Question:
If $\cos \alpha+\cos \beta=\frac{1}{3}$ and $\sin \sin \alpha+\sin \beta=\frac{1}{4}$, prove that $\cos \frac{\alpha-\beta}{2}=\pm \frac{5}{24}$
Solution:
Squaring and adding equations $\cos \alpha+\cos \beta=\frac{1}{3}$ and $\sin \alpha+\sin \beta=\frac{1}{4}$, we get
$\Rightarrow 1+1+2(\cos \alpha \times \cos \beta+\sin \alpha \times \sin \beta)=\frac{25}{144}$
$\Rightarrow 2+2 \cos (\alpha-\beta)=\frac{25}{144} \quad(\because \cos (A-B)=\cos A \times \cos B+\sin A \times \sin B)$
$\Rightarrow \cos (\alpha-\beta)=-\frac{263}{288}$ ....(1)
Now,
$\cos ^{2}\left(\frac{\alpha-\beta}{2}\right)=\frac{1+\cos (\alpha-\beta)}{2}$
$=\frac{1-\frac{263}{288}}{2} \quad[\operatorname{From}(1)]$
$=\frac{25}{576}$
$=\pm \frac{5}{24}$