Simplify cos(2arctan(x))

  1. 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. jcsd
  3. CompuChip

    CompuChip 4,296
    Science Advisor
    Homework Helper

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

    As for the simplification, in such cases you could try to do some factorization (e.g. [itex]x^2 - 1 = (x + 1)(x - 1)[/itex]) 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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