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Thread: y=arctan(x) , show that....

  1. #1
    Senior Member Twig's Avatar
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    y=arctan(x) , show that....

    Hi

    I am not sure about my solution, if itīs done "in the right way". Anyway, here it comes:

    $\displaystyle \mbox{ Let } y=arctan(x) \, \mbox{ , show that } cos^{2}(y)=\frac{1}{1+x^{2}} $

    I did:

    $\displaystyle tan(y)=x \Longleftrightarrow \frac{sin(y)}{cos(y)}=x \Rightarrow cos(y)=\frac{sin(y)}{x} $

    Now raising both sides to the second power...

    $\displaystyle cos^{2}(y)=\frac{1-cos^{2}(y)}{x^{2}} \Longleftrightarrow cos^{2}(y)=\frac{1}{1+x^{2}} $

    Thanks!
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  2. #2
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    Given: $\displaystyle y = arctan(x)$
    Show that: $\displaystyle \cos^{2}(y)=\frac{1}{1+x^{2}}$

    Looks right to me. But here's another way, just in case:
    Draw a right triangle. Label one of the acute angles y. Since
    $\displaystyle y = arctan(x)$,
    $\displaystyle \tan y = x = \frac{x}{1}$.
    Label the opposite side x and the adjacent side 1. Find the hypotenuse (s) by using the Pythagorean theorem:
    $\displaystyle \begin{aligned}
    s^2 &= 1^2 + x^2 \\
    s &= \sqrt{1 + x^2}
    \end{aligned}$
    Cosine is adjacent over hypotenuse, so
    $\displaystyle \cos y = \frac{1}{\sqrt{1 + x^2}}$.
    Squaring both sides gives you
    $\displaystyle \cos^2 y = \frac{1}{1 + x^2}$.


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  3. #3
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    Hello, Twig!

    Your work looks fine to me . . .


    Let: $\displaystyle y\:=\:\arctan x$, show that: .$\displaystyle \cos^2\!y\:=\:\frac{1}{1+x^2}$
    Here's my approach . . .


    We have: .$\displaystyle y \:=\:\arctan x\quad\Rightarrow\quad \tan y \:=\:x $

    Square both sides: .$\displaystyle \tan^2\!y \:=\:x^2$

    $\displaystyle \text{Add 1 to both sides: }\;\underbrace{1 + \tan^2\!y}_{\text{This is }\sec^2\!y} \:=\:1+x^2 \quad\Rightarrow\quad \sec^2\!y \:=\:1+x^2$

    Take reciprocals: .$\displaystyle \cos^2\!y \:=\:\frac{1}{1+x^2}$

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