I OBM

We have $\sin b-\sin a=2\sin \frac{b-a}{2}\cos \frac{b+a}{2}=2\sin \frac{b-a}{2}<2(b-a)/2=b-a$.

The second case is trivial because both $x$ and $-\sin x$ are increasing in the interval $[\pi ,3\pi /2]$.