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  4. (( cos θ+i sin θ/ sin θ+i cos θ))8+((1+ cos θ-i sin θ/1+ cos θ+i sin θ))16=

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Q. $\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{8}+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16}=$

TS EAMCET 2020

Solution:

We have
$\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{8}+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16}$
$=\left(\frac{\cos \theta+i \sin \theta}{i(\cos \theta-i \sin \theta)}\right)^{8}+\left(\frac{2 \cos ^{2} \frac{\theta}{2}-i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^{16}$
$=\frac{1}{i^{8}}\left(\frac{\cos \theta+i \sin \theta}{\cos \theta-i \sin \theta}\right)^{8}+\left(\frac{\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}\right)^{16}$
$=\frac{\cos 8 \theta+i \sin 8 \theta}{\cos 8 \theta-i \sin 8 \theta}+\frac{\cos 8 \theta-i \sin 8 \theta}{\cos 8 \theta+i \sin 8 \theta}$
$=(\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta)^{-1}+\left(\cos 8 \theta-i \sin 8 \theta(\cos 8 \theta+i \sin 8 \theta)^{-1}\right.$
$=(\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta+i(\sin 8 \theta)+(\cos 8 \theta-i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta)$
$=(\cos 8 \theta+i \sin 8 \theta)^{2}+(\cos 8 \theta-i \sin 8 \theta)^{2}$
$=\cos ^{2} 8 \theta+i^{2} \sin ^{2} 8 \theta+2 i \cos 8 \theta \sin 8 \theta+\cos ^{2} 8 \theta+i^{2} \sin ^{2} 8 \theta-2 i \cos 8 \theta \sin 8 \theta$
$=2\left(\cos ^{2} 8 \theta-\sin ^{2} 8 \theta\right)=2 \cos 16 \theta$