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Triple angle of tangent

  1. Sep 8, 2006 #1
    hey
    sorry to distrub you, but i was surfing the net for the triple angle of tangent trig function but could not find it so i decided to use infomation i knew to solve it. i would like to know if what i have done is correct, and if it can be simplified further, thanks. What i have done is as follows:

    [tex]\
    \begin{array}{c}
    \tan 3A = \tan \left( {2A + A} \right) \\
    = \frac{{\tan 2A + \tan A}}{{1 - \tan 2A\tan A}} \\
    = \frac{{\left( {\frac{{2\tan A}}{{1 - \tan ^2 A}}} \right) + \tan A}}{{1 - \left( {\frac{{2\tan A}}{{1 - \tan ^2 A}}} \right) \cdot \tan A}} \\
    = \frac{{\frac{{2\tan A + \left( {\tan A\left( {1 - \tan ^2 A} \right)} \right)}}{{1 - \tan ^2 A}}}}{{\frac{{\left( {1 - \tan ^2 A} \right) - 2\tan ^2 A}}{{1 - \tan ^2 A}}}} \\
    = \frac{{2\tan A + \left( {\tan A\left( {1 - \tan ^2 A} \right)} \right)}}{{\left( {1 - \tan ^2 A} \right) - 2\tan ^2 A}} \\
    = \frac{{2\tan A + \tan A - \tan ^3 A}}{{1 - \tan A - 2\tan ^2 A}} \\
    = \frac{{3\tan A - \tan ^3 A}}{{1 - \tan A - 2\tan ^2 A}} \\
    \end{array}
    \[/tex]

    thank you,
    Pavadrin
     
  2. jcsd
  3. Sep 9, 2006 #2
    I randomly chose A=23 and your formula didn't produce the same value as tan(69). On your 4th to 5th step you dropped the ^2 term on one of the tangents in the denominator
     
  4. Sep 9, 2006 #3
    okay thanks for checking. ill try top fix the problem and re-post
     
  5. Sep 9, 2006 #4
    okay i finally got some time to solve it again. is the triple angle of tangent equal to this:

    [tex]
    \frac{{\frac{{2\tan A}}{{1 - \tan ^2 A}} + \tan A}}{{1 - \frac{{2\tan ^2 A}}{{1 - \tan ^2 A}}}}
    [/tex]

    and is that as simple as what i can get it?
     
  6. Sep 9, 2006 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Multiply both numerator and denominator by 1- tan2A to get
    [tex]\frac{2tanA+ tanA(1- tan^2 A)}{1- tan^2 A- 2tan^2A}[/tex]
    That's essentially what you have in the fifth line of your original calculation. Then
    [tex]tan 3A= \frac{3tan A- tan^3 A}{1- 3tan^2 A}[/itex]

    As vsage said (although it was between your fifth and sixth lines by my count, not fourth and fifth) one tan2 A accidently became tan A.
     
  7. Sep 9, 2006 #6
    okay thanks for the reliy and the correction, ill try to be more careful next time
     
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