1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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

    User Avatar

    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

    User Avatar

    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

    User Avatar
    Homework Helper

    Nah, Giga's make more sense AND no pen and paper required :tongue:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook