Okay, let's try to figure out this problem. So, we have an Asymptote diagram with points R, P, Q, and S. The question states that sin(angle RPQ) is 7/25, and we need to find cos(angle RPS). Hmm, angles RPQ and RPS... Let me visualize the diagram first.

From the Asymptote code, it looks like S is at (-2,0), Q is at (2,0), P is at (0,0), and R is somewhere above the x-axis. The code says R is obtained by rotating the point (1.5,0) around P by an angle whose sine is 7/25. So, angle RPQ is the angle at point P between points R and Q. Similarly, angle RPS is the angle at point P between points R and S.

Wait, so angle RPQ is between PR and PQ, and angle RPS is between PR and PS. Since S is to the left of P, and Q is to the right, that makes sense. So, maybe these angles are adjacent or something? Let me see.

First, let me recall that in a right triangle, sine is opposite over hypotenuse. Since sin(angle RPQ) = 7/25, that might mean that in triangle RPQ, the side opposite angle RPQ is 7, and the hypotenuse is 25. But wait, is triangle RPQ a right triangle? Let me check.

Looking at the Asymptote code, R is obtained by rotating (1.5,0) around P by an angle with sin 7/25. So, the original point is (1.5,0), which is 1.5 units to the right of P. After rotation, R is somewhere above the x-axis. The rotation angle has sin(theta) = 7/25, so the y-coordinate of R would be 1.5 * sin(theta) and the x-coordinate would be 1.5 * cos(theta). But since it's rotated around P, the coordinates of R would be (1.5*cos(theta), 1.5*sin(theta)). But the Asymptote code uses rotate(aSin(7/25)), which might be rotating by theta where sin(theta) = 7/25. Therefore, cos(theta) would be sqrt(1 - (7/25)^2) = sqrt(576/625) = 24/25. So, R's coordinates are (1.5*(24/25), 1.5*(7/25)) = (36/25, 21/25). Wait, but in the Asymptote code, Q is at (2,0), S is at (-2,0), and P is at (0,0). So, R is at (36/25, 21/25). Let me confirm that.

Yes, 1.5 is 3/2. So, 3/2 * cos(theta) is (3/2)*(24/25) = 72/50 = 36/25. Similarly, 3/2 * sin(theta) is (3/2)*(7/25) = 21/25. So, R is at (36/25, 21/25). Got it.

So, points are:

- P: (0,0)

- Q: (2,0)

- S: (-2,0)

- R: (36/25, 21/25)

Now, angle RPQ is the angle at P between points R and Q. Let me confirm that. So, angle RPQ is the angle between PR and PQ. Since PQ is from P(0,0) to Q(2,0), which is along the x-axis. PR is from P(0,0) to R(36/25, 21/25). So, the angle between PR and PQ (which is along the x-axis) has sine 7/25. That makes sense because the y-coordinate of R is 21/25, which is the opposite side of the angle. So, sin(theta) = opposite/hypotenuse = (21/25)/|PR|.

Wait, but |PR| is the distance from P to R. Let's compute that. PR's coordinates are (36/25, 21/25), so the distance is sqrt((36/25)^2 + (21/25)^2) = sqrt((1296 + 441)/625) = sqrt(1737/625) = sqrt(1737)/25. Hmm, but 1737 is 9*193, so sqrt(9*193)/25 = 3*sqrt(193)/25. So, |PR| is 3*sqrt(193)/25. Then sin(theta) = (21/25)/(3*sqrt(193)/25) = 21/(3*sqrt(193)) = 7/sqrt(193). Wait, but the problem states sin(angle RPQ) = 7/25. But according to this, it's 7/sqrt(193). Hmm, that seems contradictory. Maybe I made a mistake.

Wait, perhaps the rotation was meant to create a triangle where sin(theta) is 7/25, so maybe the rotation is such that the opposite side is 7 and hypotenuse is 25. Let me think. If the rotation angle has sin(theta) = 7/25, then the y-coordinate of R is 7, and the hypotenuse (the length PR) is 25. Then, the adjacent side would be 24, so cos(theta) = 24/25. But in the Asymptote code, they rotated (1.5,0) by that angle. But 1.5 isn't 25. Hmm, maybe the code is scaling the rotation. Let me check.

Wait, the original point is (1.5,0), which is 1.5 units from P. Then, rotating it by theta where sin(theta) = 7/25 would give a new point R. So, the coordinates of R would be (1.5*cos(theta), 1.5*sin(theta)) = (1.5*(24/25), 1.5*(7/25)) = (36/25, 21/25). So, the distance PR is sqrt((36/25)^2 + (21/25)^2) = sqrt((1296 + 441)/625) = sqrt(1737)/25. As before. But if we consider that sin(theta) = 7/25, then the opposite side is 7, hypotenuse is 25, so adjacent side is 24. So, in that case, the coordinates of R would be (24,7) if PR is 25. But in the code, PR is 36/25 and 21/25, which is scaled down by a factor of 1.5/25? Wait, maybe not. Hmm, perhaps the problem statement uses a different scaling. Wait, maybe the Asymptote code is just an illustration, and the actual coordinates are not important. Maybe the key is that sin(angle RPQ) = 7/25, so we can use that to find other trigonometric ratios.

So, regardless of the coordinates, angle RPQ has sin(theta) = 7/25. Let me consider triangle PRQ, where angle at P is theta. Then, in triangle PRQ, opposite side is RQ, which is 7, hypotenuse is PR, which is 25. Wait, but RQ is not necessarily the opposite side. Wait, angle RPQ is at P, between PR and PQ. So, in triangle PRQ, angle at P is theta, with sin(theta) = 7/25. Then, the sides: PQ is along the x-axis from P(0,0) to Q(2,0), so PQ is length 2. PR is length 25? Wait, but in the Asymptote code, PR is sqrt((36/25)^2 + (21/25)^2) = sqrt(1737)/25 ≈ 1.35. So, maybe the Asymptote code is just a specific diagram, but the actual problem is general. Hmm, perhaps the Asymptote code is just an example, but the key is that in the problem, sin(angle RPQ) = 7/25, and we need to find cos(angle RPS). So, maybe we can ignore the specific coordinates and just use trigonometric relations.

Let me try to approach this step by step.

First, angle RPQ is at point P, between points R and Q. So, triangle PRQ has angle at P with sin(theta) = 7/25. Let's denote angle RPQ as theta. So, sin(theta) = 7/25, which implies that the opposite side over hypotenuse is 7/25. But in triangle PRQ, the sides adjacent to angle theta would be PR and PQ. Wait, PQ is one side, PR is another, and RQ is the third side. So, using the definition of sine in triangle PRQ: sin(theta) = opposite / hypotenuse. Wait, but in triangle PRQ, angle theta is at P, so the opposite side would be RQ, and the hypotenuse would be... Wait, is triangle PRQ a right triangle? The problem doesn't specify that. Hmm, so maybe I need to use the Law of Sines or Cosines here.

Wait, but angle RPS is another angle at point P, between points R and S. Since S is to the left of P, and Q is to the right, angle RPS is the angle between PR and PS. So, points S and Q are on opposite sides of P. Therefore, angles RPQ and RPS are adjacent angles at point P, forming a linear pair if R is in a specific position. Wait, but depending on where R is, these angles could be adjacent or not. Wait, in this case, S is at (-2,0), Q is at (2,0), so if R is above the x-axis, then angles RPS and RPQ are on either side of the y-axis? Wait, no, angle RPS is the angle between PR and PS. Since S is at (-2,0), and P is at (0,0), PS is the line from P(0,0) to S(-2,0), which is along the negative x-axis. Similarly, PQ is along the positive x-axis. So, angle RPS is the angle between PR and the negative x-axis. Similarly, angle RPQ is the angle between PR and the positive x-axis. Therefore, angles RPS and RPQ are supplementary? Wait, because together they make a straight line through point P. So, if angle RPQ is theta, then angle RPS would be 180 - theta? Wait, but that can't be, because in the Asymptote code, angle RPS is drawn from PR to PS, which is a different angle. Wait, maybe not. Wait, let me think.

Wait, if angle RPQ is the angle between PR and PQ, and angle RPS is the angle between PR and PS, and since PQ and PS are on opposite sides of P (PQ is to the right, PS is to the left), then angles RPQ and RPS are actually adjacent angles at P, adding up to 180 degrees? No, because they are on opposite sides, but the angle between PR and PQ is theta, and the angle between PR and PS would be 180 - theta? Wait, not necessarily. Wait, if PR is in some direction, then the angle between PR and PQ (to the right) is theta, and the angle between PR and PS (to the left) would be 180 - theta. But since PQ and PS are colinear but opposite, the angles would add up to 180. Wait, but in reality, angle RPS is measured between PR and PS, which is the same as the angle between PR and the negative x-axis. So, if angle RPQ is theta (between PR and positive x-axis), then angle RPS is 180 - theta? Wait, not necessarily. Wait, if PR is in the upper half-plane, then the angle between PR and positive x-axis is theta, and the angle between PR and negative x-axis would be 180 - theta. So, yes, angle RPS would be 180 - theta. But in that case, cos(angle RPS) would be cos(180 - theta) = -cos(theta). But the problem states that sin(angle RPQ) = 7/25, so theta has sin(theta) = 7/25. Then, cos(theta) would be 24/25 (since sin^2 + cos^2 = 1). Therefore, cos(angle RPS) would be -24/25. But that seems too straightforward. However, the problem asks for cos(angle RPS), which is -24/25. But the answer is supposed to be boxed, so maybe that's it? Wait, but let me check again.

Wait, but in the Asymptote code, point R is at (36/25, 21/25), which is above the x-axis. So, angle RPQ is the angle between PR and PQ, which is the angle between the vector PR (from P to R) and the vector PQ (from P to Q). Similarly, angle RPS is the angle between PR and PS, where PS is from P to S. Since PQ is along the positive x-axis and PS is along the negative x-axis, the angles RPQ and RPS are supplementary. Therefore, angle RPS = 180 - angle RPQ. Therefore, cos(angle RPS) = cos(180 - angle RPQ) = -cos(angle RPQ). Since angle RPQ has sin(theta) = 7/25, then cos(theta) = 24/25. Therefore, cos(angle RPS) = -24/25. So, the answer is -24/25. Wait, but the problem states "cos(angle RPS)", so maybe we need to confirm if the angle is indeed 180 - theta.

Wait, let me verify with coordinates. Let's compute vectors PR and PS. PR is from P(0,0) to R(36/25, 21/25), so the vector is (36/25, 21/25). PS is from P(0,0) to S(-2,0), so the vector is (-2,0).

Angle RPS is the angle between vectors PR and PS. To find this angle, we can use the dot product formula:

cos(angle RPS) = (PR · PS) / (|PR| |PS|)

Compute PR · PS: (36/25)(-2) + (21/25)(0) = -72/25 + 0 = -72/25.

|PR| is sqrt((36/25)^2 + (21/25)^2) = sqrt(1296 + 441)/25 = sqrt(1737)/25.

|PS| is sqrt((-2)^2 + 0^2) = 2.

Therefore, cos(angle RPS) = (-72/25) / (sqrt(1737)/25 * 2) = (-72/25) / (2 sqrt(1737)/25) = (-72)/(2 sqrt(1737)) = -36/sqrt(1737).

Wait, but according to this calculation, cos(angle RPS) is -36/sqrt(1737). But according to the previous reasoning, it should be -24/25. These two results are different. So, there must be an error in my previous reasoning.

Wait, perhaps the angle RPS is not supplementary to angle RPQ. Let me check again.

In the Asymptote code, Q is at (2,0), S is at (-2,0), P is at (0,0). R is at (36/25, 21/25). So, angle RPQ is the angle between PR and PQ. Vector PR is (36/25, 21/25), vector PQ is (2,0). The angle between them is computed as theta, with sin(theta) = 7/25. Let's verify that.

Compute sin(theta): The angle between PR and PQ. The cross product in 2D is |PR x PQ| = |(36/25)(0) - (21/25)(2)| = |0 - 42/25| = 42/25. The magnitude of PR is sqrt((36/25)^2 + (21/25)^2) = sqrt(1737)/25. The magnitude of PQ is 2. Then, sin(theta) = |PR x PQ| / (|PR| |PQ|) = (42/25) / (sqrt(1737)/25 * 2) = (42/25) / (2 sqrt(1737)/25) = 42 / (2 sqrt(1737)) = 21 / sqrt(1737). Simplify sqrt(1737): 1737 = 9*193, so sqrt(1737) = 3 sqrt(193). Therefore, sin(theta) = 21/(3 sqrt(193)) = 7/sqrt(193). But the problem states sin(theta) = 7/25. Therefore, there is a discrepancy here. This suggests that either the Asymptote code is illustrative but not to scale, or my interpretation is wrong.

Wait, maybe the Asymptote code is just a diagram, and the actual problem is a general problem where sin(angle RPQ) = 7/25, and we need to find cos(angle RPS). So, perhaps the coordinates in the Asymptote code are just for drawing, and the actual problem is independent of those coordinates. Therefore, maybe I should ignore the specific coordinates and just use the given sine value.

Let me try that approach.

Given that sin(angle RPQ) = 7/25. Let's denote angle RPQ as theta, so sin(theta) = 7/25. Then, we need to find cos(angle RPS). Let's denote angle RPS as phi. Now, in the diagram, points S and Q are on opposite sides of P. Therefore, angles RPQ and RPS are adjacent angles at point P, but they are not supplementary unless PR is in a specific position. Wait, but in reality, depending on where R is, the angles could be on either side. However, in the Asymptote code, R is above the x-axis, so angle RPQ is above the x-axis between PR and PQ, and angle RPS would be above the x-axis between PR and PS. Wait, but PS is to the left, so angle RPS is between PR and PS, which is also above the x-axis. Therefore, both angles RPQ and RPS are above the x-axis. Wait, but then they are not supplementary. Hmm, maybe they are adjacent but not supplementary. Wait, in the diagram, since S is to the left and Q is to the right, if R is above the x-axis, then angle RPQ is between PR and PQ (right side), and angle RPS is between PR and PS (left side). Therefore, the sum of angle RPQ and angle RPS is the angle between PR and PS plus the angle between PR and PQ, which is the total angle around point P? No, wait, the total around point P is 360 degrees, but in this case, we are only considering the angles on the same side. Wait, perhaps not. Wait, maybe angle RPS is the angle between PR and PS, which is on the left side, while angle RPQ is on the right side. So, angle RPS and angle RPQ are on opposite sides of PR. Therefore, if we consider the angles around point P, the total angle around P is 360 degrees, but the angles in the diagram are only parts of that. Therefore, angle RPS and angle RPQ are not necessarily supplementary or related in a specific trigonometric way. Hmm, this is confusing.

Alternatively, maybe we can consider triangle PRS and triangle PQR. Wait, but without knowing more sides or angles, it's difficult. Wait, but perhaps if we consider the coordinates of R, Q, and S, we can compute vectors PR, PQ, and PS, then compute the angles.

But since the problem gives sin(angle RPQ) = 7/25, and asks for cos(angle RPS), maybe we can use the Law of Sines or Law of Cosines in some triangle. Let's think.

Suppose we consider triangle PRQ. In triangle PRQ, we know sin(angle RPQ) = 7/25. Let's denote angle RPQ as theta. Then, in triangle PRQ, sides opposite to angles: side opposite to theta is RQ. So, by the Law of Sines, RQ / sin(theta) = PQ / sin(angle PRQ) = PR / sin(angle PQR). But we don't know the other angles or sides. Similarly, in triangle PRS, we can apply the Law of Sines or Cosines. Wait, but we need to relate angles RPS and RPQ.

Alternatively, maybe we can express angle RPS in terms of angle RPQ. Wait, if we can find some relationship between the angles, perhaps using coordinates or vectors.

Alternatively, since we have coordinates in the Asymptote code, maybe we can compute the exact value. Let's try that.

From the Asymptote code, R is at (36/25, 21/25), P is at (0,0), S is at (-2,0). So, vector PR is (36/25, 21/25), vector PS is (-2,0). The angle between PR and PS is angle RPS. To find cos(angle RPS), compute the dot product of PR and PS divided by the product of their magnitudes.

Dot product PR · PS = (36/25)(-2) + (21/25)(0) = -72/25 + 0 = -72/25.

|PR| = sqrt((36/25)^2 + (21/25)^2) = sqrt((1296 + 441)/625) = sqrt(1737)/25.

|PS| = sqrt((-2)^2 + 0^2) = 2.

Therefore, cos(angle RPS) = (-72/25) / (sqrt(1737)/25 * 2) = (-72/25) / (2 sqrt(1737)/25) = (-72)/(2 sqrt(1737)) = -36/sqrt(1737).

But sqrt(1737) = sqrt(9*193) = 3 sqrt(193). Therefore, cos(angle RPS) = -36/(3 sqrt(193)) = -12/sqrt(193). Rationalizing the denominator, we get -12 sqrt(193)/193. But according to the problem statement, sin(angle RPQ) = 7/25. Let's check if this is consistent.

Angle RPQ is the angle between PR and PQ. Vector PQ is (2,0). Vector PR is (36/25, 21/25). The dot product PR · PQ = (36/25)(2) + (21/25)(0) = 72/25. The magnitude of PR is sqrt(1737)/25, as before. The magnitude of PQ is 2. Therefore, cos(theta) = (72/25) / (sqrt(1737)/25 * 2) = 72/(2 sqrt(1737)) = 36/sqrt(1737) = 12/sqrt(193). Which is 12 sqrt(193)/193. Then, sin(theta) can be found using sin(theta) = sqrt(1 - cos^2(theta)) = sqrt(1 - (144*193)/193^2) = sqrt(1 - 144/193) = sqrt(49/193) = 7/sqrt(193). Which is approximately 7/13.89, which is approximately 0.505, which is roughly 7/25 = 0.28. Wait, but according to the problem statement, sin(theta) = 7/25, which is approximately 0.28. But here, we get sin(theta) = 7/sqrt(193) ≈ 7/13.89 ≈ 0.505. This is a discrepancy. Therefore, the coordinates in the Asymptote code must be scaled. Because in the Asymptote code, PR is (36/25, 21/25), which gives sin(theta) ≈ 0.505, but the problem states sin(theta) = 7/25 ≈ 0.28. Therefore, the Asymptote code is illustrative, and we need to solve the problem using the given sin(theta) = 7/25, without relying on the coordinates.

Therefore, let's proceed with the given information: sin(angle RPQ) = 7/25. We need to find cos(angle RPS). Let's denote angle RPQ as theta, so sin(theta) = 7/25. Then, we need to find cos(phi), where phi is angle RPS.

To relate these angles, perhaps we can use the fact that in triangle PRQ, angle at P is theta, and in triangle PRS, angle at P is phi. But without more information, it's difficult. Alternatively, perhaps we can consider the coordinates scaled such that PR has length 25, so that sin(theta) = 7/25 corresponds to the opposite side being 7. Wait, maybe that's a way to model it.

Alternatively, perhaps we can model triangle PRQ such that PR is the hypotenuse with length 25, PQ is the adjacent side with length 24, and RQ is the opposite side with length 7. Then, angle RPQ would have sin(theta) = 7/25, cos(theta) = 24/25. Then, if we can find angle RPS, perhaps it's related to angle RPQ.

But wait, in the Asymptote code, PR is not length 25, but the coordinates are scaled. So, if we scale the diagram such that PR is length 25, then the coordinates would change. However, the problem is given in terms of angles and sine values, so perhaps we can ignore the specific coordinates and just use the given sine value to compute the cosine.

Alternatively, perhaps angle RPS is the complement or supplement of angle RPQ. Wait, but in the Asymptote code, angle RPS is not necessarily related. So, maybe we need to use some trigonometric identities or laws.

Wait, another approach: Let's consider the vectors. Suppose we have point P at the origin. Let’s denote vector PR as **a** and vector PS as **b**. Then, angle RPQ is the angle between **a** and **PQ**, which is along the x-axis. Let's denote vector PQ as **c** = (c_x, 0). Then, the angle between **a** and **c** is theta, with sin(theta) = 7/25. Then, vector PS is **b** = (-d, 0), where d is positive. Then, the angle between **a** and **b** is phi = angle RPS, and we need to find cos(phi).

To find cos(phi), we can use the dot product formula between vectors **a** and **b**. However, we need to know the relationship between **a** and **b**. But without knowing the magnitude of **a** or **b**, it's difficult. However, perhaps we can relate the magnitudes through the given sine value.

Alternatively, since we know sin(theta) = 7/25, we can find cos(theta) = 24/25. Then, if we can express cos(phi) in terms of cos(theta), perhaps using some geometric relationship.

Wait, if we consider the coordinates of R, then vector PR is (x, y), and vector PQ is (2,0). The angle between them has sin(theta) = 7/25. Therefore, the cross product |PR x PQ| = |x*0 - y*2| = | -2y | = 2|y|. The magnitude of PR is sqrt(x^2 + y^2), and magnitude of PQ is 2. Therefore, sin(theta) = 2|y| / (sqrt(x^2 + y^2)*2) = |y| / sqrt(x^2 + y^2) = 7/25. Therefore, |y| = (7/25) sqrt(x^2 + y^2). Squaring both sides: y^2 = (49/625)(x^2 + y^2). Multiply both sides by 625: 625 y^2 = 49 x^2 + 49 y^2. Then, 625 y^2 - 49 y^2 = 49 x^2. 576 y^2 = 49 x^2. Therefore, (y/x)^2 = 49/576. Therefore, y/x = ±7/24. Since in the Asymptote code, y is positive (R is above x-axis), we can take y/x = 7/24. Therefore, y = (7/24) x. So, vector PR is (x, 7x/24). Then, the magnitude of PR is sqrt(x^2 + (49x^2)/576) = sqrt( (576x^2 + 49x^2)/576 ) = sqrt(625x^2 / 576 ) = (25x)/24. Therefore, PR = (25x)/24. Let's denote x as a scalar multiple. Let’s let x = 24k, so that PR = (25*24k)/24 = 25k. Therefore, vector PR is (24k, 7k). Then, point R is at (24k, 7k). Therefore, vector PR is (24k, 7k), vector PQ is (2, 0), vector PS is (-d, 0), where d is positive.

Now, we need to find cos(angle RPS). Let's compute angle between PR and PS. Vector PR is (24k, 7k), vector PS is (-d, 0). The dot product is (24k)(-d) + (7k)(0) = -24k d. The magnitude of PR is sqrt((24k)^2 + (7k)^2) = sqrt(576k^2 + 49k^2) = sqrt(625k^2) = 25k. The magnitude of PS is sqrt((-d)^2 + 0^2) = d. Therefore, cos(phi) = (-24k d)/(25k * d) = -24/25. Therefore, cos(angle RPS) = -24/25.

Wait, that's different from the previous result where we considered the Asymptote code. But here, we derived it using the given sine value. So, this seems correct. Therefore, the answer is -24/25.

But let's check if this makes sense. Let's consider the coordinates. If PR is (24k, 7k), then vector PR has a slope of 7/24, which matches the earlier ratio. Then, vector PS is (-d, 0). The angle between PR and PS is the angle between (24k,7k) and (-d,0). The dot product is -24k d, and the magnitudes are 25k and d. So, the cosine is -24/25, which is negative. Therefore, angle RPS is obtuse, which makes sense because point S is on the opposite side of P from Q. Therefore, the angle between PR and PS is greater than 90 degrees, hence the cosine is negative. Therefore, cos(angle RPS) = -24/25.

But wait, in the problem statement, the Asymptote code draws R such that sin(angle RPQ) = 7/25. If we use the derived coordinates, PR is (24k,7k), PQ is (2,0), then angle RPQ has sin(theta) = 7/25, which matches. Then, angle RPS has cos(phi) = -24/25. Therefore, the answer should be -24/25. However, the problem asks for cos(angle RPS), and in the Asymptote code, angle RPS is drawn as an angle between PR and PS, which is on the left side. Therefore, the angle is indeed greater than 90 degrees, hence cosine negative. Therefore, the answer is -24/25.

But let me verify once again. Suppose PR is a vector (24,7), so its magnitude is 25. Then, vector PS is (-1,0) for simplicity. Then, the dot product is (24)(-1) + (7)(0) = -24. The magnitudes are 25 and 1. Therefore, cos(theta) = -24/25. Therefore, angle RPS is arccos(-24/25), which is in the second quadrant. Therefore, yes, this is consistent.

Therefore, the answer should be -24/25. However, the problem statement says "What is $\cos \angle RPS$?" So, unless there is a miscalculation here, this should be the answer. But let me check the problem statement again.

Wait, in the problem statement, it's written as "sin(angle RPQ) = 7/25". So, angle RPQ is angle at P between PR and PQ. Then, angle RPS is angle at P between PR and PS. Since S is on the opposite side of P from Q, the angle RPS is the angle between PR and the opposite direction of PQ. Therefore, if angle RPQ is theta with sin(theta) = 7/25, then angle RPS is 180 - theta, whose cosine is -cos(theta) = -24/25. Therefore, this is consistent. Therefore, the answer is -24/25.

But let me check with the coordinates again. If angle RPQ is theta, with sin(theta) = 7/25, then cos(theta) = 24/25. Then, angle RPS is 180 - theta, so cos(angle RPS) = cos(180 - theta) = -cos(theta) = -24/25. Therefore, this matches.

Alternatively, if angle RPS is supplementary to angle RPQ, then yes, the cosine would be -24/25. Therefore, regardless of the coordinates, as long as angle RPS is supplementary to angle RPQ, the cosine would be -24/25. Therefore, the answer is -24/25.

But wait, the problem didn't specify whether angle RPS is on the same side or opposite side. However, in the Asymptote code, S is on the left side of P, so angle RPS is formed by PR going to the right (R) and PS going to the left (S). Therefore, angle RPS is the angle between PR and PS, which are on opposite sides of P. Therefore, angle RPS is a reflex angle? Wait, no, in the diagram, points R, P, S are arranged such that R is above the x-axis, P is at the origin, and S is at (-2,0). Therefore, angle RPS is the angle between PR (going from P to R, which is up and right) and PS (going from P to S, which is left along the x-axis). Therefore, angle RPS is the angle between PR and the negative x-axis. Since PR is in the first quadrant, the angle between PR and the negative x-axis is indeed 180 - theta, where theta is the angle between PR and the positive x-axis (angle RPQ). Therefore, cos(angle RPS) = -cos(theta) = -24/25.

Therefore, the answer is -24/25. However, the problem asks for $\cos \angle RPS$, and in the Asymptote code, the angle is drawn as a non-reflex angle, so it's less than 180 degrees. Therefore, the answer is -24/25.

But let me check once again with the coordinates. Suppose PR is (24,7), then the angle between PR and the positive x-axis is theta, with cos(theta) = 24/25, sin(theta) = 7/25. Then, the angle between PR and the negative x-axis is 180 - theta, which has cos(180 - theta) = -cos(theta) = -24/25. Therefore, yes, cos(angle RPS) is -24/25.

Therefore, the answer is $\boxed{-\dfrac{24}{25}}$.

**Final Answer**
\boxed{-\dfrac{24}{25}}
Therefore, the $\cos \angle RPS$ is $\boxed{-\dfrac{24}{25}}$.
</think>