Question

Solve

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

Answer 1

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

$(\cos 2\theta =2{\cos }^2\theta -1)$

so,

$\Rightarrow 2{\sin }^2\beta +4\cos (\alpha +\beta )\sin \alpha \sin \beta +2{\cos }^2(\alpha +\beta )-1$

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

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

$[\because \cos (A+B)=cosAcosB-sinAsinB]$

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

$\{\because \cos (A-B)=cosAcosB+sinAsinB\}$

$\Rightarrow 2{\sin }^2\beta -1+\cos (\alpha +\beta )\times 2\times \cos (\alpha -\beta )$

$\Rightarrow 2{\sin }^2\beta -1+2\cos (\alpha +\beta )\cos (\alpha -\beta )$

$\{\because {\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)\}$

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

$\Rightarrow 2{\cos }^2\alpha -1$

$\Rightarrow \cos 2\alpha$