x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

Dec 2018
19
0
Romania
I know x real number
I need to find "x" values such that:
2arctan(x) + arcsin(2x/(1+x^2))=pi
A. (0,infinity)
B. (-infinity,-1)U[1, infinity)
C. [1, infinity) right answer
D. [-1,1]
E. [2,infinity)
I tried everything I know.
I found only x=1
I need some ideas,suggestions..
 

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Dec 2013
2,000
757
Colombia
Just spitballing:

geogebra-export.png
In the above diagram you can see that $\theta$ is equal to $\sin \frac{2x}{1+x^2}$.
Also, if the base of the triangle is length $2$, we have that $\tan \theta = x$.

What can you say about the triangle for various values of $x$?
 
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Dec 2018
19
0
Romania
I don't really know.We can find the base which is x^2 - 1
And for every x the lenghts of the triangle are Pythagorean numbers.
I don't see what I should see, unfortunately..
 

topsquark

Forum Staff
Jan 2006
11,565
3,453
Wellsville, NY
Try this.

Given \(\displaystyle x = \dfrac{1}{\sqrt{3}} , ~1, ~ \sqrt{3}\)

Then
\(\displaystyle 2 atan(x) + asn \left ( \dfrac{2x}{1 + x^2} \right )\)

gives \(\displaystyle \dfrac{2 \pi}{3} ,~ \pi, ~ \pi \)

That means you aren't using a large enough domain when you have done your calculations. The reason that this is important is because neither sine nor tangent are single valued. (ie. \(\displaystyle sin(x + 2 \pi ) = sin(x)\).)

-Dan
 
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Jun 2013
1,110
590
Lebanon
Let

$u=2$ arctan $x$ where $-\pi <u<\pi $

and

$v=\arcsin \left(\frac{2x}{1+x^2}\right)$ where $-\frac{\pi }{2}\leq v\leq \frac{\pi }{2}$

If $u<\pi /2$ then $v=\pi -u>\pi /2$ contradiction

therefore $u\geq \frac{\pi }{2}$ and $x\geq 1$
 
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Reactions: 2 people
Dec 2018
19
0
Romania
Thank you for the help!
 
Jun 2013
1,110
590
Lebanon
Thank you for the help!
you are welcome.
What I have written is not a proof, only a couple of observations.

you still need to show that the identity holds for $x \geq 1$