## Calculus: Early Transcendentals 8th Edition

$c^{2}=a^{2}+b^{2}-2ab cos \theta$
Here $x=bcos\theta$ and $y=bsin\theta$ Thus $P(x,y)=P(bcos\theta,bsin\theta)$ Use the distance formula to compute $c$. Therefore, $c=\sqrt {(a-bcos\theta)^{2}+(0-bsin \theta)^{2}}$ $=\sqrt {(a^{2}+b^{2}cos^{2}\theta-2ab cos \theta+b^{2}sin ^{2}\theta}$ $=\sqrt {(a^{2}-2ab cos \theta+b^{2}(1)}$ Hence, $c^{2}=a^{2}+b^{2}-2ab cos \theta$