What is the Cartesian form of $r-theta = -2sin^3theta+sec^2theta$ ?

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Answer 1 · 2018-02-19T14:15:36

$r-theta=-2sin^3theta+sec^2theta$ in cartesian form is

$sqrt(x^2+y^2)-tan^-1(y/x)=-2(y/sqrt(x^2+y^2))^3+(x^2+y^2)/x^2$

$r=sqrt(x^2+y^2)$
$theta=tan^-1(y/x)$

$sintheta=y/sqrt(x^2+y^2)$

$-2sin^3theta=-2(sintheta)^3=-2(y/sqrt(x^2+y^2))^3$

$sec^2theta=1/cos^2theta=1/(x^2/(x^2+y^2))=(x^2+y^2)/x^2$

Thus,
$r-theta=-2sin^3theta+sec^2theta$ in cartesian form is

$sqrt(x^2+y^2)-tan^-1(y/x)=-2(y/sqrt(x^2+y^2))^3+(x^2+y^2)/x^2$

What is the Cartesian form of r-theta = -2sin^3theta+sec^2theta r-theta = -2sin^3theta+sec^2theta ?