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Proof of an elementary but very important result

  1. Dec 14, 2011 #1

    dextercioby

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    Alright, I know I'm kind of shaky when it comes to proving elementary things rigorously.

    I'm looking for for the most elementary (in the sense of using least possible mathematical knowledge) proof that

    [tex] x < \tan x, {} {} \forall \frac{\pi}{2}>x>0 [/tex]

    Geometrically ? I can't prove it...:redface: Algebraically ? How ? Using calculus ? I wish to see a proof to that, too...

    You can also post links to freely available resources, if you know them.

    EDIT: To make sense of post #2, the [itex] \pi/2 [/itex] was missing when ILS wrote his post.
     
    Last edited: Dec 14, 2011
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  3. Dec 14, 2011 #2

    I like Serena

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    Umm :uhh:... [itex]\pi > tan \pi = 0[/itex]


    I'm not really sure what you're looking for, but here's one proof for 0<x<pi/2...

    According to the mean value theorem, for every 0<x<pi/2 there has to be a 0<c<x such that:
    tan' c = (tan x - tan 0)/(x-0).

    Since tan'c = 1/cos²(c) > 1, it follows that tan x/x > 1, and therefore x < tan x.
     
  4. Dec 14, 2011 #3

    AlephZero

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    Another calculus proof. Let [itex]f(x) = \tan x - x[/itex]

    [itex]f(0) = 0[/itex] and [itex]f(x)[/itex] is continuous on [itex][0 , \pi/2)[/itex]

    [itex]f'(x) = \sec^2 x - 1 = \tan^2 x >= 0[/itex]

    So f(x) is monotonically increasing on [itex][0 , \pi/2)[/itex]

    Neither I nor ILS proved that [itex]\tan x[/itex] is continuous on [itex][0 , \pi/2)[/itex], but that proof depends on what you think is an "elementary" definiton of trig functions.
     
  5. Dec 14, 2011 #4

    dextercioby

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    I apologize, I forgot about ranges. Now it is correctly formulated.

    @ILS Your attempt should be a calculus version. However, how do you prove that the tangent function is continuous (you need this to use the theorem) and how do you prove that the derivative of tangent is what it is ?
     
    Last edited: Dec 14, 2011
  6. Dec 14, 2011 #5

    dextercioby

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    The proof of continuity is assumed to be the one with delta and epsilon. Tangent is defined as the ratio of sine and cosine which are in term defined as sides in a right triangle inside the unit circle => the name (circular) trigonometric functions.
     
  7. Dec 14, 2011 #6

    micromass

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    The proof of what you want occurs in this khan academy video:

    http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus [Broken]

    The inequality occurs in the first 8 minutes.

    I don't think it's possible to prove this using calculus. Because, if you use calculus, you should already know about the continuity and differentiability of the tangent function. But to prove this, you use the inequality of your OP!!

    Unless of course you define the tangent as a power series. Which is possible. But then the proof is trivial. But it is less trivial that it actually corresponds to the tangent function of trigonometry.
     
    Last edited by a moderator: May 5, 2017
  8. Dec 14, 2011 #7

    dextercioby

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    Alright, Micro, I found it on You tube, too.



    I was soo easy in the end to look for it, I mean. Anyways, it's nice to bring forth things that are seldom overlooked and taken as elementary.
     
    Last edited by a moderator: Sep 25, 2014
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