How do you graph $r = - 8 cos 2 θ$ $r=-8cos2theta$ ?

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Answer 1 · 2018-07-30T14:10:13

See 8-like graph and details.

$r = - 8 cos 2 θ \in [- 8, 8],$ $r = -8 cos 2theta in [ - 8, 8 ],$ with period $\frac{2 π}{2} = π$ $( 2pi )/2 = pi$

No pixels for $< 0$ $< 0$

$\Rightarrow 2 θ \in [\frac{π}{2}, \frac{3}{2} π] U [\frac{5}{2} π, \frac{7}{2} π]$ $rArr 2theta in [ pi/2, 3/2pi ] U [ 5/2pi, 7/2pi]$

$\Rightarrow θ \in [\frac{π}{4}, \frac{3}{4} π] U [\frac{5}{4} π, \frac{7}{4} π]$ $rArr theta in [ pi/4, 3/4pi ] U [ 5/4pi, 7/4pi]$ . So,

in two periods $θ \in [0, 2 π]$ $theta in [ 0, 2pi ]$ .

Use

0 <= r = sqrt ( x^2 + y^2 ) and ( x, y ) = r ( cos theta, sin theta )# d

convert to

${(x^{2} + y^{2})}^{1.5} = - 8 (x^{2} - y^{2})$ $( x^2 + y^2 )^ 1.5 = - 8 (x^2 - y^2 )$ , using

$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ)$ $cos 2theta = cos^2theta - sin ^2theta )$

The Socratic 8-like graph is immediate.
graph{ ( x^2 + y^2 )^ 1.5 + 8 (x^2 - y^2 ) = 0[ -18 18 -9 9 ]}

How do you graph r=-8cos2thetar=−8cos2θr=-8cos2theta?