Question

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

• A. $4{\cos }^2\frac{(\alpha –\beta )}{2}$
• B. $4{\sin }^2\frac{(\alpha –\beta )}{2}$
• C. $4{\cos }^2\frac{(\alpha +\beta )}{2}$
• D. $4{\sin }^2\frac{(\alpha +\beta )}{2}$

Answer 1

The correct option is A

$4{\cos }^2\frac{(\alpha –\beta )}{2}$

Explanation for the correct option:

Solve the expression ${(\cos \alpha +\cos \beta )}^2+{(\sin \alpha +\sin \beta )}^2$

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

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

$=1+1+2\cos (\alpha –\beta )$ ; $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{=}\mathbf{1}\right]$

$=2+2\cos (\alpha –\beta )$

$=2[1+\cos (\alpha –\beta \left)\right]$

$=2\left(2{\cos }^2\frac{(\alpha –\beta )}{2}\right)$ ; $\left[\mathbf{\because }\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\left(\frac{A}{2}\right)\right]$

$=\mathbf{4}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\frac{\mathbf{(}\mathbf{\alpha }\mathbf{}\mathbf{–}\mathbf{}\mathbf{\beta }\mathbf{)}}{\mathbf{2}}$

Hence, Option ‘A’ is Correct.