## Calculus: Early Transcendentals 8th Edition

Given: $x^{2}+y^{2}=4$ and $r=2$ or $r^{2}=4$ and $x=2sin3t,y=2cos3t$ The Cartesian equation is $x^{2}+y^{2}=r^{2}$ Then $x^{2}+y^{2}=(2sin3t)^{2}+(2cos3t)^{2}$ $=4sin^{2}3t+4cos^{2}3t$ $=4(sin^{2}3t+cos^{2}3t)$ $=4(1)$ Therefore, $x^{2}+y^{2}=4$ Hence, the given statement is true.