Question

$2{\sin }^2\beta +4\cos (\alpha +\beta ).\sin \alpha \sin \beta +2{\cos }^2(\alpha +\beta )-1$is equal to

• A. $\sin 2\alpha$
• B. $\cos 2\alpha$
• C. $\tan 2\alpha$
• D. $\cot 2\alpha$

Answer 1

The correct option is B $\cos 2\alpha$

$2{\sin }^2\beta +4\cos (\alpha +\beta ).\sin \alpha \sin \beta +2{\cos }^2(\alpha +\beta )-1$
$=2{\sin }^2\beta +2\cos (\alpha +\beta )[2\sin \alpha \sin \beta +\cos (\alpha +\beta )]-1$
$=2{\sin }^2\beta +2\cos (\alpha +\beta )[2\sin \alpha \sin \beta +\cos \alpha \cos \beta -\sin \alpha \sin \beta ]-1$
$=2{\sin }^2\beta +2\cos (\alpha +\beta )\cos (\alpha -\beta )-1$
$=2{\sin }^2\beta +\cos 2\alpha +\cos 2\beta -1$
$=-\cos 2\beta +\cos 2\alpha +\cos 2\beta$
$=\cos 2\alpha$