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Math Help - Limits

  1. #1
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    Limits

    Can you please tell me a geometrical method to prove that 0 < |sin(x)| < √2(1-cos(x)) < |x| ?
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  2. #2
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    Re: Limits

    The inequality it false for x=2. The easiest way to check an inequality before attempting to prove it is to google search the function. I google searched sqrt(2)*(1-cos(x))-abs(x) to check that part of the inequality and found a value for x when it was not less than zero.
    Also, this does not belong in the new users forum
    Last edited by Shakarri; August 5th 2013 at 09:07 AM.
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  3. #3
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    Re: Limits

    Do you mean \sqrt{2}(1- cos(x)) or \sqrt{2(1- cos(x))? What you wrote is the first but, as Shakarri said, that is simply not true.
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  4. #4
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    Re: Limits

    Your inequality signs should be less than or equal. At x=0 you get 0<0<0<0
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  5. #5
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    Re: Limits

    Yeah I'm really sorry I forgot to mention that x belongs to (-pi/2 , pi/2 ) \ {0}
    And what I meant was \sqrt{2(1- cos(x))
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  6. #6
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    Re: Limits

    Well I don't know a geometric proof but to prove |sin(x)| < √(2(1-cos(x))) you can do it this way
    |sin(x)|<\sqrt{2(1-cos(x)}

    sin^2(x)<2(1-cos(x))

    0< 1+1-sin^2(x)-2cos(x)

    0< 1+cos^2(x)-2cos(x)

    0< (1-cos(x))^2
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