$\frac{\left( \cos{ heta + i\sin heta} ight)^{4}}{\left( \sin{ heta + i\cos heta} ight)^{5}}$ is equal to

Question

$\frac{\left( \cos{	heta + i\sin	heta} ight)^{4}}{\left( \sin{	heta + i\cos	heta} ight)^{5}}$ is equal to

Dr. Abubacker · Accepted Answer

(d)
$\frac{\left( \cos{	heta + i\sin	heta} ight)^{4}}{\left( \sin{	heta + i\cos	heta} ight)^{5}} = \frac{\left( \cos{	heta + i\sin	heta} ight)^{4}}{i^{5}\left( {\cos	heta}{- i\sin	heta} ight)^{5}}$
$= - i\ \left( \cos	heta + i\ \sin	heta ight)^{9}$
$= \sin{9	heta - i\cos{9	heta}}$

Q. $\frac{\left( \cos{\theta + i\sin\theta} \right)^{4}}{\left( \sin{\theta + i\cos\theta} \right)^{5}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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