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Equations with sine and cosine.

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve for |θ12|

    2. Relevant equations

    cosθ1 + cosθ2 = 0
    sinθ1 + sinθ2 = 0

    3. The attempt at a solution

    This is a silly math problem within a larger question I'm working on. I have solved for it multiple times now using different trig identities and I get different answers.

    First of all, can I set them equal to each other because they both equal 0, or do I add them both together? The latter gives an extra negative sign.

    Can I solve this just using one equation? i.e.:

    cosθ1 + cosθ2 = 0

    =2cos((θ12)/2) cos((θ12)/2)= 0

    Then can I just divide both sides by 2cos((θ12)/2) and then solve for |θ12| or is that not proper math?

    I've spent way too much time on this high school level question....
     
  2. jcsd
  3. Mar 29, 2012 #2
    You can set them equal to each other as they are both equal to 0 (it works as well with every other number, variabele, and the like). Moreover your method should work fine, though you can only divide by 2cos((θ1+θ2)/2) as the cos is not equal to 0 (So you should find more solutions).

    Though, because it is quite nasty when there are more variables in one sinus or cosinus, I would advise you to put first all θ_1 in one side of the equation and θ_2 on the other:

    Then you will get this:
    << Solution deleted by Moderator >>
     
    Last edited by a moderator: Mar 29, 2012
  4. Mar 29, 2012 #3

    berkeman

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    Staff: Mentor

    Welcome to the PF, BishopPair. Please remember the part of the Rules link at the top of the page that says the OP has to do the bulk of the work on schoolwork-type questions.
     
  5. Mar 29, 2012 #4

    berkeman

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    Staff: Mentor

    The easiest way to solve this is graphically. Sketch the sin(theta) and cos(theta) graphs on the same horizontal axis. What difference |θ1-θ2| will always give you 0 for the two equations?

    I don't think you can set them equal, because the two equations have "+" signs. The two quantities have to be the opposite of each other.
     
  6. Mar 29, 2012 #5
    cosθ1 + sinθ1 = -cos θ2 - sinθ2

    There is no identity to go from here.

    EDIT:

    But where did you get that the difference equals zero? If the difference equals zero, then that is the answer and I am done. I don't need actual numbers for theta, I just need an expression so that I can pick an arbitrary value for either of them and fix the other value.

    I just used the result of θ12= 0 and I got what I know to be the correct answer for the rest of the problem. So how did you come up with that, using equations?

    EDIT 2: I just used these trig identities to do it, and the method I outlined above.

    I'm also an idiot and the problem is supposed to be cosθ1 - cosθ2= 0 and so on.

    Using the above:

    cosθ1 - cosθ2= 0

    (sum to product identity)
    -2sin((θ1+θ2)/2) sin((θ1-θ2)/2)= 0

    sin((θ1-θ2)/2)= 0 <----- this is where I'm worried about my math. Can you just single out any one component and get rid of it because you have a zero on the right, and then solve for what remains in the parenthases? If I get rid of the other one and solve for θ1+θ2 that also = 0 So I get conflicting answers.

    (θ1-θ2)/2 = 0

    (θ1-θ2) = 0

    EDIT 3: Ok, this should be simple.

    cosθ1 - cosθ2= 0
    cosθ1 = cosθ2

    Which only happens when θ1 and θ2 are equal, or negative of each other. So there are two answers. I only get the correct result in the rest of the problem when they are equal though.

    Aha! But I also have sinθ1 - sinθ2= 0

    sinθ1 = sinθ2

    Which only happens when they are equal. Thus eliminating one solution, and θ2 and θ1 must be equal, and:

    θ1-θ2=0
     
    Last edited: Mar 29, 2012
  7. Mar 30, 2012 #6

    berkeman

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    IF the angles are the same, then your equations are not always satisfied:

    cosθ1 + cosθ2 = 0
    sinθ1 + sinθ2 = 0

    But there does exist a difference between the angles that does satisfy those two equations...
     
  8. Mar 30, 2012 #7
    When? When does cosθ=/=cosθ, or sinθ=/=sinθ?

    Oh sorry, maybe you missed that I updated the problem, it's actually supposed to be a minus sign instead of a plus sign, so these are the two equations: (totally my fault)

    For the correct above equations the two angles must be equal.

    Believe it or not I passed high school trigonometry about 6 years ago... thanks for the help.
     
    Last edited: Mar 31, 2012
  9. Mar 31, 2012 #8
    Well, that certainly makes it easier!

    I was able to solve for [itex]\theta_1 + \theta_2[/itex] by using sum-to-product trig formulas and then using the symmetry of the unit circle to solve. Are you sure you weren't supposed to solve for [itex]\theta_1 + \theta_2[/itex]? At the moment I don't see how to solve for [itex]|\theta_1 - \theta_2|[/itex]
     
  10. Mar 31, 2012 #9
    cosθ1 - cosθ2 = 0
    sinθ1 - sinθ2 = 0

    Just think about when this is true within a single period. You can then add any multiple of the period to your answers. You could think about magnitude and sign separately too. Think "if theta 1 is in quadrant 1, in what quadrant must theta 2 be?". Do this for the first equation then the second equation. Then, use whatever quadrants are in both since both equations must be satisfied. After you isolate the quadrant(s), you can think in terms of magnitudes within that quadrant.
     
  11. Mar 31, 2012 #10
    cosθ1 - cosθ2 = 0
    sinθ1 - sinθ2 = 0

    Therefore:

    cosθ1 = cosθ2
    sinθ1 = sinθ2

    This only happens when θ2 and θ1 are equal. Really, you could get this from only the second equation if you wanted to since there is only once solution.

    θ21
    So θ12=0
     
  12. Mar 31, 2012 #11
    No. As for your claim that the angles must be equal, consider that sine and cosine are periodic. As for your other claim that you could arrive to this solution from only the first equation, consider that
    [tex]cos(a) = cos(-a)[/tex]
     
  13. Mar 31, 2012 #12

    SammyS

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    Gold Member

    How do you explain the following ?

    sin(π/3) = sin(2π/3) = sin(7π/3) = ...
     
  14. Mar 31, 2012 #13
    Not helpful. When exactly does sinθ=/=sinθ?

    I may as well be asking when does 2=/=2.

    Except that I said only the second equation. You get two possibilities if you only use the first equation with cosine. But sinθ1 only equals sin θ2 if they are both equal. Not if they are negatives of each other. So the second equation implies they are equal.

    So, what are you saying, I need to add in a factor of some number n χ ∏ ?

    Once again, this is irrelevant trig that is NOT THE FOCUS of an upper level science course. I need to MOVE ON ALREADY. I don't have more time to spend on this silly question.
     
  15. Mar 31, 2012 #14
    Your statement was false also for the use of the second equation. Consider that
    [tex]sin(a) = sin(\pi -a)[/tex]
     
  16. Mar 31, 2012 #15
    I get it, sine is a periodic function.

    What does this tell me about θ1- θ2?

    I will not be tested on this information. It is just the means to end of a homework problem. I don't need to learn trig, I need to move on. I have spent way too much time thinking about elementary trig that is not going to help me in this upper level science class.

    At this point I'm just assuming that the only answer I'm going to get is my "wrong" answer(which gave me the right result) so I'm probably going to just have to go with that. I have exams coming up; the ins and outs of trig are not really my focus right now.

    EDIT: I don't mean to sound angry I'm just getting kind of hyper that I have this huge assignment waiting to be done and I can't finish it because of what should be a simple problem, and now we are 15 posts in and I still don't have the answer.
     
    Last edited: Mar 31, 2012
  17. Apr 1, 2012 #16
    The function is not periodic by a factor of pi. It is periodic by a factor of 2pi. You seem to be ignoring everything I am saying except for the short portion right before I elaborate where I say "this is wrong."

    Take a look at these two pictures. When you draw one of these triangles for any given angle, you always have a portion of the triangle that lies on the x-axis and one that lies on the y-axis. The cosine of an angle gives you the portion along the x-axis. The sine gives you the portion along the y-axis. If your angle is in quadrants I or II, your y is positive, so your sine is positive. This is angles 0 + k2pi to pi +k2pi where k is an integer. Else your sign is negative. The same logic applies to cosine. If you are in quadrant I or IV, your x-axis portion is on the positive x-axis and your cosine is positive. The reason these functions are periodic is if you have an angle B (which we can assume is 45 degrees for concreteness), and you add 2pi to it, you wrapped around the entire circle back to 45 degrees in quadrant I. This line of thinking works for any multiple of 2pi. If you have 6 pi, for example, added to B, you wrapped around 3 times and ended up at B.

    Can you tell me in which quadrant(s) angle 2 can be, in general, using what I've said and these pictures, if angle 1 is in quadrant I, II, III, or IV?
     

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  18. Apr 1, 2012 #17
    I know you are trying really hard but I am having trouble following you.

    Yes.

    Yes

    Yes

    Yes

    Yes

    If angle 1 is in the first quadrant then sine can be in quadrants one or two. Cosine of angle 1 can be in quadrants one and four. The overlap happens in quadrant one.

    If angle 1 is in the third quadrant then sine of angle one can be in quadrants 3 or 4. Cosine of angle 1 can be in quadrants 2 or 3. The overlap happens in quadrant 3.

    And so on. The overlap happens in whatever quadrant the angle is in.


    Ok so you're saying if angle 1 is in quadrant 1 and cos(1)=cos(2) then the possible quadrants that angle 2 can be in is 1 and 4. If angle 1 is in quadrant 1 and sin(1)=sin(2) then the possible quadrants angle 2 can be in is 1 and 2. The overlap is only in quadrant 1, so angle 2 must also be in quadrant 1.

    Now what?
     
    Last edited: Apr 1, 2012
  19. Apr 1, 2012 #18
    Well, taking both equations together (both conditions), you have made the observation that angle 2 must be in whatever quadrant angle 1 is in. We have two steps to go from here. Next, we need to think magnitude. What angle must angle 2 be (which is now nicely confined to the same quadrant as angle 1) so that the cosines and sines have the same magnitude? Just ignore periodicity and think from 0 to 360 degrees. The final leap will be to take whatever answer you find for that question and say "But it can be true for any k2pi where k is an integer added to it."
     
  20. Apr 1, 2012 #19
    sine and cosine are equal at 45, 135, 225, 315, i.e. start at 45 and add 90 to get to the next quadrant and so on.

    How do I go from here to what angle 1 minus angle 2 is? I don't get what you are trying to prove. If angle 1 is 45, we have determined that angle 2 must be in the same quadrant. And sin and cosine are only equal at 45. So the angles equal each other, or rather they equal each other and either one could also have an integer +k2pi tacked on.

    EDIT: At this point my final answer is θ1-θ2=k2pi where k is an integer.
     
    Last edited: Apr 1, 2012
  21. Apr 2, 2012 #20
    Yeah, that looks right to me.
     
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