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Math Help - Prove this identity

  1. #1
    Member mybrohshi5's Avatar
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    Prove this identity

    Prove this identity:
    arcsin(x-1/x+1) = 2arctan(sqrt(x)) - pi/2

    First of all to prove this do i take the derivative of one side?
    My book does a horrible job at explaining any of this and then they throw problems like this in there.

    I tried taking the derivative of the right side and got

    arcsin(x-1/x+1) = 1/[sqrt(x) + (x)sqrt(x)]

    but im not sure what to do from here or even if i am doing it correctly.

    any help would be great.

    thank you
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  2. #2
    MHF Contributor
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    Quote Originally Posted by mybrohshi5 View Post
    Prove this identity:
    arcsin(x-1/x+1) = 2arctan(sqrt(x)) - pi/2

    First of all to prove this do i take the derivative of one side?
    My book does a horrible job at explaining any of this and then they throw problems like this in there.

    I tried taking the derivative of the right side and got

    arcsin(x-1/x+1) = 1/[sqrt(x) + (x)sqrt(x)]

    but im not sure what to do from here or even if i am doing it correctly.

    any help would be great.

    thank you
    Is the identity true - put x = 1!
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Danny View Post
    Is the identity true - put x = 1!
    Perhaps he meant to write:

    \arcsin\left(\frac{x-1}{x+1}\right)=2\arctan(\sqrt{x})-\frac{\pi}{2}

    because that does hold when x=1 (and everywhere in fact, because it's true!).

    We are going to show that \frac{x-1}{x+1}=\sin\left(2\arctan(\sqrt{x})-\frac{\pi}{2}\right)

    Using the sum identity, the right side becomes:

    \sin\left(2\arctan(\sqrt{x})\right)\cos(\pi/2)-\cos\left(2\arctan(\sqrt{x})\right)\sin(\pi/2)=-\cos\left(2\arctan(\sqrt{x})\right)

    Using the double angle formula ( \cos2\theta=\cos^2\theta-\sin^2\theta, with \theta=\arctan(\sqrt{x}) in this case), we can draw out the triangle and find:

    -\cos\left(2\arctan(\sqrt{x})\right)=-\left(\left(\frac{1}{\sqrt{x+1}}\right)^2-\left(\frac{\sqrt{x}}{\sqrt{x+1}}\right)^2\right)=  \frac{x}{x+1}-\frac{1}{x+1}=\frac{x-1}{x+1}

    as desired.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by mybrohshi5 View Post
    Prove this identity:
    arcsin(x-1/x+1) = 2arctan(sqrt(x)) - pi/2
    let \theta = \arctan(\sqrt{x})

    take the sine of the right side ...

    \sin \left(2\theta - \frac{\pi}{2}\right) =

    \sin(2\theta)\cos\left(\frac{\pi}{2}\right) - \cos(2\theta)\sin\left(\frac{\pi}{2}\right) =

    -\cos(2\theta) = 2\sin^2{\theta} - 1

    if \theta = \arctan(\sqrt{x}) , \sin{\theta} = \frac{\sqrt{x}}{\sqrt{1+x}}

    2\sin^2{\theta} - 1 = \frac{2x}{1+x} - \frac{1+x}{1+x} = \frac{x-1}{x+1}

    so ...

    \sin \left(2\theta - \frac{\pi}{2}\right) = \frac{x-1}{x+1}

    and ...

    \arcsin\left(\frac{x-1}{x+1}\right) = 2\theta - \frac{\pi}{2} = 2\arctan(\sqrt{x}) - \frac{\pi}{2}<br />
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