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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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Can you explain a bit more your answer? Im working with $$x=sin\phi cos\theta$$ $$y=sin\phi sin\theta $$ $$ z=cos\phi$$ –  Rakisbro Dec 24 '12 at 15:36
    
you mean, what you wrote multiplied by $\rho$ on the right-hand side, right? or is $\rho=\sqrt{x^2+y^2}$ ? (my confusion arises 'cause one usually reserves $\rho$ for cylindric coordinates.) –  c.p. Dec 24 '12 at 15:56
    
@Rakisbro I updated... –  c.p. Dec 24 '12 at 16:03
    
Now It is very clear. Thanks for your time Jorge Campos. –  Rakisbro Dec 24 '12 at 16:14

$(x^2+y^2)\cos(2\arctan(y/x))-1 = (x^2+y^2).\frac{x^2-y^2}{x^2+y^2} -1 = x^2-y^2-1$

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