\(\left(\frac{\cos \theta+\text{i}\sin \theta}{\sin \theta+\text{i}\cos \theta}\right)^8+\left(\frac{1+\cos \theta−\text{i}\sin \theta}{1+\cos \theta+\text{i}\sin \theta}\right)^{16}\) =

1
2 cos 8 θ
2
2 cos 16 θ
3
2 sin 8 θ
4
2 sin 16 θ

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