1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook