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Problem Statement
Prove that:

    \[\frac{(\cos \theta + i \sin \theta)^4}{(\sin \theta + i \cos \theta)^4} = \cos 8\theta + i \sin 8\theta\]

Solution
Let LHS be:

    \[\frac{(\cos \theta + i \sin \theta)^4}{(\sin \theta + i \cos \theta)^4}\]

We have:

    \[(\cos \theta + i \sin \theta)^4 = \cos 4\theta + i \sin 4\theta \quad \text{[De-Moivre's theorem]}\]

    \[(\sin \theta + i \cos \theta)^4 = \left[ \cos \left( \frac{\pi}{2} - \theta \right) + i \sin \left( \frac{\pi}{2} - \theta \right) \right]^4\]

Expanding:

    \[(\sin \theta + i \cos \theta)^4 & = \cos \left( 4 \cdot \frac{\pi}{2} - 4\theta \right) + i \sin \left( 4 \cdot \frac{\pi}{2} - 4\theta \right) \\ & = \cos (2\pi - 4\theta) + i \sin (2\pi - 4\theta)\]

Using identities:

    \[\cos (2\pi - A) & = \cos A, \\ \sin (2\pi - A) & = -\sin A\]

Thus:

    \[(\sin \theta + i \cos \theta)^4 = \cos 4\theta - i \sin 4\theta\]

Rewriting the values:

    \[\frac{(\cos \theta + i \sin \theta)^4}{(\sin \theta + i \cos \theta)^4} = \frac{\cos 4\theta + i \sin 4\theta}{\cos 4\theta - i \sin 4\theta}\]

By rationalization:

    \[& = \frac{(\cos 4\theta + i \sin 4\theta)(\cos 4\theta + i \sin 4\theta)}{\cos^2(4\theta) + \sin^2(4\theta)}\]

Using \cos^2(4\theta) + \sin^2(4\theta) = 1:

    \[= \cos 8\theta + i \sin 8\theta\]

Conclusion
Therefore, the proof is complete:

    \[\frac{(\cos \theta + i \sin \theta)^4}{(\sin \theta + i \cos \theta)^4} = \cos 8\theta + i \sin 8\theta\]

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