Solution: By De Moivre’s theorem, $ (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) $. Applying $ n = 5 $, the result is $ \cos(5\theta) + i \sin(5\theta) $. \boxed{\cos(5\theta) + i \sin(5\theta)}

Tue, 05 May 2026 23:47:53 +0000

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