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Proof that if cos(x+y)=cos(x), then y=2*pi*k?

  1. Feb 24, 2013 #1
    Recently I remembered a property of cosine that I learned in high school, namely that the cosine function repeats every 2[itex]\pi[/itex]k times for every integer k. I was told that naturally, then, if cos(x+y)=cos(x), then y is of the form 2[itex]\pi[/itex]k, where k is an integer. However, I was looking for a formal proof of this. Can someone help me out? Thanks.
  2. jcsd
  3. Feb 24, 2013 #2
    What is your definition of cosine and of pi?? What can we assume known??
  4. Feb 24, 2013 #3
    The described implication does not hold exclusively because [itex] cos(x) = cos(-x) = cos(x-2x) [/itex] for any x. As [itex] cos(z) = cos(w) [/itex] only when [itex] z [/itex] and [itex] w [/itex] are at the same x-coordinate of the unit circle we have [itex] z = \pm w + 2\pi k[/itex]. In the quoted case it is [itex] x = \pm (x+y) \rightarrow y = 2\pi k \vee y = -2x + 2\pi k [/itex].
    Last edited: Feb 24, 2013
  5. Mar 12, 2013 #4
    [itex]cos(a) = cos(b) \text{ iff } a = b + 2\pi k \text{ or } a = -b + 2\pi k[/itex]

    Therefore, [itex]cos(x) = cos(x+y) \text{ iff } x = (x + y) + 2\pi k \text{ or } x = -(x+y) + 2\pi k[/itex].

    This reduces to [itex]y = + 2\pi k \text{ or } y = -2x + 2\pi k[/itex].

    This proves that your friend's statement was incomplete, as y could be in the form of -2x+2πk as well


    Edit: Although, if one considers y as a constant (as opposed to a function of x), then y could only be in the form of 2πk and work for all values of x. Maybe your friend was on the something after all :p
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