# Homework Help: Interesting yet challenging proof

1. Mar 24, 2008

### Calixto

how can I show that ... arctan(1/v) = (π/2) - arctan(v) ???

2. Mar 24, 2008

### Einstienear

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

More of a trial and error method.

Einstienear.

3. Mar 24, 2008

### Calixto

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. Mar 24, 2008

### Feldoh

Do v and n represent anything in particular?

Because your equation only works when you have certain values for v and n.

5. Mar 24, 2008

### Calixto

i knew this would confuse someone... that "n" you see is π, or pi... sorry. Does that help?

6. Mar 26, 2008

### mhill

if you use the log representation for artan (1/x) and artan (x) so

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

get the accurate result.

7. Mar 26, 2008

### tiny-tim

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

8. Mar 26, 2008

### dynamicsolo

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: Mar 26, 2008
9. Mar 26, 2008

### Calixto

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. Mar 27, 2008

### tiny-tim

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

= 2i.arctan(x).

(You see how, to prove anything with arctan(x), you always convert to x = tany?)

11. Mar 28, 2008

### Gib Z

You basically want to show that

$$\arctan (1/v) + \arctan (v) = \frac{\pi}{2}$$.

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