Question

$4(\cos {330}^0-i\sin {330}^0)=4(\cos {30}^0+i\sin {30}^0)$. If this is true enter 1, else enter 0.

Answer 1

$\cos (2\pi -\theta )=\cos \theta$
$\sin (2\pi -\theta )=-\sin \theta$
$4(\cos {330}^0-i\sin {330}^0)=4(\cos {30}^0+i\sin {30}^0)$
$\therefore r=4,\theta ={30}^0$
Ans: 1