We are given the equations:

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

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

We need to show that:

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

Step 1: Squaring and Adding the Given Equations

We square both given equations:

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

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

Expanding each:

$\sin^2 \alpha + 2 \sin \alpha \sin \beta + \sin^2 \beta = a^2$

$\cos^2 \alpha + 2 \cos \alpha \cos \beta + \cos^2 \beta = b^2$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

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

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

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

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

Step 2: Finding $\cos (\alpha + \beta)$

Using the identity:

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

From the expanded squared equations:

$\cos \alpha \cos \beta = \frac{b^2 - 1}{2}, \quad \sin \alpha \sin \beta = \frac{a^2 - 1}{2}$

Subtracting:

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

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

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

Thus, we have proven:

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

Would you like a more detailed explanation or have any questions?

Related Questions:

1. Prove that $\cos(A + B) = \cos A \cos B - \sin A \sin B$ using the unit circle.
2. Derive $\sin(A + B) = \sin A \cos B + \cos A \sin B$ using trigonometric identities.
3. Show that $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
4. If $\sin A = 3/5$ and $\cos B = 4/5$, find $\cos(A + B)$.
5. Verify the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$.

Tip:

Always remember the fundamental trigonometric identities, especially the sum and difference formulas, as they are crucial in proving and simplifying trigonometric expressions.