What is the Cartesian form of $4 - r = sin (θ) - 2 {cos}^{2} (θ)$ $4-r = sin(theta) - 2 cos^2(theta)$ ?

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What is the Cartesian form of $4 - r = sin (θ) - 2 {cos}^{2} (θ)$ $4-r = sin(theta) - 2 cos^2(theta)$ ?

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Answer 1 · 2018-06-26T12:30:10

$6 x^{2} + 4 y^{2} = (x^{2} + y + y^{2}) \sqrt{x^{2} + y^{2}}$ $6x^2+4y^2=(x^2+y+y^2)sqrt(x^2+y^2)$

Explanation:

Polar to Cartesian relations:
$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$
$θ = arctan (\frac{y}{x})$ $theta=arctan(y/x)$

Substitute these in:
$4 - r = sin θ - 2 {cos}^{2} θ$ $4-r=sintheta-2cos^2theta$
$4 - \sqrt{x^{2} + y^{2}} = sin arctan (\frac{y}{x}) - 2 {cos}^{2} arctan (\frac{y}{x})$ $4-sqrt(x^2+y^2)=sinarctan(y/x)-2cos^2arctan(y/x)$

We can simplify the RHS; we need to express sine and cosine in terms of tangent.

Sine

$tan x = \frac{sin x}{cos x} = \frac{sin x}{\sqrt{1 - {sin}^{2} x}}$ $tanx=sinx/cosx=sinx/sqrt(1-sin^2x)$
$(1 - {sin}^{2} x) {tan}^{2} x = {sin}^{2} x$ $(1-sin^2x)tan^2x=sin^2x$
${tan}^{2} x = {sin}^{2} x (1 + {tan}^{2} x)$ $tan^2x=sin^2x(1+tan^2x)$
$sin x = \frac{tan x}{\sqrt{1 + {tan}^{2} x}}$ $sinx=tanx/sqrt(1+tan^2x)$

So $sin arctan x = \frac{x}{\sqrt{1 + x^{2}}}$ $sinarctanx=x/sqrt(1+x^2)$

Cosine

$tan x = \frac{sin x}{cos x} = \frac{\sqrt{1 - {cos}^{2} x}}{cos x}$ $tanx=sinx/cosx=sqrt(1-cos^2x)/cosx$
${cos}^{2} x {tan}^{2} x = 1 - {cos}^{2} x$ $cos^2xtan^2x=1-cos^2x$
${cos}^{2} x (1 + {tan}^{2} x) = 1$ $cos^2x(1+tan^2x)=1$
$cos x = \frac{1}{\sqrt{1 + {tan}^{2} x}}$ $cosx=1/sqrt(1+tan^2x)$

So $cos arctan x = \frac{1}{\sqrt{1 + x^{2}}}$ $cosarctanx=1/sqrt(1+x^2)$

Substitute

So
$sin arctan (\frac{y}{x}) = \frac{\frac{y}{x}}{\sqrt{1 + \frac{y^{2}}{x^{2}}}} = \frac{y}{\sqrt{x^{2} + y^{2}}}$ $sinarctan(y/x)=(y/x)/sqrt(1+y^2/x^2)=y/sqrt(x^2+y^2)$
and
${cos}^{2} arctan (\frac{y}{x}) = {⎡ ⎢ ⎢ ⎣ \frac{1}{\sqrt{1 + \frac{y^{2}}{x^{2}}}} ⎤ ⎥ ⎥ ⎦}^{2} = {[\frac{x}{\sqrt{x^{2} + y^{2}}}]}^{2} = \frac{x^{2}}{x^{2} + y^{2}}$ $cos^2arctan(y/x)=[1/sqrt(1+y^2/x^2)]^2=[x/sqrt(x^2+y^2)]^2=x^2/(x^2+y^2)$

We have after Polar to Cartesian substitution
$4 - \sqrt{x^{2} + y^{2}} = sin arctan (\frac{y}{x}) - 2 {cos}^{2} arctan (\frac{y}{x})$ $4-sqrt(x^2+y^2)=sinarctan(y/x)-2cos^2arctan(y/x)$

which then becomes
$4 - \sqrt{x^{2} + y^{2}} = \frac{y}{\sqrt{x^{2} + y^{2}}} - \frac{2 x^{2}}{x^{2} + y^{2}}$ $4-sqrt(x^2+y^2)=y/sqrt(x^2+y^2)-(2x^2)/(x^2+y^2)$

Put over a common denominator:
$4 (x^{2} + y^{2}) - {(x^{2} + y^{2})}^{\frac{3}{2}} = y \sqrt{x^{2} + y^{2}} - 2 x^{2}$ $4(x^2+y^2)-(x^2+y^2)^(3/2)=ysqrt(x^2+y^2)-2x^2$
$6 x^{2} + 4 y^{2} = (x^{2} + y + y^{2}) \sqrt{x^{2} + y^{2}}$ $6x^2+4y^2=(x^2+y+y^2)sqrt(x^2+y^2)$

We can square out the root to create a polynomial in $x$ $x$ and $y$ $y$ , which is a more conventional way to express this, but less tidy. It also risks creating surplus roots to the equation that weren't roots of the original.

${(6 x^{2} + 4 y^{2})}^{2} = {(x^{2} + y + y^{2})}^{2} (x^{2} + y^{2})$ $(6x^2+4y^2)^2=(x^2+y+y^2)^2(x^2+y^2)$

$36 x^{4} + 48 x^{2} y^{2} + 16 y^{2} = (x^{4} + 2 x^{2} y + 2 x^{2} y^{2} + y^{2} + 2 y^{3} + y^{4}) (x^{2} + y^{2})$ $36x^4+48x^2y^2+16y^2=(x^4+2x^2y+2x^2y^2+y^2+2y^3+y^4)(x^2+y^2)$

$36 x^{4} + 48 x^{2} y^{2} + 16 y^{2} = x^{6} + 2 x^{4} y + 2 x^{4} y^{2} + x^{2} y^{2} + 2 x^{2} y^{3} + x^{2} y^{4} + x^{4} y^{2} + 2 x^{2} y^{3} + 2 x^{2} y^{4} + y^{4} + 2 y^{5} + y^{6}$ $36x^4+48x^2y^2+16y^2= x^6+2x^4y+2x^4y^2+x^2y^2+2x^2y^3+x^2y^4+ x^4y^2+2x^2y^3+2x^2y^4+y^4+2y^5+y^6$

Collect terms and group them according to the degree of each monomial:

$(x^{6} + 3 x^{4} y^{2} + 3 x^{2} y^{4} + y^{6}) + (2 x^{4} y + 4 x^{2} y^{3} + 2 y^{5}) + (- 36 x^{4} - 47 x^{2} y^{2} + y^{4}) + (- 16 y^{2}) = 0$ $(x^6+3x^4y^2+3x^2y^4+y^6)+(2x^4y+4x^2y^3+2y^5)+(-36x^4-47x^2y^2+y^4)+(-16y^2)=0$

Notice that we can factorise a couple of the groups, a process that does some returning along the path taken:

${(x^{2} + y^{2})}^{3} + 2 y {(x^{2} + y^{2})}^{2} - 36 x^{4} - 47 x^{2} y^{2} + y^{4} - 16 y^{2} = 0$ $(x^2+y^2)^3+2y(x^2+y^2)^2-36x^4-47x^2y^2+y^4-16y^2=0$

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