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Math Help - Equation

  1. #1
    Junior Member
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    Equation

    Hi people,

    I want to show that sin(Arctanx)=\frac{x}{\sqrt{1+x^2}}, can you help me please???
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  2. #2
    MHF Contributor alexmahone's Avatar
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    let tan^{-1}x=y

    x=tan y

    sec^2 y=1+tan^2 y

    sec^2 y=1+x^2

    sec y=\sqrt{1+x^2}

    sin y=tan y cos y

    sin y=\frac{tan y}{sec y}

    sin y=\frac{x}{\sqrt{1+x^2}}

    sin(tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by lehder View Post
    Hi people,

    I want to show that sin(Arctanx)=\frac{x}{\sqrt{1+x^2}}, can you help me please???
    For a (imho) somewhat more intuitive solution consider a right triangle with acute angle \alpha, opposite leg x and adjacent leg 1: By construction we have \arctan(x)=\alpha. But by Pythagoras's theorem we also have that the length of the hypotenuse must be \sqrt{1+x^2}. By definition of \sin in a right triangle we find that \sin(\alpha)=\frac{x}{\sqrt{1+x^2}}, i.e. the ratio of opposite leg to hypotenuse in that right triangle.
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