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

[suckmyfish1]

Recently I remembered a property of cosine that I learned in high school, namely that the cosine function repeats every 2\pik times for every integer k. I was told that naturally, then, if cos(x+y)=cos(x), then y is of the form 2\pik, where k is an integer. However, I was looking for a formal proof of this. Can someone help me out? Thanks.

 

[micromass]

What is your definition of cosine and of pi?? What can we assume known??

 

[Ferramentarius]

Recently I remembered a property of cosine that I learned in high school, namely that the cosine function repeats every 2\pik times for every integer k. I was told that naturally, then, if cos(x+y)=cos(x), then y is of the form 2\pik, where k is an integer. However, I was looking for a formal proof of this. Can someone help me out? Thanks.

The described implication does not hold exclusively because cos(x) = cos(-x) = cos(x-2x) for any x. As cos(z) = cos(w) only when z and w are at the same x-coordinate of the unit circle we have z = \pm w + 2\pi k. In the quoted case it is x = \pm (x+y) \rightarrow y = 2\pi k \vee y = -2x + 2\pi k.

 

[Someone2841]

Recently I remembered a property of cosine that I learned in high school, namely that the cosine function repeats every 2\pik times for every integer k. I was told that naturally, then, if cos(x+y)=cos(x), then y is of the form 2\pik, where k is an integer. However, I was looking for a formal proof of this. Can someone help me out? Thanks.

cos(a) = cos(b) \text{ iff } a = b + 2\pi k \text{ or } a = -b + 2\pi k

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

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

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

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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