simplify the expression:
cos(2arctanx)
would the answer be 2x/sqrt(x^{2}+1)?
im not really sure if im heading in the right direction or not....
I think you are on the right track
simplify cos(2*arctan(x)) - Wolfram|Alpha
$\displaystyle cos(2\theta)= 2 cos(\theta)sin(\theta)$
And, if $\displaystyle \theta= arctan(x)$, then $\displaystyle tan(\theta)= x$.
We can represent that by a right triangle having angle $\displaystyle \theta$, "opposite side" x, and "near side" 1. By the Pythagorean theorem the hypotenuse has length $\displaystyle \sqrt{1+ x^2}$. So what are $\displaystyle cos(\theta)$ and $\displaystyle sin(\theta)$?
I'm pretty sure that $\displaystyle \displaystyle \begin{align*} \cos{2\theta} \end{align*}$ is NOT $\displaystyle \displaystyle \begin{align*} 2\cos{\theta}\sin{\theta} \end{align*}$. That would be $\displaystyle \displaystyle \begin{align*} \sin{2\theta} \end{align*}$.
Rather, $\displaystyle \displaystyle \begin{align*} \cos{2\theta} \equiv 2\cos^2{\theta} - 1 \end{align*}$.