In the diagram below, we have $\sin \angle RPQ = \frac{7}{25}$.  What is $\cos \angle RPS$?

[asy]

pair R,P,Q,SS;

SS = (-2,0);

P = (0,0);

Q = (2,0);

R = rotate(aSin(7/25))*(1.5,0);

dot("$S$",SS,S);

dot("$Q$",Q,S);

dot("$R$",R,N);

dot("$P$",P,S);

draw(Q--SS);

draw(P--R);

[/asy]
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 $\cos\angle RPS = -\cos\angle RPQ = \boxed{-\frac{24}{25}}$.