(A) We know that $\sin 4\theta = 2\sin 2\theta \cos 2\theta$. Using the double angle formulas $\sin 2\theta = 2\sin \theta \cos \theta$ and $\cos 2\th

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$4\sin \theta (1 - 2\sin^2 \theta )\sqrt{1 - \sin^2 \theta}$

(A) We know that $\sin 4\theta = 2\sin 2\theta \cos 2\theta$. Using the double angle formulas $\sin 2\theta = 2\sin \theta \cos \theta$ and $\cos 2\theta = 1 - 2\sin^2 \theta$,we get: $\sin 4\theta = 2(2\sin \theta \cos \theta)(1 - 2\sin^2 \theta)$. Since $\cos \theta = \sqrt{1 - \sin^2 \theta}$ (assuming $\cos \theta > 0$),we can write: $\sin 4\theta = 4\sin \theta (1 - 2\sin^2 \theta)\sqrt{1 - \sin^2 \theta}$.