Another Explanation (5): Given:
\[
\sin \alpha + \sin \beta = a
\]
\[
\cos \alpha + \cos \beta = b
\]
Using the sum-to-product identities:
\[
\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}
\]
\[
\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}
\]
Let:
\[
S = \frac{\alpha + \beta}{2}
\]
\[
D = \frac{\alpha - \beta}{2}
\]
Then:
\[
a = 2 \sin S \cos D
\]
\[
b = 2 \cos S \cos D
\]
Dividing the first by the second:
\[
\frac{a}{b} = \frac{2 \sin S \cos D}{2 \cos S \cos D} = \frac{\sin S}{\cos S} = \tan S
\]
So:
\[
\sin S = \frac{a}{\sqrt{a^2 + b^2}}
\]
\[
\cos S = \frac{b}{\sqrt{a^2 + b^2}}
\]
Now, from the expressions:
\[
a = 2 \sin S \cos D
\]
\[
b = 2 \cos S \cos D
\]
Square both:
\[
a^2 = 4 \sin^2 S \cos^2 D
\]
\[
b^2 = 4 \cos^2 S \cos^2 D
\]
Add:
\[
a^2 + b^2 = 4 \cos^2 D (\sin^2 S + \cos^2 S) = 4 \cos^2 D \times 1 = 4 \cos^2 D
\]
Therefore:
\[
\cos^2 D = \frac{a^2 + b^2}{4}
\]
\[
\cos D = \frac{\sqrt{a^2 + b^2}}{2}
\]
Now, \(\cos (\alpha - \beta) = \cos 2D = 2 \cos^2 D - 1\):
\[
\cos (\alpha - \beta) = 2 \left( \frac{\sqrt{a^2 + b^2}}{2} \right)^2 - 1 = 2 \times \frac{a^2 + b^2}{4} - 1 = \frac{a^2 + b^2}{2} - 1
\]
**Final answer:**
\(\boxed{\frac{a^2 + b^2 - 2}{2}}\)
