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I need convert this spherical expression, to a rectangular form (specific surface). $\rho^2\cos(2\phi)-1=0$ Thanks for a while.

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If $\phi$ is the polar and $\theta$ the azimuthal coordinate, using double angle trigonometric identity your surface is described by $\rho^2\cos(2\phi)=\rho^2\cos^2(\phi)-\rho^2\sin^2(\phi)=1,$i.e. $\rho^2\cos^2(\phi)-\rho^2\sin^2(\phi)(\sin^2\theta+\cos^2\theta)=1$ which transforming to Cartesian coordinates yields the hyperboloid $z^2-(y^2+x^2)=1$.

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    Now It is very clear. Thanks for your time Jorge Campos.2012-12-24
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$(x^2+y^2)\cos(2\arctan(y/x))-1 = (x^2+y^2).\frac{x^2-y^2}{x^2+y^2} -1 = x^2-y^2-1$