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  4. 2 sin 2 β+4 cos (α+β) ⋅ sin α ⋅ sin β+ cos 2(α+β) is equal to

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Q. $2 \sin ^2 \beta+4 \cos (\alpha+\beta) \cdot \sin \alpha \cdot \sin \beta+\cos 2(\alpha+\beta)$ is equal to

Trigonometric Functions

Solution:

$ 2 \sin ^2 \beta+4 \cos (\alpha+\beta) \sin \alpha \sin \beta+\cos 2(\alpha+\beta) $
$=2 \sin ^2 \beta+4(\cos \alpha \cos \beta-\sin \alpha \sin \beta) \sin \alpha \sin \beta $
$ +(\cos 2 \alpha \cos 2 \beta-\sin 2 \alpha \sin 2 \beta) $
$= 2 \sin ^2 \beta+4 \sin \alpha \cos \alpha \sin \beta \cos \beta-4 \sin ^2 \alpha \sin ^2 \beta $
$+\cos 2 \alpha \cos 2 \beta-\sin 2 \alpha \sin 2 \beta $
$= 2 \sin ^2 \beta+\sin 2 \alpha \sin 2 \beta-4 \sin ^2 \alpha \sin ^2 \beta$
$ +\cos 2 \alpha \cos 2 \beta-\sin 2 \alpha \sin 2 \beta$
$= (1-\cos 2 \beta)-\left(2 \sin ^2 \alpha\right)\left(2 \sin ^2 \beta\right)+\cos 2 \alpha \cos 2 \beta $
$= (1-\cos 2 \beta)-(1-\cos 2 \alpha)(1-\cos 2 \beta)+\cos 2 \alpha \cos 2 \beta$
$= \cos 2 \alpha$