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For all x∈]0;2pi[ : tan(x) > x

  1. Dec 23, 2009 #1
    Can someone prove this?

    For all x∈]0;2pi[ : tan(x) > x
  2. jcsd
  3. Dec 23, 2009 #2


    Staff: Mentor

    No, because it isn't true on the interval you gave. The tangent function is undefined at x = pi/2 and x = 3pi/2. For all x in (pi/2, pi), tan(x) < x, and the same is true for the interval (3pi/2, 2pi).
  4. Dec 23, 2009 #3
    No, 0 and 2pi don't belong to the interval (see the ] [ ), so it is true.
    We've seen a similar example with sin x: for all x > 0: sin x < x
    But with tan x I can't prove it.
  5. Dec 23, 2009 #4


    User Avatar
    Science Advisor

    But [itex]\pi/2[/itex] and [itex]3\pi/2[/itex] do belong to that interval! That's what Mark44 was saying. And, in fact, for [itex]\pi/2< x< \pi[/itex], tan(x) is negative and can't possibly be larger than x!

    You need to reduce to [itex]]0, \pi/2[[/itex] in order to have tan(x)< x.
    Look at the function f(x)= tan(x)- x. It's derivative is sec2(x)- 1 and since sec(x)> 1 for all x in [itex]]0, \pi/2[[/itex], that derivative is positive.
  6. Dec 24, 2009 #5
    Oh sorry I made a mistake.. it had to be: For all x∈]0;pi/2[ : tan(x) > x

    I'm sorry, you were both right
    Thanks for the answer!
  7. Dec 24, 2009 #6
    I'm very sorry, I made a mistake. You were right

    It had to be: For all x∈]0;pi/2[ : tan(x) > x

  8. Dec 24, 2009 #7


    Staff: Mentor

    See HallsOfIvy's post #4.
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