Question

If $x=r\sin A\cos C,y=r\sin A\sin C$ and $z=r\cos A$, prove that $r^2=x^2+y^2+z^2$.

Answer 1

$x=r\sin A\cos C$

$y=r\sin A\sin C$

$\therefore x^2+y^2=r^2{\sin }^{2}A{\cos }^{2}C+r^2{\sin }^{2}A{\sin }^{2}C$

$\therefore x^2+y^2=r^2{\sin }^{2}A({\cos }^{2}C+{\sin }^{2}C)$

$\therefore x^2+y^2=r^2{\sin }^{2}A$

$\therefore x^2+y^2+z^2=r^2{\sin }^{2}A+r^2{\cos }^{2}A$

$\therefore x^2+y^2+z^2=r^2({\sin }^2+{\cos }^{2}A)$

$\therefore x^2+y^2+z^2=r^2$

Hence proved