Question

Prove the following.
$(\sin \theta +\cos \theta {)}^n=\cos n(\frac{1}{2}\pi -\theta )+i\sin n(\frac{1}{2}\pi -\theta )$

Answer 1

We shall use De-moivre's Theorem :
$(cos\theta +isin\theta )=cosn\theta +isinn\theta .$
$(sin\theta +icos\theta {)}^n$
$=\left[cos\right(\pi /2-\theta )+isin(\pi /2-\theta ){]}^n$
$=cosn(\pi /2-\theta )+isinn(\pi /2-\theta ).$