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Math Help - Evaluating arcsec at infinity...?

  1. #1
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    Evaluating arcsec at infinity...?

    arcsec(2) - arcsec(1) + arcsec(\infty) - arcsec(2)
    No idea how to procede, I see the first and last terms cancel out, my book says arcsec of 1 comes out to zero and infinity comes out to pi-halfs. How?
    Last edited by Nervous; January 10th 2013 at 05:57 AM.
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  2. #2
    Super Member ebaines's Avatar
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    Re: Evaluating arcsec at infinity...?

    The arcsecant of 1 is 0 because sec(0) =  \frac 1 {\cos(0)} = \frac 1 1 = 1.

    The arcsecant of  \infty is  \frac {\pi} 2 because sec(\pi /2) = \frac 1 {\cos (\pi/2)} = \frac 1 0 = \infty
    Last edited by ebaines; January 10th 2013 at 08:15 AM.
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  3. #3
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    Re: Evaluating arcsec at infinity...?

    Just to be pedantic- what ebaines really means \sec(\pi/2)= \lim_{x\to \pi/2}\frac{1}{cos(x)}= \frac{1}{\lim_{x\to\pi/2} cos(x)}= \infty.

    (I just refuse to say that " \frac{1}{0}= \infty"!)
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  4. #4
    Super Member ebaines's Avatar
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    Re: Evaluating arcsec at infinity...?

    Go ahead - be pedantic! It keeps us honest.
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