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Interesting yet challenging proof

  • Thread starter Calixto
  • Start date
  • #1
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how can I show that ... arctan(1/v) = (π/2) - arctan(v) ???
 

Answers and Replies

  • #2
well,try giving and playing with the variables untill you get a result.

More of a trial and error method.

Einstienear.
 
  • #3
16
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I know I've been doing that, moving around variables and using the unit circle and right triangles, but I cannot seem to come across a substantial reason why they are equal. Any thoughts?
 
  • #4
1,341
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Do v and n represent anything in particular?

Because your equation only works when you have certain values for v and n.
 
  • #5
16
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i knew this would confuse someone... that "n" you see is π, or pi... sorry. Does that help?
 
  • #6
188
1
if you use the log representation for artan (1/x) and artan (x) so

[tex] artan(x)= (2i)^{-1}(log(1+ix)-log(1-ix)) [/tex] and the same replacing x--> 1/x you

get the accurate result.
 
  • #7
tiny-tim
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how can I show that ... arctan(1/v) = (π/2) - arctan(v) ???
Are you allowed to use x = arctan(v) => v = tanx ? :smile:
 
  • #8
dynamicsolo
Homework Helper
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I'm going to point significantly (*points significantly*) to my signature. The clue is given by tiny-tim (and that is what they want you to use): make a diagram of a right triangle with x as one of the non-right angles and use v, written as v/1 , as the ratio of the sides that would come from finding the tangent of angle x (label the sides of the triangle appropriately).

Now, in the same triangle, what angle has a tangent of 1/v ? What is the relationship between that angle and angle x ?

(And, with all due respect to mhill, while that relationship is true, the math is probably way beyond what is being done in Calixto's course...)
 
Last edited:
  • #9
16
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Thanks, I understand it now... But could you explain some more what you wrote about the logs and stuff? Just maybe explain where that comes from so I can impress my teacher :)
 
  • #10
tiny-tim
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If x = tany, then y = arctanx, and so:

log(1+ix)-log(1-ix) = log[(1+ix)/(1-ix)]

= log[(cosy + isiny)/(cosy - isiny)]

= log[e^{2iy}]

= 2iy

:smile: = 2i.arctan(x). :smile:

(You see how, to prove anything with arctan(x), you always convert to x = tany?)
 
  • #11
Gib Z
Homework Helper
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You basically want to show that

[tex]\arctan (1/v) + \arctan (v) = \frac{\pi}{2}[/tex].

Draw a right angled triangle, with the smaller sides length 1 and v. What does [itex]\arctan v[/itex] represent here?
 

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