Question

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

• A. $\sin \left(2\alpha \right)$
• B. $\cos \left(2\beta \right)$
• C. $\cos \left(2\alpha \right)$
• D. $\sin \left(2\beta \right)$

Answer 1

The correct option is C

$\cos \left(2\alpha \right)$

Explanation for correct option:

Step-1: Apply formula, $\cos \left(\alpha +\beta \right)=\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)$

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

$$\begin{gathered} =2{\sin }^2\left(\beta \right)+4\left\{\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)\right\}\sin \left(\alpha \right)\sin \left(\beta \right)+\left\{\cos \left(2\alpha \right)\cos \left(2\beta \right)-\sin \left(2\alpha \right)\sin \left(2\beta \right)\right\} \\ =2{\sin }^2\left(\beta \right)+4\left\{\cos \left(\alpha \right)\cos \left(\beta \right)\sin \left(\alpha \right)\sin \left(\beta \right)-{\sin }^2\left(\alpha \right){\sin }^2\left(\beta \right)\right\}+\cos \left(2\alpha \right)\cos \left(2\beta \right)-\sin \left(2\alpha \right)\sin \left(2\beta \right) \\ =2{\sin }^2\left(\beta \right)+\sin \left(2\alpha \right)\sin \left(2\beta \right)-4{\sin }^2\left(\alpha \right){\sin }^2\left(\beta \right)+\cos \left(2\alpha \right)\cos \left(2\beta \right)-\sin \left(2\alpha \right)\sin \left(2\beta \right)\mathbf{}\left[\because sin2\theta =2sin\theta cos\theta \right] \\ =2{\sin }^2\left(\beta \right)-4{\sin }^2\left(\alpha \right){\sin }^2\left(\beta \right)+\cos \left(2\alpha \right)\cos \left(2\beta \right) \end{gathered}$$

Step-2: Apply formula, $\cos \left(2x\right)=1-2{\sin }^2\left(x\right)$

$\begin{array}{rcl}L.H.S& =& \left(1-\cos \left(2\beta \right)\right)-\left(2{\sin }^2\left(\alpha \right)\right)\left(2{\sin }^2\left(\beta \right)\right)+\cos \left(2\alpha \right)\cos \left(2\beta \right)\\ & =& \left(1-\cos \left(2\beta \right)\right)-\left(1-\cos \left(2\alpha \right)\right)\left(1-\cos \left(2\beta \right)\right)+\cos \left(2\alpha \right)\cos \left(2\beta \right)\\ & =& 1-\cos \left(2\beta \right)-1+\cos \left(2\beta \right)+\cos \left(2\alpha \right)-\cos \left(2\alpha \right)\cos \left(2\beta \right)+\cos \left(2\alpha \right)\cos \left(2\beta \right)\\ & =& \cos \left(2\alpha \right)\end{array}$

Therefore, correct answer is option $\left(C\right)$