How do you graph $r = \sqrt{cos (2 \cdot θ)}$ $r=sqrt(cos(2*theta))$ ?

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How do you graph $r = \sqrt{cos (2 \cdot θ)}$ $r=sqrt(cos(2*theta))$ ?

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Answer 1 · 2016-03-01T07:59:21

Hence it is in polar coordinates we have that

$x = r \cdot cos (θ)$ $x=r*cos(theta)$ and $y = r \cdot sin (θ)$ $y=r*sin(theta)$

so we have that $r^{2} = (x^{2} + y^{2})$ $r^2=(x^2+y^2)$

and

$cos (2 θ) = {cos}^{2} θ - {sin}^{2} θ = {(\frac{x}{r})}^{2} - {(\frac{y}{r})}^{2}$ $cos(2theta)=cos^2theta-sin^2theta=(x/r)^2-(y/r)^2$

From our given relation we have that

$r = \sqrt{cos (2 θ)} \Rightarrow r^{2} = cos (2 θ) \Rightarrow r^{2} = (\frac{1}{r^{2}}) (x^{2} - y^{2}) \Rightarrow r^{4} = (x^{2} - y^{2}) \Rightarrow {(x^{2} + y^{2})}^{2} = x^{2} - y^{2}$ $r=sqrt(cos(2theta))=>r^2=cos(2theta)=> r^2=(1/r^2)(x^2-y^2)=>r^4=(x^2-y^2)=> (x^2+y^2)^2=x^2-y^2$

The graph of ${(x^{2} + y^{2})}^{2} = x^{2} - y^{2}$ $(x^2+y^2)^2=x^2-y^2$ is

enter image source here

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How do you graph $r = \sqrt{cos (2 \cdot θ)}$ $r=sqrt(cos(2*theta))$ ?

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