# help simplify the expression

• Oct 1st 2012, 04:16 PM
pnfuller
help simplify the expression
simplify the expression:
cos(2arctanx)

im not really sure if im heading in the right direction or not....
• Oct 1st 2012, 04:52 PM
pickslides
Re: help simplify the expression
• Oct 1st 2012, 05:01 PM
HallsofIvy
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 $sin(\theta)$?
• Oct 1st 2012, 05:08 PM
pnfuller
Re: help simplify the expression
i know that but is my answer right?
Quote:

Originally Posted by HallsofIvy
[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 $sin(\theta)$?

• Oct 1st 2012, 05:44 PM
Prove It
Re: help simplify the expression
Quote:

Originally Posted by HallsofIvy
$cos(2\theta)= 2 cos(\theta)sin(\theta)$

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*}.