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Question of the basic properties of tan

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data

    I've solved a problem in my engineering homework to a point where I have the following data:
    ψ = arctan(3) = 71.57 (degrees) , inverse tangent of 3
    δ = 30 degrees
    solve for gamma and chi (χ):

    tan(2gamma) = tan(2ψ)cos(δ)
    sin(2χ) = sin(2ψ)sin(δ)

    2. Relevant equations

    tan(2gamma) = tan(2ψ)cos(δ)
    sin(2χ) = sin(2ψ)sin(δ)

    3. The attempt at a solution

    My problem rests with the gamma solution. When I draw it (using other data in the problem), it is an angle larger than 70 degrees at least. When I solve for gamma with a calculator I get gamma = -16.5°.

    my process:
    tan(2ψ) = -.75
    (-.75)cos(30) = -0.6495
    arctan(-0.6495) = -33°
    gamma = -33/2 = -16.5°

    I know that there are properties of tangent that you need to take into consideration, but I just cannot remember them, and I am sure that is why I get this negative angle.

    Or, did I do this correctly, and I just need to redraw my problem?

    EDIT: the answer in the back of the book is gamma = 73.5 degrees
     
    Last edited: Sep 19, 2013
  2. jcsd
  3. Sep 19, 2013 #2

    vela

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    Without knowing the context of the problem, it's hard to say what's going on, but I don't think it's a coincidence that 73.5+16.5 = 90.
     
  4. Sep 19, 2013 #3
    I, too, noticed that they summed to 90. Well, the entire problem, minus the EE info, is:

    I have an equation that makes an elipse:

    E = x[10cos(wt + 30°)] + y[30cos(wt)]

    We can make wt our variable. x and y are the vector directions, I believe. (they have hats ^ above them)

    gamma is the angle that the major axis of the elipse makes with the x-axis. When I draw the elipse, I can clearly see that said gamma is near positive 70°)

    ψ is the angle that the amplitudes of the x and y components make with the x-axis (arctan(30/10) = 71.57)

    and chi, χ, is an angle that is hard to describe, but not pertinent to what I need help with.
     
  5. Sep 19, 2013 #4

    vela

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    The tangent function has a period of 180°, so when you take the arctan, you can add 180° to what the calculator gives you to get another solution.
     
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