
#1
Aug1308, 04:10 AM

P: 4

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



#2
Aug1308, 04:19 AM

Sci Advisor
HW Helper
P: 4,301

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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