Question

$(\frac{\cos \theta +i\sin \theta }{\sin \theta +i\cos \theta }{)}^4$ is equals.

• A. $\cos 6\theta +i\sin 6\theta$
• B. $\sin 8\theta -i\cos 8\theta$
• C. $\cos 8\theta +i\sin 8\theta$
• D. none

Answer 1

The correct option is B $\sin 8\theta -i\cos 8\theta$

${\left(\frac{\cos \theta +i\sin \theta }{\sin \theta +i\cos \theta }\right)}^{4}Letz=\frac{\cos \theta +i\sin \theta }{\sin \theta +i\cos \theta }onrationalisingz\frac{(\cos \theta +i\sin \theta )(\sin \theta -i\cos \theta )}{{\sin }^2\theta -i^2{\cos }^2\theta }=2\cos \theta \sin \theta -i\left({\cos }^2\theta -{\sin }^2\theta \right)=(\sin 2\theta -i\cos 2\theta )z^4={(\sin 2\theta -i\cos 2\theta )}^4\therefore z^4=(\sin 8\theta -i\cos 8\theta )$