Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Identity: arctan(1/x) = arcot(x) or arccot(1/x) = arctan(x)

  1. Jan 14, 2014 #1
    I've been looking for this identity: arctan(1/x) = arcot(x) or arccot(1/x) = arctan(x)

    After just visually inspecting this to be true, I have been unable to find any formal proofs for it.
    Any references would be great!
     
  2. jcsd
  3. Jan 14, 2014 #2
    The cotangent function is defined as the reciprocal of the tangent function.
     
  4. Jan 15, 2014 #3
    The identity is true for positive values of ##x## but not for negative.

    I would suggest starting from the definitions of those functions; [tex]y=\arctan x\iff \tan y=x \text{ and }y\in (-\frac{\pi}{2},\frac{\pi}{2})[/tex] and [tex]y=\text{arc}\cot x\iff \cot y=x \text{ and }y\in (0,\pi)[/tex] along with the trig identity suggested by Chestermiller.
     
  5. Jan 15, 2014 #4
    Is there a reason cot and tan defined on different ranges?
     
  6. Jan 15, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes. Since tangent and arctangent are periodic functions, in order to talk about "well-defined" inverses (since [itex]tan(\pi/4)= tan(5\pi/4)= 1[/itex] should arctan(1) be [itex]\pi/4[/itex] or [itex]5\pi/4[/itex]?), we have to restrict the range of tangent and cotangent and so restrict the domain of arctangent and arccotangent. The usual convention is to restrict to the largest interval containing 0 on which the function is one-to-one. For the tangent that is [itex]-\pi/2[/itex] to [itex]\pi/2[/itex]. Since [itex]cot(x)= tan(\pi/2- x)[/itex], the corresponding domain for cotangent is 0 to [itex]\pi[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Identity: arctan(1/x) = arcot(x) or arccot(1/x) = arctan(x)
  1. Area under ArcTan[x] (Replies: 10)

  2. F(x) = x/x+1 (Replies: 3)

Loading...