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Are pi/2+/-2kpi and -pi/2+/-2kpi equal

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Are pi/2+/-2kpi and -pi/2+/-2kpi equal

    2. Relevant equations
    3. The attempt at a solution

    K runs over all integers (i.e. ...-2, -1, 0, 1, 2, ...)

    1.

    Is this sufficient to say pi/2+/-2kpi ≠-pi/2+/-2kpi ?

    pi(1/2 + 2k) & pi(-1/2 + 2k) aren't equal.

    pi(1/2 - 2k) & pi(-1/2 - 2k) aren't equal.

    But,

    2.

    Does this even matter:

    pi(1/2 + 2k) & pi(1/2 - 2k) are equal.

    pi(-1/2 + 2k) & pi(-1/2 - 2k) are equal.

    In other words, does it strictly have to be pi/2+2kpi = -pi/2+2kpi or pi/2-2kpi = -pi/2-2kpi for pi/2+/-2kpi = -pi/2+/-2kpi or can we also go with pi/2+2kpi = -pi/2-2kpi or pi/2-2kpi = -pi/2+2kpi?

    Thanks.
     
  2. jcsd
  3. Jun 4, 2012 #2
    Hi solve!! :biggrin:

    It seems you didnt start using LaTeX yet :tongue2: I dont understand what you wrote there...

    [tex]\frac{\pi}{2}+\frac{1}{-2k\pi} = -\frac{\pi}{2}+\frac{1}{-2k\pi}[/tex]

    Is that what you were trying to say? :uhh:

    Hint : Use LaTeX editor for ease.. http://www.codecogs.com/latex/eqneditor.php :wink:


    Edit : Oh wait, I think I understand.

    This?

    [tex]\frac{\pi}{2}\pm 2k\pi = \frac{-\pi}{2} \pm 2k\pi[/tex]

    If so, they are unequal. The [itex]2k\pi[/itex] on one side will cancel out the [itex]2k\pi[/itex] on the other side for every k, giving

    [tex]\frac{\pi}{2} = -\frac{\pi}{2}[/tex]

    Which is obviously untrue.
     
    Last edited: Jun 4, 2012
  4. Jun 4, 2012 #3
    What up, Infinitum!

    Just can't get around to learning this LaTex. One of these days, for real.

    pi/2+ and -2kpi= -pi/2+ and-2kpi
     
  5. Jun 4, 2012 #4
    Oh I just edited my post, and it looks like I guessed right. :smile:
     
  6. Jun 4, 2012 #5
    Now this perfectly stratifies my question. Thanks, Infinitum.
     
  7. Jun 4, 2012 #6
    Hi Solve, Hi Infinitum.
    Indeed, they are not equal, but the justification for it ins incorrect.
    You are comparing two infinite sets, and the way to do it is to prove that some member of one set does not belong to the other set.
    Your reasoning is wrong, imagine if the statement was:
    0 ± 2k= 2 ± 2k
    you would have come to 0 ≠ 2 and therefore concluded that 0 ± 2k ≠ 2 ± 2k, which would be wrong.

    Back to your example, you just have to show that some element, say, for k=0 in the first set, ∏/2 doesn't belong to the other set.
    That would translate into finding some k in Z such that ∏/2=-∏/2 + 2k∏
    that would then lead you to conclude that k must be 1/2 which doesn't belong to Z, therefore ∏/2 can't belong to the set defined by -∏/2 ± 2k∏, and that is why the two sets are different.
    (at least this is enough to show they are different, but with little more effort you can also show that they are also completely disjoint, that is no element of any set belong to the other set)
     
  8. Jun 4, 2012 #7
    Hi oli!

    I dont see it :frown: Is there any k for which this relation i.e [itex]0 \pm 2k = 2 \pm 2k[/itex] is true? After all, whatever k you substitute into the LHS, the same k would go to the RHS giving 0 ≠ 2. I agree with your argument when we have an equation such as

    [tex]0 \pm 2m = 2 \pm 2n[/tex]

    Where m and n both belong to integers.
     
  9. Jun 4, 2012 #8
    Hi Infinitum
    0±2k = 2±2k does not mean "for each/every k this equality must hold"
    0±2k (I will say + now because it's the same thing if k belongs to Z instead of N) should in fact be written (in (bad) english because I don't know LateX) this way

    "The set of numbers such that any member of it can be written as n=0+2k, where k belongs to Z"
    and in the same way, the other set is
    "The set of numbers such that any member of it can be written as n=2+2k, where k belongs to Z"

    Those two sets are identical, indeed, for any member of the first set, you found some k, well, this member will also be in the other set using k-1

    Saying this in another way, the first set looks like this:
    { -∞ , ..., -4(=0+-2*2), -2(=0+-1*2), 0(=0+0*2), 2(=0+1*2), ... +∞}
    The second set looks like this
    { -∞ , ..., -2(=2+-2*2), -0(=2+-1*2), 2(=2+0*2), 4(=2+1*2), ... +∞}
    They are identical.

    Is it clearer ?

    Cheers...
     
  10. Jun 4, 2012 #9
    Indeed. This is what I assumed would be expressed as,

    [tex]0 + mk = 2 + nk[/tex]

    so that it is clearer to understand that it is not the same variable k.


    I only doubt the validity of this statement:

    Just not sure, I'll have to do some reading on this.
     
  11. Jun 4, 2012 #10
    Hi Infinitum, sorry, replying again, I didn't read your post well, you said you agree with
    0±2m=2±2n
    So you do have it clear and my explanations were not necessary I guess.
    but nonetheless, this is standard mathematical way of writing sets definitions.
    There is no way solve was asked to solve a silly equation of this form
    0+2k=2+2k (with k in Z) because it is indeed silly.
    On the other hand, {0+2k} or {0+2k∏} is a very very common way to represent sets (usually sets of solutions to congruences equations) so you would be comparing sets as a whole instead of 'local' equations (so to speak)

    Cheers...
     
  12. Jun 4, 2012 #11
    Yes, yes, my bad, sorry, I noticed I had not noticed this while submitting and replied again while you were replying too. (my previous part is not a answer to this reply of yours but a second reply when I realized the confusion was not here)
     
  13. Jun 4, 2012 #12
    Agreed! I definitely wouldn't have answered the way I did, if it was denoted in set notation i.e of the form [itex]\left \{ 0+2k \right \} = \left \{ 2+2k \right \}[/itex] :smile:

    Even though this is not the general discussion forum, I must confess I chuckled while reading that. :rofl:
     
  14. Jun 4, 2012 #13
    Thinking about the origin of this problem, I can see this arising when solving inverse trig functions.
    In that case he may get two solutions, and he may have used k for each. Then he was just wondering if using both of those equations was necessary.
    And so in that case an an answer such as 2±2k would be the same as 0±2k.

    But I think it all depends on what leads to getting the two answers.
     
  15. Jun 4, 2012 #14
    Good :) this was the intended effect :)
    I just didn't know first how you were 'reading' the equation and got into unnecessary explanations which you didn't require.
    Now, the question was not too clear and probably out of context, but, for the level of difficulty of such an equation, I am convinced that if it had to be about answering it not in the context of equality of two sets, well, for once, k would have been replaced by n (just because of convention) and the question would have been
    'is there an n such that, blablah ?'
     
  16. Jun 4, 2012 #15
    Hi, people.

    This question indeed comes from a trig equation: 3cosx+ 4sinx=5.

    I am offered to solve it by using either f(x)=Rsin(x+theta) or f(x)= Rcos(x+phi). Very little explanation as to the background relationship of an equation of the form acosx+ bsinx=c and Rsin(x+theta). Nor do I have any idea where the function f(x)=Rsin(x+theta) comes from.

    Used Rsin(x+theta)=5

    R here is either 5 or -5. Theta=0.64 rad.

    If we take R=5, then x=pi/2- 0.64 +/- 2kpi

    If R=-5, then x=(3pi)/2- 0.64 +/- 2kpi

    *I just realized I originally used -pi/2 in x=(3pi)/2- 0.64 +/- 2kpi instead of, say, (3pi)/2 for some reason.

    This is where the comparison question comes from. Since this is an old problem that was filed away in "Leftovers" list, I forgot why I ignored theta=0.64 in my comparison attempts. Can't remember what I was thinking then.

    Anyway, I'd like to see how R=-5 would produce the right answer.

    Thanks.
     
    Last edited: Jun 4, 2012
  17. Jun 4, 2012 #16

    SammyS

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    As for how Rsin(x+θ) is related to 3cosx+ 4sinx :
    Using the angle addition identity for sine:

    [itex]R\sin(x+\theta)[/itex]

    [itex]=R\sin(x)\cos(\theta)+R\sin(\theta)\cos(x)[/itex]

    So that if θ is chosen so that [itex]\displaystyle \sin(\theta)=\frac{4}{5}[/itex] and [itex]\displaystyle \cos(\theta)=\frac{3}{5}\,,[/itex] then you have the desired result if R=5.

    Using R=-5, chose θ so that [itex]\displaystyle \sin(\theta)=-\frac{4}{5}[/itex] and [itex]\displaystyle \cos(\theta)=-\frac{3}{5}\,.[/itex]
     
  18. Jun 6, 2012 #17
    I get θ just by Rsinθ/Rcosθ= tanθ

    I don't quite get how I can chose θ so that [itex]\displaystyle \sin(\theta)=-\frac{4}{5}[/itex] and [itex]\displaystyle \cos(\theta)=-\frac{3}{5}\,.[/itex]

    Thanks.
     
  19. Jun 6, 2012 #18
    I assume you know what angles in the first quadrant give sin and cos to be 4/5 and 3/5 respectively.

    In which quadrant are both sine and cosine negative?
     
  20. Jun 6, 2012 #19
    cosθ=0.825 and sinθ=-0.717 ?

    And sin and cos are both negative in third quadrant.
     
  21. Jun 6, 2012 #20
    Noo!! You need to find θ in sinθ = 4/5 for the first quadrant....

    Yep!!

    Say you found θ. What would sin(180+θ) give you? :wink:
     
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