arctan

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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: $<br />$

Your code should look like
[tex]\\ \frac{-\pi}{4}= \arctan{\frac{(x+1)}{(x-1)}} +\arctan(x)\\[/tex]
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!!!