I have gotten this far: Using cos(2θ) = 1-tan^{2}θ / 1+tan^{2}θ From a previous question: Let θ= arctan(x): cos(2θ) = 1-tan^{2}(arctan(x)) / 1+tan^{2}(arctan(x)) =1-x^{2} / 1+x^{2} Where x^{2} 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?!
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.