Using De Moivre's theorem, $(\cos heta + i\sin heta)^4$ equals:

Question

Using De Moivre's theorem, $(\cos	heta + i\sin	heta)^4$ equals:

Accepted Answer

D. $\cos 4	heta + i\sin 4	heta$, multiplying the angle by 4

De Moivre: $(\cos	heta + i\sin	heta)^n = \cos n	heta + i\sin n	heta$.

Answer

$\cos	heta + i\sin	heta$, the same unchanged power here

Answer

$4(\cos	heta + i\sin	heta)$, scaled linearly without the angle factor

Answer

$\cos(4 + 	heta) + i\sin(4 + 	heta)$, adding 4 to the angle

Using De Moivre's theorem, $(\cos\theta + i\sin\theta)^4$ equals: