Question

Prove that:
$\left(\cos \alpha +\cos {\beta }^2\right)+{\left(\sin \alpha +\sin \beta \right)}^2=4{\cos }^2\left(\frac{\alpha -\beta }{2}\right)$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\left(\mathrm{cos\alpha }+\mathrm{cos\beta }\right)}^2+{\left(\mathrm{sin\alpha }+\mathrm{sin\beta }\right)}^2 \\ ={\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+2\mathrm{cos\alpha cos\beta }+{\sin }^2\mathrm{\alpha }+{\sin }^2\mathrm{\beta }+2\mathrm{sin\alpha sin\beta } \\ =({\cos }^2\mathrm{\alpha }+{\sin }^2\mathrm{\alpha })+({\cos }^2\mathrm{\beta }+{\sin }^2\mathrm{\beta })+2\left(\mathrm{cos\alpha cos\beta }+\mathrm{sin\alpha sin\beta }\right) \\ =1+1+2\cos (\mathrm{\alpha }-\mathrm{\beta }) \\ =2\left\{1+\cos (\mathrm{\alpha }-\mathrm{\beta })\right\} \\ =2\left\{2{\cos }^2\left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)\right\} \\ =4{\cos }^2\left(\frac{\text{α}-\mathrm{\beta }}{2}\right)=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$