How do you write the Cartesian Equation $r^{2} = sin 2 (θ)$ $r^2 = sin 2(theta)$ ?

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Answer 1 · 2016-05-11T08:59:29

${(x^{2} + y^{2})}^{2} - 2 x y = 0$ $(x^2+y^2)^2-2xy=0$

Use $r^{2} = x^{2} + y^{2}, sin θ = \frac{y}{r} and cos θ = \frac{x}{r}$ $r^2=x^2+y^2, sin theta = y/r and cos theta = x/r$ .

Here, $x^{2} + y^{2} = 2 sin θ cos θ = 2 \frac{x y}{r^{2}} = \frac{2 x y}{x^{2} + y^{2}}$ $x^2+y^2=2sin theta cos theta=2(xy)/r^2=(2xy)/(x^2+y^2)$ .

So, ${(x^{2} + y^{2})}^{2} - 2 x y = 0$ $(x^2+y^2)^2-2xy=0$ .

The curve is looking like $\infty$ $oo$ rotated through

through $45^{0}$ $45^0$ , in the anticlockwise sense, about the center (origin)..

The graph of the Limacon $r^{2} = cos 2 θ$ $r^2=cos 2theta$ looks like $\infty$ $oo$ .

How do you write the Cartesian Equation r2=sin2(θ)r^2 = sin 2(theta)?