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A little bit of trigonometry

  1. Sep 18, 2010 #1
    Hi, I have a problem that arose in one of my courses that I've gotten stuck on. I reduced it to two equations in two unknowns but can't get any further. I've apparently forgotten all the trigonometry I used to know.

    C = Acos(x) + Bcos(y)

    0 = Asin(x) + Bsin(y)

    where A,B,C are known but tedious to write (they are on the order of 10^(-19)).

    I'd appreciate any help you can offer. I'm sure the solution will be obvious once I see it.
     
  2. jcsd
  3. Sep 19, 2010 #2

    danago

    User Avatar
    Gold Member

    First thing that comes to my mind is to first express the equations in the following form:

    [tex]\begin{array}{l}
    \cos (y) = \frac{{C - A\cos (x)}}{B}\\
    \sin (y) = - \frac{{A\sin (x)}}{B}
    \end{array}[/tex]

    You can then use the pythagorean identity ([tex]{\sin ^2}(y) + {\cos ^2}(y) \equiv 1[/tex]) to form the following equation:

    [tex]{\left( {\frac{{A\sin (x)}}{B}} \right)^2} + {\left( {\frac{{C - A\cos (x)}}{B}} \right)^2} = 1[/tex]

    Expand it out and it should be easy to solve for x, which can then be substituted into one of the first equations to find y.
     
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