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Homework Help: Trigonometric identity for inverse tan

  1. Apr 28, 2010 #1
    Hello,

    Could you please clarify if this is correct:

    If tan^(-1)(x) = Pi/2 - tan^(-1)(1/x)

    Then if we have (ax) as the angle where a is a constant, do we get:
    tan^(-1)(ax) = Pi/2 - tan^(-1)(a/x)
    or does the constant go on the bottom with the x? i.e. or:
    tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))

    Thank you very much,
    Kat
     
  2. jcsd
  3. Apr 28, 2010 #2

    berkeman

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    Staff: Mentor

    I believe that you can answer this yourself by drawing a right triangle (use a 30-60-30 triangle), and writing the tangent in terms of the ratio of the opposite and adjacent sides. Then double the height of the opposite side....
     
  4. Apr 28, 2010 #3
    Do you think the identity should change if you replace the letter x with the letter y? Is the identity invalid for some value of y? What if y = ax? If you agree that it is still valid, what do you get when you put ax in the place of y?
     
  5. Apr 28, 2010 #4
    OK so it seems to work out as the
    tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))
    Could you please confirm that this is correct?
     
  6. Apr 28, 2010 #5

    berkeman

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    Staff: Mentor

    Which method did you use to conclude that? Giga and I have a bet going...
     
  7. Apr 28, 2010 #6
    lol how kind of you.. Jokers heheee
    The triangle one.
    They both makes sence. Thanks,
    Kat
     
  8. Apr 28, 2010 #7

    berkeman

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    Staff: Mentor

    Oh fooey. If they both make sense, then neither of us wins the bet. Oh well :smile:
     
  9. Apr 28, 2010 #8

    Mentallic

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    Homework Helper

    Nah, Giga's make more sense AND no pen and paper required :tongue:
     
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