Prove that $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$

(N/A) We use the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$.
Let $A = \frac{\pi}{4}-x$ and $B = \frac{\pi}{4}-y$.
The given expression is of the form $\cos A \cos B - \sin A \sin B$,which equals $\cos(A+B)$.
Substituting the values of $A$ and $B$:
$\cos \left[\left(\frac{\pi}{4}-x\right) + \left(\frac{\pi}{4}-y\right)\right]$
$= \cos \left[\frac{\pi}{2} - (x+y)\right]$
Since $\cos \left(\frac{\pi}{2} - \theta\right) = \sin \theta$,we get:
$= \sin(x+y)$
$= R.H.S.$