From the problem statement, we are given:

$\sin \alpha + \sin \beta = a$ $\cos \alpha + \cos \beta = b$

To find $\cos(\alpha - \beta)$, we can use the identity:

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

However, first we need to find expressions for $\cos \alpha \cos \beta$ and $\sin \alpha \sin \beta$.

Step 1: Square both equations

$(\sin \alpha + \sin \beta)^2 = a^2$ $(\cos \alpha + \cos \beta)^2 = b^2$

Step 2: Expand the squares using trigonometric identities

Using the identities for the sum of trigonometric functions: $\sin^2 \alpha + 2 \sin \alpha \sin \beta + \sin^2 \beta = a^2$ $\cos^2 \alpha + 2 \cos \alpha \cos \beta + \cos^2 \beta = b^2$

Step 3: Simplify using $\sin^2 + \cos^2 = 1$

Assuming $\alpha$ and $\beta$ are angles in the same function domain, then: $\sin^2 \alpha + \sin^2 \beta = 1 - \cos^2 \alpha + 1 - \cos^2 \beta = 2 - (\cos^2 \alpha + \cos^2 \beta)$

Therefore: $2 - (1 - \sin^2 \alpha + 1 - \sin^2 \beta) = a^2$ $2 - (2 - (b^2 - 2 \cos \alpha \cos \beta)) = a^2$

Rearranging terms, we find: $1 + \cos \alpha \cos \beta = \frac{b^2 + a^2}{2}$ $\cos \alpha \cos \beta = \frac{b^2 + a^2}{2} - 1$

Step 4: Use sum-to-product identities

From the sum-to-product identities: $2 \sin \alpha \sin \beta = a^2 - (2 - b^2 + 2 \cos \alpha \cos \beta)$ $2 \sin \alpha \sin \beta = a^2 - 2 + b^2 - (b^2 + a^2 - 2)$ $\sin \alpha \sin \beta = \frac{a^2 + b^2 - 2}{2}$

Step 5: Combine to find $\cos(\alpha - \beta)$

$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ $\cos(\alpha - \beta) = \left(\frac{b^2 + a^2}{2} - 1\right) + \left(\frac{a^2 + b^2 - 2}{2}\right)$ $\cos(\alpha - \beta) = a^2 + b^2 - 2$

This is the expression for $\cos(\alpha - \beta)$ in terms of $a$ and $b$. Would you like more details or do you have any further questions?

Here are some related questions you might find interesting:

1. What would be the value of $\sin(\alpha - \beta)$ given the same conditions?
2. How would you determine $\alpha + \beta$ using $a$ and $b$?
3. Can you derive expressions for $\tan(\alpha - \beta)$ from these conditions?
4. How would the solution change if $\alpha$ and $\beta$ were angles in different quadrants?
5. How can you verify the correctness of the expression for $\cos(\alpha - \beta)$?

Tip: Always check if $\alpha$ and $\beta$ satisfy the basic trigonometric identity $\sin^2 x + \cos^2 x = 1$ when working on problems involving multiple angle formulas. This can help identify errors in algebraic manipulation or assumptions.