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Sin(x) = 0 iff x = kpi for some k in Z

  1. Apr 8, 2007 #1
    Well the title is basically the question.
    --------
    Show that sin(x) = 0 if and only if [itex]x/\pi[/itex] is an integer.
    --------

    Some definitions.

    Let z be a complex number, then:

    [tex]\cos(z) = \frac{e^{iz} + e^{-iz}}{2}[/tex]

    [tex]\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}[/tex]

    We also have the power series definition, though I don't think it would be helpful here.

    We have already proved the following, if they may be of use (most likely c,d, e):

    (a) We have [itex]\sin^2(z) + \cos^2(z) = 1[/itex].

    (b) We have [itex]\sin'(x) = \cos(x)[/itex] and [itex]\cos'(x) = -\sin(x)[/itex].

    (c) We have [itex]\sin(-x) = -\sin(x)[/itex] and [itex]\cos(-x) = \cos(x)[/itex].

    (d) We have [itex]\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)[/itex] and [itex]\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)[/itex]

    (e) We have [itex]\sin(0) = 0[/itex] and [itex]\cos(0) = 1[/itex].

    (f) We have [itex]e^{ix} = \cos(x) + i\sin(x)[/itex] and [itex]e^{-ix} = \cos(x) - i\sin(x)[/itex]

    Also we proved in the previous part of this exercise that [itex]\cos(x+ \pi) = -\cos(x)[/itex] and [itex]\sin(x+ \pi) = -\sin(x)[/itex].

    And we define [itex]\pi = \inf\{x\in (0,\infty): sin(x) = 0\}[/itex]

    ---------

    For the problem, I think the [itex]\Leftarrow[/itex] is proved with a simple induction argument and using that [itex]\sin(-x) = -\sin(x)[/itex]. Though, if you have an interesting way to do it, please share :smile:

    However, the [itex]\Rightarrow[/itex] direction is seeming to give me some trouble. Any ideas here?

    edit... I just got the wild idea of proving the contrapositive which seems like it would work well. If we write [itex]x= k\pi + r[/itex] for some [itex]0 < r <\pi[/itex]. Then use the addition formula for sin to finally get sin(x) = 0 + sin(r). Which cannot be 0 since we know that [itex]\pi[/itex] is the smallest positive number with sin(x) = 0. Sounds good to me. Always fun to come up with an idea after posting some huge mess :smile:
     
    Last edited: Apr 8, 2007
  2. jcsd
  3. Apr 8, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    Heres an argument that may have also worked -

    for any k in Z, k*pi can be written as pi + pi + pi +pi ...k times.

    Every time we took away a pi, it would become -sin(x), then take away another pi, sin x again. Keep doing that until eventually there are no pi's left, - sin 0= sin 0 = 0.
     
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