentially two reactions you can have:
    1) Wow, why are you being so pedantic. We all know that areas work the way they do. I don't see any reason to doubt it, you can see it clearly!
    2) Hmm, this is really annoying me. I must find out more.
    Both reactions are acceptable. I agree that I am being pedantic. But it is only (2) which characterizes the mathematician.
  18. Aug 9, 2015 #17
    Hmm, very insightful post. To me it seemed straight forward because I can "feel" the geometry so to speak. Yes, I can understand your point of how geometry is not put in rigorous footing in these classes.

    So, what would be an adequate proof of lim sinx/x? Can you inbox me. I feel it will derail the intention of this thread.
  19. Aug 9, 2015 #18
    Well, it is straightforward, except for one little thing: determining the area of a circle. This is highly nontrivial! I will PM you to talk about this further. Anybody else who wishes to discuss this, feel free to PM me or to make a new thread.
  20. Aug 10, 2015 #19


    User Avatar
    Science Advisor

    Why is arc AC < |AB| + |BC|? It appears to be taken on face value. How do you prove it? Certainly not by naively calculating the length of the arc and comparing, as this would require knowing the derivatives of the trigonometric functions, which are derived from the fact that ## \lim_{x \to 0} \frac{\sin(x)}{x} = 0##. This goes to show how this proof is actually not that rigorous, it requires a solid definition of arc length, or more generally continuous curve length. There is quite a bit of additional work that will go into proving this inequality.
    Last edited: Aug 10, 2015
  21. Aug 10, 2015 #20
    Ah, I'm too busy, I just read your post. Perhaps it is not useful, but I'll read it anyway. See if there's something that I can learn from this. :smile:
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