jmc

For any angle $x$, we have $\cos (180^{\circ }-x)=-\cos x$, so $\cos \angle RPS=\cos (180^{\circ }-\angle RPQ)=-\cos \angle RPQ$.

Since ${\sin }^2\angle RPQ+{\cos }^2\angle RPQ=1$, we have ${\cos }^2\angle RPQ=1-{\left(\frac{7}{25}\right)}^2=\frac{576}{625}$. Since $\angle RPQ$ is acute, we have $\cos \angle RPQ=\frac{24}{25}$, which gives us $\cos \angle RPS=-\cos \angle RPQ=\boxed{-\frac{24}{25}}$.