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....

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- Oct 1st 2012, 04:16 PMpnfullerhelp simplify the expression
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.... - Oct 1st 2012, 04:52 PMpickslidesRe: help simplify the expression
I think you are on the right track

simplify cos(2*arctan(x)) - Wolfram|Alpha - Oct 1st 2012, 05:01 PMHallsofIvyRe: help simplify the expression
$\displaystyle cos(2\theta)= 2 cos(\theta)sin(\theta)$

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 $\displaystyle sin(\theta)$? - Oct 1st 2012, 05:08 PMpnfullerRe: help simplify the expression
- Oct 1st 2012, 05:44 PMProve ItRe: help simplify the expression
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, $\displaystyle \displaystyle \begin{align*} \cos{2\theta} \equiv 2\cos^2{\theta} - 1 \end{align*}$.