Question

How do you graph `r=sqrt(cos(2*theta))`?

Answer 1

Hence it is in polar coordinates we have that

`x=r*cos(theta)` and `y=r*sin(theta)`

so we have that `r^2=(x^2+y^2)`

and

`cos(2theta)=cos^2theta-sin^2theta=(x/r)^2-(y/r)^2`

From our given relation we have that

#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` is