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Tan[arctan(2/3) + arccos(8/17)]

  1. Jul 29, 2011 #1
    1. The problem statement, all variables and given/known data

    tan[arctan(2/3)+arccos(8/17)]

    2. Relevant equations

    tan(u+v)=[itex]\frac{\tan u+\tan v}{1-\tan u\tan v}[/itex]

    3. The attempt at a solution

    After drawing 2 triangles for arctan and arccos (both in Quadrant 1), and inserting them into the addition formula for tan, I get:
    [tex]\frac{\frac{2}{3}+\frac{15}{8}}{1-\frac{2}{3}(\frac{15}{8})}[/tex]
    is that right?
    if so, then maybe my math is off...
    I got the numerator reduced to 23/8
    and denominator of -1/4
    for a total of -23/3 ...but thats not correct.
     
  2. jcsd
  3. Jul 29, 2011 #2

    Dick

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    Re: tan[arctan(2/3)+arccos(8/17)]

    I think your math with the fractions is off. 2/3+15/8 is not equal to 23/8. Can you show why you think it is?
     
  4. Jul 29, 2011 #3
    Re: tan[arctan(2/3)+arccos(8/17)]

    Sorry, mis-typed...I meant 23/12

    2/3 =16/24
    15/8=30/24******Found my problem, this should be 45/24


    Thanks for looking at this, I knew it had to be something simple...I need to slow down.
     
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