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

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

    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..
    Attached Thumbnails Attached Thumbnails x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi-1.jpg   x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi-2.jpg  
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  2. #2
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    Re: x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

    Just spitballing:

    x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi-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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  3. #3
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    Re: x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

    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..
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  4. #4
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    Re: x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

    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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  5. #5
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    Re: x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

    Thanks from Vali and topsquark
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    Re: x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

    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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  7. #7
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    Re: x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

    Thank you for the help!
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  8. #8
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    Re: x values for 2arctan(x) + arcsin(2x/(1+x^2))=pi

    Quote Originally Posted by Vali View Post
    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 $x \geq 1$
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