Exercise 15.1.2

Proof. By definition, a point $M=(x,y)\in {ℝ}^2$ is a point of the lemniscate $L$ if and only if

If $(r,𝜃)$ are polar coordinates of $M=M(r,𝜃)$, then $x=r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,y=r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃$, thus, using ${\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃+{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃=1$, and $\cos <!--\; FUNCTION\; APPLICATION\; -->(2𝜃)={\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃)-{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃$, we obtain

The equation of the lemniscate is $r^2=\cos <!--\; FUNCTION\; APPLICATION\; -->(2𝜃)$. □