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Trigonometry assistance

  1. Aug 18, 2005 #1
    How do I go about finding the equivalent expression for cot 50 degrees?

    When I work the problem, I come up with tan 130 degrees.

    Where am I going wrong?
     
  2. jcsd
  3. Aug 18, 2005 #2

    TD

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    Homework Helper

    I rather work in radians, if you don't mind.

    [tex]50^\circ = \frac{{5\pi }}{{18}}[/tex]

    We know that:
    [tex]\cot \left( \alpha \right) = \tan \left( {\frac{\pi }{2} - \alpha } \right)[/tex]

    So:
    [tex]\cot \left( {\frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{\pi }{2} - \frac{{5\pi }}{{18}}} \right) = \tan \left( {\frac{{2\pi }}{9}} \right) = \tan \left( {40^\circ } \right)[/tex]
     
  4. Aug 18, 2005 #3
    That makes perfect sense now.....

    The light has come on, thank you
     
  5. Aug 18, 2005 #4

    TD

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    Homework Helper

    No problem :smile:
     
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