Question: Value of $\cos \left( \frac{\pi }{6} - x \right)\cos \left( \frac{\pi }{3} - y \right) - \sin \left( \frac{\pi }{6} - x \right)\sin \left( \frac{\pi }{3} - y \right)$.
sin $(x – y)$
sin $(x + y)$
cos $(x + y)$
None of these
1. **Problem Statement:** Find the value of
$$\cos \left( \frac{\pi}{6} - x \right) \cos \left( \frac{\pi}{3} - y \right) - \sin \left( \frac{\pi}{6} - x \right) \sin \left( \frac{\pi}{3} - y \right)$$
2. **Recall the cosine addition formula:**
$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$
3. **Compare the given expression:**
The expression matches the form of $\cos(A + B)$ where
$$A = \frac{\pi}{6} - x, \quad B = \frac{\pi}{3} - y$$
4. **Apply the formula:**
$$\cos \left( \frac{\pi}{6} - x \right) \cos \left( \frac{\pi}{3} - y \right) - \sin \left( \frac{\pi}{6} - x \right) \sin \left( \frac{\pi}{3} - y \right) = \cos \left( \left( \frac{\pi}{6} - x \right) + \left( \frac{\pi}{3} - y \right) \right)$$
5. **Simplify the argument:**
$$\left( \frac{\pi}{6} - x \right) + \left( \frac{\pi}{3} - y \right) = \frac{\pi}{6} + \frac{\pi}{3} - x - y = \frac{\pi}{2} - (x + y)$$
6. **So the expression becomes:**
$$\cos \left( \frac{\pi}{2} - (x + y) \right)$$
7. **Use the identity:**
$$\cos \left( \frac{\pi}{2} - \theta \right) = \sin \theta$$
8. **Therefore:**
$$\cos \left( \frac{\pi}{2} - (x + y) \right) = \sin (x + y)$$
**Final answer:**
$$\sin (x + y)$$