Question

$2\sin 2\beta + 4\cos (\alpha + \beta)\sin \alpha \sin \beta + \cos 2(\alpha + \beta)$

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

Answer 1

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

$2{\sin }^2\beta +4\cos (\alpha +\beta )\sin \alpha \sin \beta +\cos 2(\alpha +\beta )$

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

$=2{\sin }^2\beta +4\cos \alpha \cos \beta \sin \alpha \sin \beta -1{\sin }^2\alpha {\sin }^2\beta +{\cos }^2\alpha {\cos }^2\beta -\sin 2\alpha \sin 2\beta$

$=2{\sin }^2\beta +{\sin }^2\alpha {\sin }^2\beta -4{\sin }^2\alpha {\sin }^2\beta +\cos 2\alpha {\cos }^2\alpha -{\sin }^2\alpha {\sin }^2\beta$

$=(-{\cos }^2\beta )-\left(2{\sin }^2\alpha \right)\left(2{\sin }^2\beta \right)+{\cos }^2\alpha {\cos }^2\beta$

$=(1-\cos 2\beta )-(1-\cos 2\alpha )(1-\cos 2\beta )+\cos 2\alpha \cos 2\beta$

$=\cos 2\alpha$