2sin A{cos ^3}A - 2{sin ^3}Acos A = | vedclass

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(B) Given expression: $2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A$Step $1$: Factor out $2\sin A\cos A$ from the expression:$= 2\sin A\cos A(\cos^2 A - \sin

Vedclass · Accepted Answer

$\frac{1}{2}\sin 4A$

Vedclass · Answer

(B) Given expression: $2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A$Step $1$: Factor out $2\sin A\cos A$ from the expression:$= 2\sin A\cos A(\cos^2 A - \sin^2 A)$Step $2$: Use the double angle identities $\sin 2A = 2\sin A\cos A$ and $\cos 2A = \cos^2 A - \sin^2 A$:$= \sin 2A \cdot \cos 2A$Step $3$: Multiply and divide by $2$ to use the identity $\sin 2	heta = 2\sin 	heta \cos 	heta$ where $	heta = 2A$:$= \frac{1}{2}(2\sin 2A \cos 2A)$$= \frac{1}{2}\sin 4A$.

$2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A = $

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