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Quick trig

  1. May 29, 2006 #1
    [tex]\tan 2\theta = \frac{{2r + k\cos \theta }}{{2h - k\sin \theta }}[/tex]

    How to (isolate) find [tex]\theta \; {?}[/tex]
    ([itex]h,k,r >0[/itex])
     
    Last edited: May 29, 2006
  2. jcsd
  3. May 29, 2006 #2
    !Anyone?? :bugeye:
    (It's just trigonometry!)
     
  4. May 30, 2006 #3

    HallsofIvy

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    That's right, it's just trigonometry so how about you showing some idea of how you would at least try to solve the proglem!
     
  5. May 30, 2006 #4
    There is a function that is the reverse of a tangent.
     
  6. May 31, 2006 #5
    are h,k ,r independent of each other?
     
  7. May 31, 2006 #6

    HallsofIvy

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    [tex]tan(2\theta)= \frac{2tan(\theta)}{1+ tan^2(\theta)}[/tex]
    [tex]= \frac{\frac{2sin(\theta)}{cos(\theta)}}{1+ \frac{sin^2(\theta)}{cos^2(\theta)}}[/tex]
    [tex]= \frac{2sin(\theta)cos(\theta)}{sin^2(\theta)+ cos^2(\theta)}= 2sin(\theta)cos(\theta)[/tex]
    So your equation is really
    [tex]2 sin(\theta)cos(\theta)= \frac{2r+ kcos(\theta)}{2h-ksin(\theta)}[/tex]

    Multiply both sides by the denominator on the right and you will have a quadratic equation for sin([itex]\theta[/itex]).
     
  8. May 31, 2006 #7

    VietDao29

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    Err, so are you saying that: tan(2x) = 2sin(x) cos(x) = sin(2x)? :tongue:
    There's a slight error in the first step. It should be:
    [tex]\tan (2 \theta) = \frac{2 \tan \theta}{1 - \tan ^ 2 \theta}[/tex]. :)
    ---------------
    @ bomba923, have you done anything? I just wonder whether you are asking others to help you or you are just challenging people...
     
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