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Math Help - working with inverse trig functions

  1. #1
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    working with inverse trig functions

    Hey, all.

    In another thread, someone was asking a question which involved solving this equation for x:

    \frac{\arctan{x}}{\arccos{x}}=\frac{\sqrt{1-x^2}}{1+x^2}

    How would one go about tackling something like that? Can it even be done?

    I'm not asking about this equation specifically. Wikipedia has a few helpful identities, but nothing which I can use to solve the above equation for x.

    Thoughts?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by hatsoff View Post
    Hey, all.

    In another thread, someone was asking a question which involved solving this equation for x:

    \frac{\arctan{x}}{\arccos{x}}=\frac{\sqrt{1-x^2}}{1+x^2}

    How would one go about tackling something like that? Can it even be done?

    I'm not asking about this equation specifically. Wikipedia has a few helpful identities, but nothing which I can use to solve the above equation for x.

    Thoughts?
    I don't think we can find an exact answer. We can approximate x, though.

    One way [using calculus] would be to apply the Newton-Raphson Method: x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

    where f(x)=\sqrt{1-x^2}\cos^{-1}(x)-(1+x^2)\tan^{-1}(x)

    I hope this helps!

    --Chris
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by hatsoff View Post
    Hey, all.

    In another thread, someone was asking a question which involved solving this equation for x:

    \frac{\arctan{x}}{\arccos{x}}=\frac{\sqrt{1-x^2}}{1+x^2}

    How would one go about tackling something like that? Can it even be done?

    I'm not asking about this equation specifically. Wikipedia has a few helpful identities, but nothing which I can use to solve the above equation for x.

    Thoughts?
    You should start by using a graphical method, so you have some idea where the solutions are and how many there are. In this case the domain we are interested in is [-1, 1) and the root is near 0.6.

    Such equations usually do not have simple solutions, but there is a little trick that can be employed to find them even if you can't by manipulating the equations. If you calculate the solution numerically to high precision you can feed it into the inverse symbolic calculator to see if it can identify probable forms for the solution, then test them to see if they are in fact exact solutions to the equations.

    In this case using a solution correct to 15 digits turns up no candidates on ISC.


    RonL
    Last edited by CaptainBlack; August 9th 2008 at 12:48 AM.
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