1. **State the problem:** Use De Moivre's theorem to show that $$\frac{\sin 4\theta}{\sin \theta} = 8 \cos 3\theta - 4 \cos \theta.$$\n\n2. **Recall De Moivre's theorem:** For any integer $n$, $$(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta.$$\n\n3. **Express $\sin 4\theta$ using complex exponentials:**\nUsing Euler's formula, $\sin 4\theta = \operatorname{Im}[(\cos \theta + i \sin \theta)^4]$.\nExpand $(\cos \theta + i \sin \theta)^4$ using the binomial theorem:\n$$ (\cos \theta + i \sin \theta)^4 = \sum_{k=0}^4 \binom{4}{k} (\cos \theta)^{4-k} (i \sin \theta)^k. $$\n\n4. **Calculate the imaginary part:**\nOnly terms with odd $k$ contribute to the imaginary part. For $k=1$ and $k=3$:\n- $k=1$: $\binom{4}{1} (\cos \theta)^3 (i \sin \theta)^1 = 4 \cos^3 \theta (i \sin \theta)$\n- $k=3$: $\binom{4}{3} (\cos \theta)^1 (i \sin \theta)^3 = 4 \cos \theta (i^3 \sin^3 \theta) = 4 \cos \theta (i^3 \sin^3 \theta)$\nSince $i^3 = i^2 \cdot i = (-1) \cdot i = -i$, this term becomes $-4 i \cos \theta \sin^3 \theta$.\n\nSum of imaginary parts:\n$$ 4 \cos^3 \theta \sin \theta - 4 \cos \theta \sin^3 \theta = 4 \sin \theta (\cos^3 \theta - \cos \theta \sin^2 \theta). $$\n\n5. **Simplify the expression inside parentheses:**\nRecall $\sin^2 \theta = 1 - \cos^2 \theta$, so\n$$ \cos^3 \theta - \cos \theta \sin^2 \theta = \cos^3 \theta - \cos \theta (1 - \cos^2 \theta) = \cos^3 \theta - \cos \theta + \cos^3 \theta = 2 \cos^3 \theta - \cos \theta. $$\n\n6. **Therefore,**\n$$ \sin 4\theta = 4 \sin \theta (2 \cos^3 \theta - \cos \theta) = 4 \sin \theta \cos \theta (2 \cos^2 \theta - 1). $$\n\n7. **Divide both sides by $\sin \theta$ (assuming $\sin \theta \neq 0$):**\n$$ \frac{\sin 4\theta}{\sin \theta} = 4 \cos \theta (2 \cos^2 \theta - 1). $$\n\n8. **Use triple-angle identity for cosine:**\nRecall that $$ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta. $$\nRewrite $4 \cos \theta (2 \cos^2 \theta - 1)$ as\n$$ 8 \cos^3 \theta - 4 \cos \theta = (4 \cos^3 \theta - 3 \cos \theta) + (4 \cos^3 \theta - \cos \theta) = \cos 3\theta + (4 \cos^3 \theta - \cos \theta). $$\nBut more directly, note that\n$$ 8 \cos^3 \theta - 4 \cos \theta = 8 \cos^3 \theta - 4 \cos \theta. $$\nExpress $8 \cos^3 \theta$ as $2(4 \cos^3 \theta)$ and use $\cos 3\theta$ identity:\n$$ 8 \cos^3 \theta - 4 \cos \theta = 2(4 \cos^3 \theta - 3 \cos \theta) + 2 \cos \theta = 2 \cos 3\theta + 2 \cos \theta. $$\nThis is not matching the target expression, so let's try another approach.\n\n9. **Rewrite $8 \cos 3\theta - 4 \cos \theta$ using the triple-angle formula:**\n$$ 8 \cos 3\theta - 4 \cos \theta = 8 (4 \cos^3 \theta - 3 \cos \theta) - 4 \cos \theta = 32 \cos^3 \theta - 24 \cos \theta - 4 \cos \theta = 32 \cos^3 \theta - 28 \cos \theta. $$\n\n10. **Compare with the expression from step 7:**\nWe have $$ \frac{\sin 4\theta}{\sin \theta} = 4 \cos \theta (2 \cos^2 \theta - 1) = 8 \cos^3 \theta - 4 \cos \theta. $$\nThis is different from $32 \cos^3 \theta - 28 \cos \theta$. So the original problem likely has a typo or requires a different approach.\n\n11. **Alternative approach: Use sum-to-product formulas:**\nRecall that $$ \sin 4\theta = 2 \sin 2\theta \cos 2\theta $$ and $$ \sin 2\theta = 2 \sin \theta \cos \theta. $$\nSo, $$ \sin 4\theta = 2 (2 \sin \theta \cos \theta) \cos 2\theta = 4 \sin \theta \cos \theta \cos 2\theta. $$\nDivide both sides by $\sin \theta$:\n$$ \frac{\sin 4\theta}{\sin \theta} = 4 \cos \theta \cos 2\theta. $$\n\n12. **Express $\cos 2\theta$ in terms of $\cos \theta$:**\n$$ \cos 2\theta = 2 \cos^2 \theta - 1. $$\nSo,\n$$ \frac{\sin 4\theta}{\sin \theta} = 4 \cos \theta (2 \cos^2 \theta - 1) = 8 \cos^3 \theta - 4 \cos \theta. $$\n\n13. **Use triple-angle identity for cosine:**\n$$ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta. $$\nRewrite $8 \cos^3 \theta - 4 \cos \theta$ as\n$$ 8 \cos^3 \theta - 4 \cos \theta = 2 (4 \cos^3 \theta - 3 \cos \theta) + 2 \cos \theta = 2 \cos 3\theta + 2 \cos \theta. $$\n\n14. **Final conclusion:**\nThe expression $$ \frac{\sin 4\theta}{\sin \theta} = 8 \cos 3\theta - 4 \cos \theta $$ is not correct as stated. The correct identity is\n$$ \frac{\sin 4\theta}{\sin \theta} = 8 \cos^3 \theta - 4 \cos \theta = 2 \cos 3\theta + 2 \cos \theta. $$\n\n**Answer:**\n$$ \boxed{\frac{\sin 4\theta}{\sin \theta} = 8 \cos^3 \theta - 4 \cos \theta = 2 \cos 3\theta + 2 \cos \theta.} $$