Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$.

We use the trigonometric identity: $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$.
Let $A = \frac{3 \pi}{4} + x$ and $B = \frac{3 \pi}{4} - x$.
Then,$L.H.S. = \cos \left(\frac{3 \pi}{4} + x\right) - \cos \left(\frac{3 \pi}{4} - x\right)$.
Applying the identity:
$= -2 \sin \left(\frac{(\frac{3 \pi}{4} + x) + (\frac{3 \pi}{4} - x)}{2}\right) \sin \left(\frac{(\frac{3 \pi}{4} + x) - (\frac{3 \pi}{4} - x)}{2}\right)$
$= -2 \sin \left(\frac{\frac{6 \pi}{4}}{2}\right) \sin \left(\frac{2x}{2}\right)$
$= -2 \sin \left(\frac{3 \pi}{4}\right) \sin x$
Since $\sin \left(\frac{3 \pi}{4}\right) = \sin \left(\pi - \frac{\pi}{4}\right) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$= -2 \times \frac{1}{\sqrt{2}} \times \sin x$
$= -\sqrt{2} \sin x = R.H.S.$