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Interesting yet challenging proof

  1. Mar 24, 2008 #1
    how can I show that ... arctan(1/v) = (π/2) - arctan(v) ???
     
  2. jcsd
  3. Mar 24, 2008 #2
    Kind of a hint: v = v/1 = 1/(1/v). You just need to remember the definition of tan
     
  4. Mar 24, 2008 #3
    I'm sorry, I don't understand what you did with the v. And what about the pi/2 part?
     
  5. Mar 24, 2008 #4
    In words, arctan(x) is the angle formed by a right triangle whose side opposite the angle and adjacent sides have ratio x/1. What is the measure of the other non-right angle in this hypothetical triangle, and what would the tangent of this other angle be?
     
  6. Mar 24, 2008 #5
    Here's how I think of it... I rearranged the equation so that arctan(1/v) + arctan(v) = π/2

    And I understand how the tan of, say ø = 1/v, making the tan of, say ß = v. That part makes sense. The part I'm getting confused on is the π/2, and what that does to the equation.
     
  7. Mar 25, 2008 #6
    what's the sum of the angles in a triangle? Trig functions assume that one angle is 90 degrees. If one angle is [itex]\theta[/itex], then what's the other? Draw a triangle to aid you in this. Make one such that [itex]arctan(v)=\theta[/itex]. What's the other angle? What's the tangent of this other angle?
     
  8. Mar 25, 2008 #7
    Ok, the other angle must be 180 - (90+ø), correct? And the tangent of this angle would be v... right?

    I'm still not seeing this proving business.
     
  9. Mar 25, 2008 #8
    No! For the other angle, which is the opposite side and which is the adjacent?
     
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