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Proof Involving MVT?

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove for all real x and y that
    |sinx - siny| <= |x-y|


    2. Relevant equations
    It's a question from the Mean Value Theorem/Rolle's Theorem section.


    3. The attempt at a solution
    Honestly, I've tried. It looks somewhat similar to the triangle inequality (I think?), but truth be told I can't get anywhere with this.



    I'd appreciate if anyone could give me a hand and point me in the proper direction. Thanks!
     
  2. jcsd
  3. Jan 6, 2010 #2

    jgens

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    Gold Member

    Well, what does the mean value theorem say?
     
  4. Jan 7, 2010 #3
    Reading the hotlink I still fail to see the connection with MVT to the problem. Could the "real values x and y" have anything to do with the (a, b) interval?
     
  5. Jan 7, 2010 #4

    jgens

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    Gold Member

    Well, what do you think?
     
  6. Jan 7, 2010 #5
    Well, my guess is it does but I still fail to see any connection.

    [itex]| \sin \displaystyle x- \sin \displaystyle y| = 2 \left| \cos \left( \frac{\displaystyle x+ \displaystyle y}{2} \right) \sin \left( \frac{\displaystyle x- \displaystyle y}{2} \right) \right| [/itex]

    Am I getting anywhere with this?
     
  7. Jan 7, 2010 #6
    How about you let [tex] x = b[/tex] and [tex] y = a[/tex]?
     
  8. Jan 7, 2010 #7
    Hmm, ok, thank you.

    So I get now

    |sinb - sina|
    __________ <= 1
    | b - a|

    Which is similar to the theorem f(b)-f(a) / b-a ?

    If it is, this basically says that the equation I came up with (which is the derivative) is equal to 1?

    Can I work on this further by taking the derivative and getting cosb - cosa?


    Sorry if I'm coming off as a bit.. stubborn. This question is really confusing me.
     
  9. Jan 7, 2010 #8
    You're close.
    It's not exactly the derivative. I mean, it is the derivative at point c between a and b.

    So, we have the equation in the form of
    [tex]
    \frac{\sin b - \sin a}{b - a} = \cos c
    [/tex]
    As for your derivative idea, no. Derivative of sin a would be 0 because sin a is a constant. Instead, consider the following:
    What do we know about [tex]\cos[/tex]'s range?
     
  10. Jan 7, 2010 #9
    Aha! I see it now. The range of cos is between 1 and -1, so it cannot exceed 1... I think I see that now.

    Just one question, how did you find "cosc"?

    Thank you very much.
     
  11. Jan 7, 2010 #10
    Once more, look up "Mean Value Theorem". It's all in there.
     
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