# Simplify cos(2arctan(x))

1. Aug 13, 2008

### steph124355

I have gotten this far:

Using cos(2θ) = 1-tan2θ / 1+tan2θ From a previous question:

Let θ= arctan(x):

cos(2θ) = 1-tan2(arctan(x)) / 1+tan2(arctan(x))

=1-x2 / 1+x2

Where x2 cannot equal 0 or a negative number

Have I done this in the right way and if so is this as far as I can simplify it?!

2. Aug 13, 2008

### CompuChip

You did the simplification correctly, but usually we give the domain in terms of x, not its square. And why is $x^2$ not allowed to be zero? $x^2$ can never be negative, but how about x?

As for the simplification, in such cases you could try to do some factorization (e.g. $x^2 - 1 = (x + 1)(x - 1)$) and cancel something in the top and bottom, though you'll find in this case that such an approach doesn't help you very much.

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