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Homework Help: 1 + 2cos(2x+ π/3)

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the zeroes of

    f(x) = 1 + 2cos(2x+ π/3)

    Range: (0 ; π)


    3. The attempt at a solution

    1 + 2cos(2x + π/3) = 0

    hence, cos(2x + π/3) = -1/2

    hence, 2x + π/3 = 2π/3

    and, x = π/6 which is 30 degrees

    can someone help me find the other 0 ?

    I don't know how to!

    Thanks a lot!

    (the answer is x=π/2 but i dont know how to get there!)
     
    Last edited: Jun 2, 2010
  2. jcsd
  3. Jun 2, 2010 #2

    Char. Limit

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    Gold Member

    First, I don't quite know how you went from your first step to your second, even though it does turn out to be correct. A bit of explanation would be most appreciated.

    Second, have you tried using the sum-of-angles formula to expand your cosine out? It might make it easier, I think...
     
  4. Jun 2, 2010 #3
    Ok Char. Limit,

    I edited my answer and showed the transition step.

    What do you mean by "the sum of angles formula" ?
     
  5. Jun 2, 2010 #4
    Well, if one looks at the unit circle, one can see that 2*pi / 3 is one place where
    cos(y) = -1/2. However, there are more places on the unit circle in which
    cos(y) = -1/2. Finally, one can also add 2*pi*n to the result, and end up at
    the same place.
     
  6. Jun 2, 2010 #5

    Char. Limit

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    But he restricted x to be between zero and pi...

    The sum-of-angles formula is as follows:

    [tex]cos(x+y)=cos(x)cos(y)-sin(x)sin(y)[/tex]

    I have no idea how to prove it, but it is a theorem, I believe.
     
  7. Jun 2, 2010 #6
    Yes, but still, one can manually inspect the formulas to see when
    [tex]
    2x + \pi /3 = 2\pi /3 +2\pi n
    [/tex]
    , and after that see if x is inside the interval between 0 and pi.
     
  8. Jun 2, 2010 #7
    Yes, my question ultimately is

    How does one determine these other places where cos(y) = -1/2 ?
     
  9. Jun 2, 2010 #8
    If you are familiar with the unit circle, you have that the x-coordinate on the circle
    is cos(x), while the y-coordinate is sin(x). For a picture, please see
    http://en.wikipedia.org/wiki/Unit_circle
    In the picture, you can see that cos(x) = -1/2 means: at which place at the unit circle
    is the x-coordinate equal to -1/2? And there are two places that can happen at in the circle. Does this help?
     
  10. Jun 2, 2010 #9
    Well, wouldn't you just have to add pi/4 for the next zero and then 3pi/4 and keep going until you're beyond your range in this case? Since the function is pretty much a squashed together cos curve (twice as many zeros as a normal one) that is half it's amplitude up the y-axis, it's amplitude being 2, that is out of phase of your standard cos curve by pi/3. I'm not sure I'm visualising it right, but I guess you could always sketch it up on some spare sheet.
     
  11. Jun 2, 2010 #10

    Mark44

    Staff: Mentor

    Correction: cos(y) = -1/2 for y = 2pi/3 or for y = 4pi/3.
    Replace y by 2x + pi/3 and you should get two values of x in (0, pi).
     
    Last edited: Jun 2, 2010
  12. Jun 2, 2010 #11

    Mark44

    Staff: Mentor

    It actually makes it much harder.
     
  13. Jun 2, 2010 #12

    Char. Limit

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    I stand corrected then.
     
  14. Jun 2, 2010 #13
    How do you know this?
     
  15. Jun 2, 2010 #14

    Cyosis

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    Mark gave you the values where the cosine is -1/2 in the third quadrant and 1/2 in the fourth quadrant. You're interested in the values where cosine is -1/2, not 1/2.

    The first thing you should do is draw a unit circle. Then sketch the places where the cosine is equal to -1/2 and find the appropriate angles. In which two quadrants does your solution lay?
     
  16. Jun 2, 2010 #15
    How does one find the place where cosine is equal to -1/2?

    and as you said, the solution is in the third quadrant and the first solution (Pi/6) is in the first.
     
  17. Jun 2, 2010 #16

    Mark44

    Staff: Mentor

    Due to a mental lapse, one of the values I gave was incorrect. The values of y should have been in the second and third quadrants, but not in the fourth. I have corrected my earlier response.
     
  18. Jun 2, 2010 #17

    Cyosis

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    No, [itex] \cos \pi/6=\frac{1}{2}\sqrt{3}[/itex]. Secondly I am not asking you to draw the position where cos is -1/2 in the unit circle exactly. The use in the unit circle is that you can see in which quadrant the solutions lay. When you draw a line radially outwards from the origin you can form a triangle. Which side of that triangle corresponds to the cosine and which one to the sine? To answer your question you want to know where that side of the triangle is -1/2.
     
  19. Jun 2, 2010 #18

    Mark44

    Staff: Mentor

    There are just a handful of angles that you are expected to know exactly: 0, pi/6, pi/4, pi/3, pi/2, plus their complements and supplements. If you plot these angles in the unit circle, you can use symmetry to show that cos(y) = 1/2 for y = pi/3 and y = 5pi/3, and that cos(y) = -1/2 for y = 2pi/3 or y = 4pi/3.
     
  20. Jun 2, 2010 #19
    Ok! thanks, i kind of get it now!
     
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