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Math Help - Simplifying Inverse Trigonometric Functions

  1. #1
    jhb
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    Simplifying Inverse Trigonometric Functions

    Hello all,

    well, I'm not sure how to go about simplifying the following problems

    tan(2arctan(x))

    and

    cos(2arcsin(x))

    I know I should let theta = 2arctan(x) => tan(theta) = 2x .... I think. However, from that point forward I'm not sure what to do.

    I have several others, however these two are the ones that I believe will enable me to figure out the remainder of the problems.

    Any and all help is greatly appreciated. Thank you for your time.


    jhb
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  2. #2
    Super Member Random Variable's Avatar
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     \tan(2u) = \frac {2 \tan(u)}{1-\tan^{2}(u)}

    so  \tan(2 \arctan(x)) = \frac {2 \tan(\arctan(x))}{1 - \tan^{2}(\arctan(x))}  = \frac {2x}{1 -x^{2}}
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  3. #3
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    Quote Originally Posted by jhb View Post
    Hello all,

    well, I'm not sure how to go about simplifying the following problems

    tan(2arctan(x))

    and

    cos(2arcsin(x))

    I know I should let theta = 2arctan(x) => tan(theta) = 2x .... I think. However, from that point forward I'm not sure what to do.

    I have several others, however these two are the ones that I believe will enable me to figure out the remainder of the problems.

    Any and all help is greatly appreciated. Thank you for your time.


    jhb
    <br />
\cos{2\sin^{-1}{x}} = \cos{2\theta} = 1 - 2\sin^2{\theta}

    since i let  \sin^{-1}{x} = \theta \implies x = \sin{\theta}

     \cos{2\sin^{-1}{x}}  =  1 - 2x^2
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  4. #4
    jhb
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    awesome. thank you for clearing that up for me... I see what I was missing.
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