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Question

Question: \[2\sin A\cos^{3}A - 2\sin^{3}A\cos A =\]...

2sin⁡Acos⁡3A−2sin⁡3Acos⁡A=2\sin A\cos^{3}A - 2\sin^{3}A\cos A =

A

sin⁡4A\sin 4A

B

12sin⁡4A\frac{1}{2}\sin 4A

C

14sin⁡4A\frac{1}{4}\sin 4A

D

None of these

Answer

12sin⁡4A\frac{1}{2}\sin 4A

Explanation

Solution

2sin⁡Acos⁡3A−2sin⁡3Acos⁡A2 \sin A \cos ^ { 3 } A - 2 \sin ^ { 3 } A \cos A

=2sin⁡Acos⁡A(cos⁡2A−sin⁡2A)= 2\sin A\cos A(\cos^{2}A - \sin^{2}A)

=2sin⁡Acos⁡Acos⁡2A=sin⁡2Acos⁡2A=12sin⁡4A= 2\sin A\cos A\cos 2A = \sin 2A\cos 2A = \frac{1}{2}\sin 4A.