Calculus: Early Transcendentals 8th Edition

$r^2cos\,2\theta=4$
We are given that $x^2-y^2=4$. Using the cartesian-to-polar substitutions of $x=rcos\theta$ and $y=rsin\theta$, we can rewrite the expression as $(rcos\theta)^2-(rsin\theta)^2=4$. This can be rewritten as $r^2cos^2\theta-r^2sin^2\theta=4$. Factoring out an $r^2$, we get $r^2(cos^2\theta-sin^2\theta)=4$. Using the double-angle identity for cosine, we get $r^2cos\,2\theta=4$.