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Math Help - 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:

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

    I did:

     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...

    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
    Super Member
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    Given: y = arctan(x)
    Show that: \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
    y = arctan(x),
    \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:
    \begin{aligned}<br />
s^2 &= 1^2 + x^2 \\<br />
s &= \sqrt{1 + x^2}<br />
\end{aligned}
    Cosine is adjacent over hypotenuse, so
    \cos y = \frac{1}{\sqrt{1 + x^2}}.
    Squaring both sides gives you
    \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: y\:=\:\arctan x, show that: . \cos^2\!y\:=\:\frac{1}{1+x^2}
    Here's my approach . . .


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

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

    \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: . \cos^2\!y \:=\:\frac{1}{1+x^2}

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