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Sin function

  1. Mar 22, 2008 #1
    [SOLVED] sin function

    1. The problem statement, all variables and given/known data
    My complex analysis book says, "from calculus, [itex]\sin \theta \leq \theta[/itex] for [itex]0 \leq \theta \leq \pi/2 [/itex]." Could someone please give me a better reason why that is true?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 22, 2008 #2
    well, the only way i know how to prove that [tex]|sin(x)|<|x|[/tex] is with the aid of geometry. YOu need to construct a circle with radius R=1, and denote wit OAB a triangle, and with AB the arch. Denote also by x the angle that the two sides, the radiuses of the circle enclose. And you will see the validity of this. This is true only when the angle is measured in radians. Because also remember that a portion of the arch of the circle, call it acrAB is actually the measure of the angle in radians, if we draw two lines that pass through the two points in the circle A,B.
    Last edited: Mar 22, 2008
  4. Mar 22, 2008 #3


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    You could also consider the function f(x)=x-sin(x) and it's derivative.
  5. Mar 22, 2008 #4
    Very nice.
  6. Mar 23, 2008 #5

    Gib Z

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    It should also be clear if you use the series definition for sine.
  7. Mar 23, 2008 #6
    Drawing their graphs will also make it clear.
  8. Mar 23, 2008 #7


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    … chord-length < arc-length …

    Hi ehrenfest! :smile:

    The intuitive explanation is: 2sinx = chord-length (a straight line :smile: ), which is less than 2x = arc-length (not a straight line :frown: ), for 2x < π.

    (This is sutupidmath's explanation, of course, with the details left out.)
  9. Mar 23, 2008 #8
    Its not clear. For that to be true, you need to know that the sum of all the terms after the first is negative and has magnitude less than the first term in the series. Why is that true?
    Last edited: Mar 23, 2008
  10. Mar 23, 2008 #9
    Neither to me...
  11. Mar 24, 2008 #10

    Gib Z

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    My bad. The things you stated there are sometimes used when deriving the series in the first place and showing the Taylor Remainder term goes to zero, but not always. I wrongly assumed you learned it the same way I did.
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