A) \[\sin 4A\]
B) \[\frac{1}{2}\sin 4A\]
C) \[\frac{1}{4}\sin 4A\]
D) None of these
Correct Answer: B
Solution :
\[2\sin A{{\cos }^{3}}A-2{{\sin }^{3}}A\cos A\]\[=2\sin A\cos A({{\cos }^{2}}A-{{\sin }^{2}}A)\] \[=2\sin A\cos A\cos 2A=\sin 2A\cos 2A=\frac{1}{2}\sin 4A\].You need to login to perform this action.
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