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Trig Equation

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the following equation, where 0 [tex]\leq[/tex]u [tex]\leq[/tex] 2pi [360 degrees]

    2sinucosu = 0

    2. Relevant equations



    3. The attempt at a solution

    (2sinucosu/2) = (0/2)

    sinucosu = 0
     
  2. jcsd
  3. Nov 18, 2008 #2
    Try seeing if there's a specific trig identity to help you :wink:
     
  4. Nov 19, 2008 #3
    sin2u?

    tried that as well, and not sure what to do.
     
  5. Nov 19, 2008 #4

    Mark44

    Staff: Mentor

    If 2 sin(u)cos(u) = 0, then sin(2u) = 0. What can you say about 2u? You should several values for 2u, and several for u.
     
  6. Nov 19, 2008 #5
    You gotta read this step by step alright no peeping into the next step.

    1st step
    Refer to the rule
    Sin 2A=2 SinA CosA

    Your range of 2U
    0 [tex]\leq[/tex]u [tex]\leq[/tex] 2pi
    0 [tex]\leq[/tex]2u [tex]\leq[/tex] 4pi
     
  7. Nov 19, 2008 #6
    going off of what he was saying, imagine for a second that the 2 wasn't there, so that you have sin(u)=0. For confusions sake, let's call it n now. So you've got sin(n)=0 Essentially saying the sign of some angle "n" equals 0 right? You should be able to figure out what that angle n is. If you think of it like that and can solve for n, then all you have to do now is think of n=2u, and that i think you can figure out =). Hope that wasn't too revealing =X
     
  8. Nov 19, 2008 #7

    HallsofIvy

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    So either sin u= 0 or cos u= 0. What values if y satisfy those?
     
  9. Nov 19, 2008 #8
    haha.. tats true we nid consider tat factor..

    Therefore u = 2pi
     
  10. Nov 20, 2008 #9
    Well.... I got as far as sinu = 0 . I honestly have no idea how to continue it further, since no special triangle can get you to equal 0. On a sin graph with a period of 2pi though, sin touches 0 at pi and 2pi
     
  11. Nov 20, 2008 #10

    HallsofIvy

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    Apparently you didn't consider it. sin(u)= 0 for more values than 2pi, cos(u)= 0 for other values.
     
  12. Nov 20, 2008 #11

    HallsofIvy

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    Yes, exactly. And since your problem says "[itex]0\le u\le 2\pi[/itex], those are two solutions. Now, what values of u make cos(u)= 0?
     
  13. Nov 20, 2008 #12
    pi/2 and 3pi/2

    I have a question though, how would I show my work for it? It's worth 6 marks...would I sketch the graph for each of them or? :)
     
  14. Nov 20, 2008 #13
    Also, is using cosu necessary for the equation too? If so, changing it to sin2u = 0 is irrelevant. Since sin2u = 0 and diving sin2u by 2 and 0 by 2, giving us sinu = 0 , then would cos not be apart of this?
     
  15. Nov 21, 2008 #14

    HallsofIvy

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    No sin(au)/a is NOT equal to sin(u).

    If xy= 0, then either x= 0 or y= 0. Surely you know that?
     
  16. Nov 21, 2008 #15

    Mark44

    Staff: Mentor


    You should get the same solutions either way:
    By solving sin(2u) = 0
    By solving sin(u)cos(u) = 0
     
  17. Nov 21, 2008 #16

    HallsofIvy

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    But NOT by "dividing sin 2u= 0 by 2" to get sin u= 0!
     
  18. Nov 24, 2008 #17
    u=theta

    sin2u=0 , is the double angle identity. Sorry but cos has nothing to do with the answer.

    2pi/2 = pi pi is the period.
     
  19. Nov 24, 2008 #18
    u=theta

    sin2u=0 , is the double angle identity. Sorry but cos has nothing to do with the answer.

    (2pi/2) = pi ---->pi is the period.
     
  20. Nov 24, 2008 #19
    Mark44 was right

    [tex]sin(2u) = 0[/tex] is the same thing as [tex]sin(u)cos(u) = 0[/tex]. You will get the same answers either way, it is just a matter of preference.
     
  21. Nov 24, 2008 #20
    Hmm, well my answers were (pi/2), (3pi/2), 2pi, pi
     
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