Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    [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


    User Avatar
    Homework Helper

    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...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook