Question

Find the value of $2{\sin }^2\beta +4\cos (\alpha +\beta )\sin \alpha \sin \beta +\cos \left[2\right(\alpha +\beta \left)\right]$.

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

Answer 1

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

$a=2{\sin }^2\beta +4\cos (\alpha +\beta )\sin \alpha \sin \beta +\cos 2(\alpha +\beta )$
$\Rightarrow a=2{\sin }^2\beta +2\cos (\alpha +\beta )\left(2\sin \alpha \sin \beta \right)+\cos 2(\alpha +\beta )$
$\Rightarrow a=1-\cos 2\beta +2\cos (\alpha +\beta )(\cos (\alpha -\beta )-\cos (\alpha +\beta ))+\cos 2(\alpha +\beta )$
...$\{\because \cos 2A=1-2{\sin }^{2}A\&2\sin A\sin B=\cos (A-B)-\cos (A+B)\}$
$\Rightarrow a=1-\cos 2\beta +\left(2\cos (\alpha +\beta )\cos (\alpha -\beta )\right)-\left(2{\cos }^2(\alpha +\beta )\right)+\cos 2(\alpha +\beta )$
$\Rightarrow a=1-\cos 2\beta +\left(2\cos (\alpha +\beta )\cos (\alpha -\beta )\right)-\left(2{\cos }^2(\alpha +\beta )\right)+\cos 2(\alpha +\beta )$
$\Rightarrow a=1-\cos 2\beta +(\cos 2\alpha +\cos 2\beta )-(1+\cos 2(\alpha +\beta ))+\cos 2(\alpha +\beta )$
...$\{\because \cos 2A=2{\cos }^{2}A-1\&2\cos A\cos B=\cos (A+B)+\cos (A-B)\}$
$\therefore a=\cos 2\alpha$
Hence, option 'A' is correct.