Trigonometry: arctan equation?

XYZ313 · Nov 7, 2011

Can someone please let me know how to find x in

π/4=arctan(x+arctan(x+...)) ?

micromass · Nov 7, 2011

Let me do a similar one:

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

Can you do your problem now along similar lines??

XYZ313 · Nov 7, 2011

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 · Nov 7, 2011

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

XYZ313 · Nov 7, 2011

lurflurf: and then what?

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

XYZ313 · Nov 7, 2011

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

XYZ313 · Nov 7, 2011

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

lurflurf · Nov 7, 2011

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

XYZ313 · Nov 7, 2011

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 · Nov 8, 2011

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

XYZ313 · Nov 8, 2011

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

spamiam · Nov 8, 2011

XYZ313 said: ↑

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 · Nov 8, 2011

spamiam: right, sorry, I'm really dumb

x = 1 - π/4

Thank you all for bearing with me...

XYZ313 · Nov 8, 2011

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

spamiam · Nov 8, 2011

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