The value of 2 sin


The value of $2 \sin ^{2} B+4 \cos (A+B) \sin A \sin B+\cos 2(A+B)$ is

(a) 0

(b) cos 3A

(c) cos 2A

(d) none of these


(c) cos 2A

We have,

$2 \sin ^{2} B+4 \cos (A+B) \sin A \sin B+\cos 2(A+B)$

$=1-\cos 2 B+\cos 2(A+B)+4 \cos (A+B) \sin A \sin B$

$=1+(\cos 2(A+B)-\cos 2 B)+4 \cos (A+B) \sin A \sin B$

$=1-2 \sin A \sin (A+2 B)+4 \cos (A+B) \sin A \sin B$

$\left[\because \cos C-\cos D=-2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}\right]$

$=1-2 \sin A[\sin (A+2 B)-2 \sin B \cos (A+B)]$

$=1-2 \sin A[\sin (A+2 B)-\{\sin (B+A+B)+\sin (B-(A+B))\}]$

$[\because 2 \sin C \cos D=\sin (C+D)+\sin (C-D)]$

$=1-2 \sin A[\sin (A+2 B)-\{\sin (A+2 B)+\sin (-A)\}]$

$=1-2 \sin A[\sin A]$

$=1-2 \sin ^{2} A$

$=\cos 2 A$

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