How do you convert $r = 2 {(cos (θ))}^{2}$ $r=2(cos(theta))^2$ to rectangular form?

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How do you convert $r = 2 {(cos (θ))}^{2}$ $r=2(cos(theta))^2$ to rectangular form?

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Answer 1 · 2016-03-26T11:27:03

$\Rightarrow {(x^{2} + y^{2})}^{3} = 4 x^{4}$ $=>(x^2+y^2)^3=4x^4$

Explanation:

Multiplying both sides by $r^{2}$ $r^2$ we have
$r^{3} = 2 {(r cos θ)}^{2}$ $r^3=2(rcostheta)^2$
Now $r = {(x^{2} + y^{2})}^{\frac{1}{2}}$ $r=(x^2+y^2)^(1/2)$ and $r cos θ = x$ $rcostheta= x$
So
${(x^{2} + y^{2})}^{\frac{3}{2}} = 2 x^{2}$ $(x^2+y^2)^(3/2)=2x^2$
Squaring both sides
$\Rightarrow {(x^{2} + y^{2})}^{3} = 4 x^{4}$ $=>(x^2+y^2)^3=4x^4$ this is the rectangular form

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How do you convert $r = 2 {(cos (θ))}^{2}$ $r=2(cos(theta))^2$ to rectangular form?

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Explanation:

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How do you convert $r = 2 {(cos (θ))}^{2}$ $r=2(cos(theta))^2$ to rectangular form?