How to Prove arctan(1/v) + arctan(v) Equals π/2 Using a Triangle?

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Homework Help Overview

The discussion revolves around proving the identity arctan(1/v) + arctan(v) = π/2 using geometric interpretations, particularly involving right triangles and trigonometric relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore various methods, including manipulating variables, using the unit circle, and considering right triangles to understand the relationship between the angles represented by arctan(1/v) and arctan(v).

Discussion Status

Some participants have offered insights into using right triangles to visualize the problem, while others have introduced complex logarithmic representations. There is ongoing exploration of different interpretations and approaches without a clear consensus on the best method.

Contextual Notes

There are mentions of specific values for variables and constraints related to the context of the problem, as well as references to course content that may influence the discussion.

Calixto
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how can I show that ... arctan(1/v) = (π/2) - arctan(v) ?
 
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well,try giving and playing with the variables until you get a result.

More of a trial and error method.

Einstienear.
 
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?
 
Do v and n represent anything in particular?

Because your equation only works when you have certain values for v and n.
 
i knew this would confuse someone... that "n" you see is π, or pi... sorry. Does that help?
 
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.
 
Calixto said:
how can I show that ... arctan(1/v) = (π/2) - arctan(v) ?

Are you allowed to use x = arctan(v) => v = tanx ? :smile:
 
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:
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
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
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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