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Thread: Equation

  1. #1
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    Equation

    Hi people,

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

    $\displaystyle x=tan y$

    $\displaystyle sec^2 y=1+tan^2 y$

    $\displaystyle sec^2 y=1+x^2$

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

    $\displaystyle sin y=tan y cos y$

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

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

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