2015/119) Prove that $(sin\theta+i\cos\theta)^8=\cos8\theta-i\sin8\theta$

$(\sin\theta+i\cos\theta)^8 = i^8(\cos\theta - i \sin \theta)^8$
$=(e^{-i\theta})^8= e^{-i8\theta}$
$=\cos 8\theta - i \sin 8\theta$