# arctan

• January 16th 2013, 09:43 AM
Boo
arctan
Dear All!
Pleas,e coudl someone help me to get
f(x)=-pi/4 for x<1
and
f(x)=arctan2+arctan3 for x>1

$f(x)=arctan\frac{x+1}{x-1}+arctanx$
Many thanks!!!
• January 16th 2013, 10:13 AM
Barioth
Re: arctan
you have

$\\ \frac{-\pi}{4}= \arctan{\frac{(x+1)}{(x-1)}} +\arctan(x)\\$
you put Tan everywhere
$\\\tan\frac{-\pi}{4} = \frac{(x+1)}{(x-1)} +(x)\\-1 = \frac{(x+1)}{(x-1)} +(x)\\0 = \frac{(x+1}{x-1} +(x) + 1$

Solve the quadratic formula, you end up with x = 0 and x = (-1)
• January 16th 2013, 10:59 AM
Plato
Re: arctan
Quote:

Originally Posted by Barioth
you have

$\\ \frac{-\pi}{4}= \arctan{\frac{(x+1)}{(x-1)}} +\arctan(x)\\$
you put Tan everywhere
$\\\tan\frac{-\pi}{4} = \frac{(x+1)}{(x-1)} +(x)\\-1 = \frac{(x+1)}{(x-1)} +(x)\\0 = \frac{(x+1}{x-1} +(x) + 1$

Sorry about the <br/> I have no idea why these keep apearing :(

As you can I got rid of the <br/> in your post.
That happens if a linefeed in the code. This like this: $
$

$$\\ \frac{-\pi}{4}= \arctan{\frac{(x+1)}{(x-1)}} +\arctan(x)\\$$
you put Tan everywhere
[TEX]\\\tan\frac{-\pi}{4} = \frac{(x+1)}{(x-1)} +(x)\\-1 = \frac{(x+1)}{(x-1)} +(x)\\0 = \frac{(x+1}{x-1} +(x) + 1[/TEX]
• January 16th 2013, 11:00 AM
Barioth
Re: arctan
Thanks for the info!
• January 17th 2013, 08:38 AM
Boo
Re: arctan
Hello!
Many thanks!
But, I should not know in advance that the result is pi/4! I have to calculate it...how?
P.s. Please, can U tell me how did U got rid of arctans putting tan in front of arctan?
Many thanks!!!