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$\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16}=$
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Verified Answer
The correct answer is:
$2 \cos 16 \theta$
We have,
$\begin{aligned}
& \left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16} \\
& =\left(\frac{\cos \theta+i \sin \theta}{i(\cos \theta-i \sin \theta)}\right)^8+\left(\frac{2 \cos ^2 \frac{\theta}{2}-i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^{16} \\
& =\frac{1}{i^8}\left(\frac{\cos \theta+i \sin \theta}{\cos \theta-i \sin \theta}\right)^8+\left(\frac{\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}\right)^{16}
\end{aligned}$
$\begin{aligned} & =\frac{\cos 8 \theta+i \sin 8 \theta}{\cos 8 \theta-i \sin 8 \theta}+\frac{\cos 8 \theta-i \sin 8 \theta}{\cos 8 \theta+i \sin 8 \theta} \\ & =(\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta)^{-1} \\ & \quad+\left(\cos 8 \theta-i \sin 8 \theta(\cos 8 \theta+i \sin 8 \theta)^{-1}\right. \\ & =(\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta+i(\sin 8 \theta)+ \\ & \quad(\cos 8 \theta-i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta) \\ & =(\cos 8 \theta+i \sin 8 \theta)^2+(\cos 8 \theta-i \sin 8 \theta)^2 \\ & =\cos ^2 8 \theta+i^2 \sin ^2 8 \theta+2 i \cos 8 \theta \sin 8 \theta+\cos ^2 8 \theta \\ & \quad+i^2 \sin ^2 8 \theta-2 i \cos 8 \theta \sin 8 \theta \\ & =2\left(\cos ^2 8 \theta-\sin ^2 8 \theta\right)=2 \cos 16 \theta\end{aligned}$
$\begin{aligned}
& \left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16} \\
& =\left(\frac{\cos \theta+i \sin \theta}{i(\cos \theta-i \sin \theta)}\right)^8+\left(\frac{2 \cos ^2 \frac{\theta}{2}-i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^{16} \\
& =\frac{1}{i^8}\left(\frac{\cos \theta+i \sin \theta}{\cos \theta-i \sin \theta}\right)^8+\left(\frac{\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}\right)^{16}
\end{aligned}$
$\begin{aligned} & =\frac{\cos 8 \theta+i \sin 8 \theta}{\cos 8 \theta-i \sin 8 \theta}+\frac{\cos 8 \theta-i \sin 8 \theta}{\cos 8 \theta+i \sin 8 \theta} \\ & =(\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta)^{-1} \\ & \quad+\left(\cos 8 \theta-i \sin 8 \theta(\cos 8 \theta+i \sin 8 \theta)^{-1}\right. \\ & =(\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta+i(\sin 8 \theta)+ \\ & \quad(\cos 8 \theta-i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta) \\ & =(\cos 8 \theta+i \sin 8 \theta)^2+(\cos 8 \theta-i \sin 8 \theta)^2 \\ & =\cos ^2 8 \theta+i^2 \sin ^2 8 \theta+2 i \cos 8 \theta \sin 8 \theta+\cos ^2 8 \theta \\ & \quad+i^2 \sin ^2 8 \theta-2 i \cos 8 \theta \sin 8 \theta \\ & =2\left(\cos ^2 8 \theta-\sin ^2 8 \theta\right)=2 \cos 16 \theta\end{aligned}$
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