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Trig Help

  1. Jan 10, 2007 #1
    I have the problem:

    0 = 3[cos(35)*cos(A) - sin(35)*sin(A)] - cos(A)
    where A is alpha...my unknown degree.

    somehow that turns into this:

    tan(A) = [cos(35) - 1/3] / sin(35)

    I am not drawing the connection or seeing how that is happening....

    Could you help???? THANKS
     
  2. jcsd
  3. Jan 10, 2007 #2
    Show some work. Hint: look up some trig identities and start playing around.
     
  4. Jan 10, 2007 #3
    i tried... this isn't a hw problem... its in the text book. It jumps from that first step to the next one just saying "Then we can see:" ... and I cant see that
     
  5. Jan 10, 2007 #4
    Hint: divide everything by cos A first.
     
  6. Jan 11, 2007 #5

    VietDao29

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    Homework Helper

    Ok, in fact, the book does skip some steps:
    [tex]0 = 3[\cos (35 ^ o) \cos A - \sin(35 ^ o) - \sin A] - \cos A[/tex]
    Now divide both sides by cos A, we have:
    [tex]0 = \frac{3[\cos (35 ^ o) \cos A - \sin(35 ^ o) - \sin A] - \cos A}{\cos A} = 3 \cos (35 ^ o) - 1 - 3 \sin (35 ^ o) \tan A[/tex]
    [tex]\Leftrightarrow 3 \cos (35 ^ o) - 1 = 3 \sin (35 ^ o) \tan A[/tex]
    Divide both sides by 3, we have:
    [tex]\Leftrightarrow \cos (35 ^ o) - \frac{1}{3} = \sin (35 ^ o) \tan A[/tex]
    Now, divide everything by sin(35o), we have:
    [tex]\Leftrightarrow \frac{\cos (35 ^ o) - \frac{1}{3}}{\sin (35 ^ o)} = \tan A[/tex]
    Can you get this? :)
     
    Last edited: Jan 11, 2007
  7. Jan 11, 2007 #6
    Yes, thank you very much
     
  8. Jan 11, 2007 #7
    VietDao, how did you get your second line?
     
  9. Jan 13, 2007 #8

    Gib Z

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    Homework Helper

    He divided both sides by 3, if you mean the 2nd part of that line, He just added 3 sin 35 tan A to both sides.
     
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