|Aug13-08, 04:10 AM||#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?!
|Aug13-08, 04:19 AM||#2|
Blog Entries: 5
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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