To solve the system of trigonometric equations:

$\cos(x) - \cos(y) = a \quad \text{(1)}$ $\sin(x) - \sin(y) = b \quad \text{(2)}$

we can use trigonometric identities and manipulation.

Step 1: Use sum-to-product identities

We know the following sum-to-product identities:

$\cos(A) - \cos(B) = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ $\sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$

Step 2: Apply these identities to the given equations

For equation (1): $\cos(x) - \cos(y) = -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = a$ For equation (2): $\sin(x) - \sin(y) = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = b$

Now we have the system: $-2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = a$ $2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = b$

Step 3: Solve for $\sin\left(\frac{x-y}{2}\right)$

Divide the two equations: $\frac{-2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)}{2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)} = \frac{a}{b}$ Simplifying: $\frac{- \sin\left(\frac{x+y}{2}\right)}{\cos\left(\frac{x+y}{2}\right)} = \frac{a}{b}$ This gives: $- \tan\left(\frac{x+y}{2}\right) = \frac{a}{b}$ So: $\tan\left(\frac{x+y}{2}\right) = -\frac{a}{b}$

Step 4: Solve for $\frac{x+y}{2}$

Thus: $\frac{x+y}{2} = \tan^{-1}\left(-\frac{a}{b}\right)$ So: $x + y = 2 \tan^{-1}\left(-\frac{a}{b}\right)$

Step 5: Solve for $\sin\left(\frac{x-y}{2}\right)$

Substitute $\tan\left(\frac{x+y}{2}\right)$ back into one of the original equations (for simplicity, the second one): $2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = b$ Since we know $\frac{x+y}{2}$, we can solve for $\sin\left(\frac{x-y}{2}\right)$.

Finally, after solving $\sin\left(\frac{x-y}{2}\right)$ and $\frac{x+y}{2}$, we can find both $x$ and $y$ by solving the system: $x = \frac{x+y}{2} + \frac{x-y}{2}$ $y = \frac{x+y}{2} - \frac{x-y}{2}$

Would you like me to continue with detailed calculations or clarifications?

Further Questions:

1. What are the possible solutions for $x$ and $y$ in different quadrants?
2. How do we handle cases where $\tan^{-1}(-a/b)$ gives multiple solutions?
3. Can these equations have a geometric interpretation on the unit circle?
4. How do the signs of $a$ and $b$ affect the solutions?
5. What happens if $a = 0$ or $b = 0$?

Tip: When working with trigonometric identities, sum-to-product formulas are often useful in simplifying differences between functions.