From polar to cartesian equation

how do i convert the polar eq. (r^2)(cos2Θ)=16

so r^2 = x^2 + y^2 and rcosΘ = x

then x^2 + y^2 = (16)/(2x)

is it done like that?

16/2x does not make sense, this means you have said cos2Θ = 2x, this is not true.

(1/2)(1 + cos2Θ) = cos^2Θ --> cos2Θ = 2 cos^2 Θ - 1
(1/2)(1 - cos2Θ) = sin^2 Θ --> cos2Θ = 1 - 2 sin^2 Θ

Use whichever one you want, for instance we can write:

r^2 cos2Θ = r^2 (2 cos^2Θ - 1) = 2 r^2 cos^2 Θ - r^2

or

r^2 cos2Θ = r^2 (1 - 2 sin^2 Θ) = r^2 - 2 r^2 sin^2 Θ

so we can write either

2 r^2 cos^2 Θ - r^2 = 16

or

r^2 - 2r^2 sin^2 Θ = 16

they are equivalent.

For example, if we choose to go with the second one, then notice r^2 = x^2 + y^2, and y = r sin Θ, so our equation is

x^2 + y^2 - 2 y^2 = 16

x^2 - y^2 = 16

This is a hyperbola.

Notice if we chose the other equation that we got from a different identity: 2 r^2 cos^2 Θ - r^2 = 16, we would have

2 x^2 - x^2 - y^2 = 16

x^2 - y^2 = 16

which is the same thing.

I do not follow what you did, but I do not see a way it is equivalent to what I wrote, so I do not think it is done like that (however it is that you "did it" that is! :) ).

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@YahooUser: I think you quoted an identity for sine, not cosine, i.e. sin 2Θ = 2 cosΘ sin Θ, I do not believe there is an identity like that for cos2Θ

cos2Θ = 2sinΘcosΘ

r^2 (cos2Θ) = 2xy =16

Study polar coordinates. It doesn't look like you understand them... and do some practice. It is simple stuff.