Angle Help!15 = arctan(2/x) - arctan (1/x)

  1. 1. The problem statement, all variables and given/known data
    Basically, solve for x
    15 = arctan(2/x) - arctan (1/x)

    2. Relevant equations
    tan (A-B) = (Tan A -Tan B) / (1+Tan A*Tan B)

    3. The attempt at a solution
    I really tried everything.
    My first step was to:
    Let y = arctan (2/x)
    Therefore, tan y = 2/x
    Similarly, u = 1/x
    Then, tan (y-u) = (Tan y -Tan u) / (1+Tan u*Tan y) = 15
    15 = (2/x-(1/x) / (1+(2/x^2)
    15 = (1/x) / (x^2 + 2 / x^2)
    15 = (x) / (x^2+2)
    15x^2 + 30 - x = 0
    Which has no real roots :(
    But, with guess and check, it's around 0.65
     
  2. jcsd
  3. rock.freak667

    rock.freak667 6,228
    Homework Helper

    You need to post an attempt before we can help you. Start by taking the tangent of both sides.
     
  4. Yeah, sorry. I just misclicked the first time
     
  5. Alright, sorry guys to waste your time, but I believe I figured it out. Thanks for the hint of "tanning" both sides.
    Instead of 15, it's supposed to be tan 15.
    So that,
    x/(x^2+2) = tan 15
    x = 0.64
    x = 3.08 (approximately)

    Thanks for the help!
     
  6. Mentallic

    Mentallic 3,701
    Homework Helper

    If you rearrange that equation you'll get the quadratic

    [tex]x^2-\frac{1}{tan(15)}x+2=0[/tex]

    There are no real solutions to this quadratic.
     
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