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  1. #1
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    help simplify the expression

    simplify the expression:
    cos(2arctanx)

    would the answer be 2x/sqrt(x2+1)?

    im not really sure if im heading in the right direction or not....
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  2. #2
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    Re: help simplify the expression

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  3. #3
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    Re: help simplify the expression

    cos(2\theta)= 2 cos(\theta)sin(\theta)

    And, if \theta= arctan(x), then tan(\theta)= x.

    We can represent that by a right triangle having angle \theta, "opposite side" x, and "near side" 1. By the Pythagorean theorem the hypotenuse has length \sqrt{1+ x^2}. So what are cos(\theta) and sin(\theta)?
    Last edited by HallsofIvy; October 1st 2012 at 05:42 PM.
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    Re: help simplify the expression

    i know that but is my answer right?
    Quote Originally Posted by HallsofIvy View Post
    [tex]cos(2\theta)= 2 cos(\theta)sin(\theta)[/itex]

    And, if \theta= arctan(x), then tan(\theta)= x.

    We can represent that by a right triangle having angle \theta, "opposite side" x, and "near side" 1. By the Pythagorean theorem the hypotenuse has length \sqrt{1+ x^2}. So what are cos(\theta) and sin(\theta)?
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  5. #5
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    Re: help simplify the expression

    Quote Originally Posted by HallsofIvy View Post
    cos(2\theta)= 2 cos(\theta)sin(\theta)

    And, if \theta= arctan(x), then tan(\theta)= x.

    We can represent that by a right triangle having angle \theta, "opposite side" x, and "near side" 1. By the Pythagorean theorem the hypotenuse has length \sqrt{1+ x^2}. So what are cos(\theta) and sin(\theta)?
    I'm pretty sure that \displaystyle \begin{align*} \cos{2\theta} \end{align*} is NOT \displaystyle \begin{align*} 2\cos{\theta}\sin{\theta} \end{align*}. That would be \displaystyle \begin{align*} \sin{2\theta} \end{align*}.

    Rather, \displaystyle \begin{align*} \cos{2\theta} \equiv 2\cos^2{\theta} - 1 \end{align*}.
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