Question

If $x=a\cos \theta -b\sin \theta$ and $y=a\sin \theta +b\cos \theta$ what is the value of $\frac{x^2+y^2}{a^2+b^2}$ ?

Answer 1

$x^2+y^2={(a\cos \theta -b\sin \theta )}^2+{(a\sin \theta +b\cos \theta )}^2$
$=a^2{\cos }^2\theta +b^2{\sin }^2\theta -2ab\sin \theta \cos \theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta +2ab\sin \theta \cos \theta$
$=a^2({\sin }^2\theta +{\cos }^2\theta )+b^2({\sin }^2\theta +{\cos }^2\theta )$
$=a^2+b^2$
$\therefore \frac{x^2+y^2}{a^2+b^2}=1$