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

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3.Trigonometrical Ratios, Functions and Identities

easy

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

A

$\sin 4A$

B

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

C

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

D

इनमें से कोई नहीं

Solution

(b) $2\sin A{\cos ^3}A – 2{\sin ^3}A\cos A$

$ = 2\sin A\cos A({\cos ^2}A – {\sin ^2}A)$

$ = 2\sin A\cos A\cos 2A $

$= \sin 2A\cos 2A $

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

Standard 11

Mathematics

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