# Trigonometry: arctan equation?

• XYZ313
In summary, the conversation is about finding the value of x in π/4=arctan(x+arctan(x+...)) by using similar methods to the problem x=\sqrt{2+\sqrt{2+\sqrt{2+...}}}. It involves taking the tangent of both sides and using the tangent addition rule. The final solution is x = 1 - π/4.

#### XYZ313

Can someone please let me know how to find x in

π/4=arctan(x+arctan(x+...)) ?

Let me do a similar one:

Let $x=\sqrt{2+\sqrt{2+\sqrt{2+...}}}$. Then $x^2-2=x$.

Can you do your problem now along similar lines??

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...

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

lurflurf: and then what?

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

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

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

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

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

then

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

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

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

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.

spamiam: right, sorry, I'm really dumb

x = 1 - π/4

Thank you all for bearing with me...

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

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