Solution
$$4\cos\theta\ \sin^3\theta\ -4\ \sin\theta\ \cos^3\theta\ =4\ \sin\theta\ \cos\theta\ \left(\sin^2\theta\ -\cos^2\theta\ \right)=-4\ \sin\theta\ \cos\theta\ \left(\cos^2\theta\ -\sin^2\theta\ \right)=-4\ \sin2\theta\ \cos2\theta\ =-\sin4\theta\ $$
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