Question

If ${(\cos \alpha +\cos \beta )}^2+{(\sin \alpha +\sin \beta )}^2=\lambda {\cos }^2\frac{(\alpha -\beta )}{2}$ write the value of $\lambda$ ?

Answer 1

${\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta +{\sin }^2\alpha +si{n}^2\beta +2\sin \alpha \sin \beta$

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

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

$=2+2\cos (\alpha -\beta )$ $(\therefore \cos x=2{\cos }^2\frac{x}{2}-1)$

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

$=4{\cos }^2\frac{(\alpha -\beta )}{2}\Rightarrow \lambda =4$