Trigonometry: arctan equation?

[XYZ313]

Can someone please let me know how to find x in

π/4=arctan(x+arctan(x+...)) ?

 

[micromass]

Let me do a similar one:

Let $x=\sqrt{2+\sqrt{2+\sqrt{2+...}}}$. Then $x^2-2=x$.
Solving the quadratic equation gives us the answer.

Can you do your problem now along similar lines??

 

[XYZ313]

micromass: isn't it the use of inverse operator (tan in this case)?
like z=arctan(x+arctan(x+...)) and make the use of tanz?

and then use tan addition rule? I am really confused...

Can you explain further please?

 

[lurflurf]

π/4=arctan(x+arctan(x+...))
take tan of both sides
tan(π/4)=tan(arctan(x+arctan(x+...)) )

 

[XYZ313]

lurflurf: and then what?

tan(π/4) = x + arctan(x+arctan(x+...)) ?

 

[XYZ313]

what is tan(tanx)? Is above equation some kind of sum of geometric progress?

 

[XYZ313]

is it so easy that no one is going even to comment?

 

[lurflurf]

tan(π/4)=tan(arctan(x+arctan(x+...)) )
1=x+arctan(x+arctan(x+...))

 

[XYZ313]

lurflurf: thank you, this is obvious and not the solution

then

tan(1-x) = x + arctan(x+arctan(x+...))

and where it is leading?

 

[lurflurf]

arctan(x+arctan(x+...)) ?

 

[XYZ313]

if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread...

 

[spamiam]

if I knew what "arctan(x+arctan(x+...))" equals to, I wouldn't be posting this thread...

What? You said what this equals in your first post. It's x that is the unknown.

 

[XYZ313]

spamiam: right, sorry, I'm really dumb

x = 1 - π/4

Thank you all for bearing with me...

 

[XYZ313]

This is dumbest question I have ever asked. Hope this is never going to happen again...

 

[spamiam]

You're not dumb--this question is difficult until you see the trick to use.