Task B-3.4. If $\cos \alpha+\cos \beta=\frac{1}{2}$ and $\sin \alpha+\sin \beta=\frac{1}{4}$, what is the value of $\cos (\alpha-\beta)$?
## Solution.

Square each of the equations $\cos \alpha+\cos \beta=\frac{1}{2}, \sin \alpha+\sin \beta=\frac{1}{4}$.

Then we have

$$
\begin{aligned}
& \cos ^{2} \alpha+2 \cos \alpha \cos \beta+\cos ^{2} \beta=\frac{1}{4} \\
& \sin ^{2} \alpha+2 \sin \alpha \sin \beta+\sin ^{2} \beta=\frac{1}{16}
\end{aligned}
$$

By adding these two equations, we get

$$
1+2 \cos (\alpha-\beta)+1=\frac{5}{16}
$$

From here,

$$
\begin{aligned}
& 2 \cos (\alpha-\beta)=-\frac{27}{16} \\
& \cos (\alpha-\beta)=-\frac{27}{32}
\end{aligned}
$$