<Question>
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]
</Question>

<think>
<Plan_1>
Okay, let me try to figure out how to find cos(angle RPS) given that sin(angle RPQ) is 7/25. Hmm, first, I need to understand the diagram described by the Asymptote code. Let me parse that.

<prompt_1.1> Understand the diagram described by the Asymptote code and identify the coordinates of points R, P, Q, and S. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The Asymptote code defines points R, P, Q, and S. P is at (0,0), Q is at (2,0), S is at (-2,0), and R is constructed by rotating the point (1.5,0) around P by an angle whose sine is 7/25. So, R is somewhere above the x-axis, and the 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. The question is asking for the cosine of that angle.

First, let me visualize this. Points P, Q, and S are on the x-axis, with P at the origin, Q at (2,0), and S at (-2,0). Point R is somewhere in the upper half-plane (since it's rotated above the x-axis). The angle RPQ is the angle between PR and PQ, and angle RPS is the angle between PR and PS. Since Q and S are on opposite sides of P, these angles are likely supplementary or something else? Wait, not necessarily. Let me think.

Given that sin(angle RPQ) = 7/25, we can find cos(angle RPQ) first. Since sin^2(theta) + cos^2(theta) = 1, cos(theta) would be sqrt(1 - (7/25)^2) = sqrt(1 - 49/625) = sqrt(576/625) = 24/25. So cos(angle RPQ) is 24/25.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But how does that help us find cos(angle RPS)?

Wait, angle RPQ is the angle between PR and PQ, and angle RPS is the angle between PR and PS. Since Q and S are on opposite sides of P, these two angles might be related. Let me consider the positions.

<prompt_2.1> Analyze the relationship between angles RPQ and RPS based on the positions of Q and S relative to P. </prompt_2.1>
-----

But let me verify this with coordinates. Let's try to compute the coordinates of R.

<prompt_2.2> Verify the solution by computing the coordinates of point R and calculating cos(angle RPS) using vector dot product. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
If I stand at point P, angle RPQ is the angle looking from R to P to Q, which is on the right side (since Q is at (2,0)). Angle RPS is looking from R to P to S, which is on the left side. So, these two angles might add up to the total angle around point P between R and the negative x-axis? Wait, maybe not. Let me think again.

Alternatively, since Q is at (2,0) and S is at (-2,0), then PQ is the vector from P to Q, which is (2,0), and PS is the vector from P to S, which is (-2,0). PR is the vector from P to R, which is somewhere in the upper half-plane. The angle RPQ is between PR and PQ, and angle RPS is between PR and PS. Since PQ and PS are in opposite directions along the x-axis, the angles RPQ and RPS might be related such that angle RPQ + angle RPS = 180 degrees? No, that can't be, because R is on one side.

Wait, no. If you have a point R above the x-axis, then from point P, the angle between PR and PQ (which is on the right) is angle RPQ, and the angle between PR and PS (which is on the left) is angle RPS. Since PQ and PS are in opposite directions, the sum of angles RPQ and RPS should be equal to the angle between PQ and PS, which is 180 degrees? But that doesn't make sense because R is a single point. Wait, perhaps angle RPQ and angle RPS are on different sides of PR. Let me draw a mental picture.

Imagine point P at the origin. Point Q is to the right at (2,0), point S is to the left at (-2,0), and point R is somewhere above the x-axis. The angle RPQ is the angle between the line PR and PQ (which is along the positive x-axis). The angle RPS is the angle between PR and PS (which is along the negative x-axis). So, if you consider PR, angle RPQ is the angle between PR and the positive x-axis, and angle RPS is the angle between PR and the negative x-axis. Therefore, angle RPQ + angle RPS = 180 degrees? Because if you add the angle from PR to PQ (positive x-axis) and from PR to PS (negative x-axis), they should sum up to a straight line, which is 180 degrees. So, angle RPS = 180° - angle RPQ. Therefore, cos(angle RPS) = cos(180° - angle RPQ) = -cos(angle RPQ).
</execution_2.1>

<execution_2.2>
Since R is obtained by rotating the point (1.5, 0) around P (the origin) by an angle whose sine is 7/25. So, the original point (1.5, 0) is on the x-axis. Rotating it by angle theta (where sin(theta) = 7/25) would give the coordinates of R as (1.5*cos(theta) - 0*sin(theta), 1.5*sin(theta) + 0*cos(theta)) = (1.5*cos(theta), 1.5*sin(theta)). Since sin(theta) = 7/25, cos(theta) is 24/25. Therefore, R is at (1.5*(24/25), 1.5*(7/25)) = (3.6/25*1.5? Wait, 1.5 is 3/2. So 1.5 * 24/25 = (3/2)*(24/25) = 72/50 = 36/25 = 1.44. Similarly, 1.5 * 7/25 = (3/2)*(7/25) = 21/50 = 0.42. So R is at (36/25, 21/50).

Now, let's compute vectors PR and PS. Wait, angle RPS is at point P, between points R, P, and S. So vectors are PR and PS. Wait, actually, angle RPS is the angle at P between points R, P, S. So the angle between vectors PR and PS. Similarly, angle RPQ is the angle at P between R, P, Q, so vectors PR and PQ.

Given that, since we have coordinates for all points, maybe we can compute the vectors and then use the dot product formula to find the cosine of angle RPS.

First, let's note the coordinates:

P = (0,0)

R = (36/25, 21/50)

Q = (2,0)

S = (-2,0)

So vector PR is R - P = (36/25, 21/50)

Vector PS is S - P = (-2, 0)

So the angle between PR and PS is angle RPS. To find cos(angle RPS), we can use the dot product formula:

cos(theta) = (PR · PS) / (|PR| |PS| )

Compute PR · PS:

(36/25)(-2) + (21/50)(0) = (-72/25) + 0 = -72/25

|PR| is sqrt( (36/25)^2 + (21/50)^2 )

Let's compute that:

(36/25)^2 = (1296)/(625)

(21/50)^2 = (441)/(2500)

So |PR| = sqrt(1296/625 + 441/2500) = sqrt( (1296*4 + 441)/2500 ) = sqrt( (5184 + 441)/2500 ) = sqrt(5625/2500) = sqrt(225/100) = 15/10 = 3/2

That's interesting, |PR| is 3/2.

|PS| is sqrt( (-2)^2 + 0^2 ) = sqrt(4) = 2

Therefore, cos(theta) = (-72/25) / ( (3/2)*2 ) = (-72/25) / 3 = (-72/25)*(1/3) = -24/25

So, cos(angle RPS) is -24/25.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But wait, is that correct?

Wait, if angle RPQ is the angle between PR and PQ, and angle RPS is the angle between PR and PS, then since PQ and PS are in opposite directions, the angles RPQ and RPS would be supplementary? Let me think. If you have two angles that are on opposite sides of a line from a common vertex, their sum would be 180 degrees if they are linear pairs. But in this case, since R is above the x-axis, the angles RPQ and RPS are on different sides. Wait, actually, if you imagine standing at point P, looking towards Q (to the right), angle RPQ is the angle above the line of sight to Q. Then angle RPS is the angle below the line of sight to S (to the left). But since R is a single point above the x-axis, maybe angles RPQ and RPS are not supplementary. Hmm, maybe my initial thought was wrong.

Alternatively, perhaps angle RPQ and angle RPS are measured from PR to PQ and from PR to PS, but since PQ and PS are in opposite directions, the total angle around point P would be angle RPQ + angle RPS = 180°, but that might not necessarily be true because R is a fixed point.

Wait, perhaps a better approach is to use coordinates and vectors. Let's assign coordinates as per the Asymptote code.

<prompt_3.1> Re-evaluate the relationship between angles RPQ and RPS using vector analysis and consider the position of point R. </prompt_3.1>
-----

Based on execution_2.2:
But wait, angles in triangles can't have negative cosine values? Wait, no, cosine can be negative. For example, if the angle is greater than 90 degrees, cosine is negative.
So, if angle RPS is greater than 90 degrees, its cosine can be negative. But let's see if this makes sense.

Given that R is above the x-axis, and S is to the left of P, the angle RPS is the angle between PR and PS. Since PR is going up to the left, and PS is going straight left, the angle between them could be acute or obtuse. Depending on where R is.

<prompt_3.2> Verify the solution by analyzing the position of point R and the angle RPS. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
The Asymptote code defines P at (0,0), Q at (2,0), S at (-2,0), and R is obtained by rotating the point (1.5,0) by an angle whose sine is 7/25. So, the original point (1.5,0) is on the x-axis. Rotating this point by an angle theta (where sin(theta) = 7/25) around P (0,0) would give the new coordinates of R as (1.5*cos(theta) - 0*sin(theta), 1.5*sin(theta) + 0*cos(theta)) = (1.5*cos(theta), 1.5*sin(theta)). Since sin(theta) = 7/25, we can find cos(theta) as 24/25 (since sin^2 + cos^2 = 1). Therefore, R has coordinates (1.5*(24/25), 1.5*(7/25)) = (3.6/25*1.5? Wait, 1.5 is 3/2. So 1.5 * 24/25 = (3/2)*(24/25) = 72/50 = 36/25 = 1.44. Similarly, 1.5 * 7/25 = (3/2)*(7/25) = 21/50 = 0.42. So R is at (36/25, 21/50).
</execution_3.1>

<execution_3.2>
Wait, in the Asymptote code, R is constructed by rotating the point (1.5,0) by an angle with sin(theta) = 7/25. Since sin(theta) is positive, the angle is either in the first or second quadrant. However, the Asymptote code uses rotate(aSin(7/25)), which would give an angle in the first quadrant (since arcsin(7/25) is between 0 and pi/2). Therefore, R is above the x-axis, in the second quadrant? Wait, no. If you rotate the point (1.5,0) around P (the origin) by an angle theta with sin(theta) = 7/25, then the coordinates of R would be (1.5*cos(theta), 1.5*sin(theta)). But if theta is in the first quadrant, then cos(theta) is positive, so R would be in the first quadrant. But in the Asymptote code, S is at (-2,0), so maybe R is actually in the second quadrant? Wait, the Asymptote code draws Q--SS (which is S) and P--R. If R is constructed by rotating (1.5,0) by arcsin(7/25), which is a small angle, then R would be in the first quadrant. But the diagram has S at (-2,0). Wait, perhaps there's a mistake here.

Wait, the Asymptote code defines R as rotate(aSin(7/25))*(1.5,0). The rotate function in Asymptote rotates a point around the origin (since P is at (0,0)). The angle arcsin(7/25) is an angle in the first quadrant, so rotating (1.5,0) by that angle would place R in the first quadrant. However, the diagram has S at (-2,0), which is to the left of P. So if R is in the first quadrant, then angle RPS would be the angle between PR (going to the first quadrant) and PS (going to the left). That angle would be greater than 90 degrees, which aligns with the cosine being negative. So that makes sense.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
So, angle RPQ is the angle between PR and PQ. Let's compute that angle to verify.

<prompt_4.1> Verify the calculated coordinates of point R and consider the angle RPQ. </prompt_4.1>
-----

Based on execution_3.2:
Alternatively, perhaps there's a way to do this without coordinates, using trigonometric identities.

Given that sin(angle RPQ) = 7/25, so angle RPQ = arcsin(7/25).

<prompt_4.2> Explore an alternative method using trigonometric identities to find cos(angle RPS). </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
First, vector PQ is from P to Q, which is (2,0). Vector PR is from P to R, which is (36/25, 21/50). The angle between these two vectors is angle RPQ. Let's compute cos(angle RPQ) using the dot product:

PR · PQ = (36/25)(2) + (21/50)(0) = 72/25 + 0 = 72/25

|PR| = sqrt( (36/25)^2 + (21/50)^2 ) = sqrt(1296/625 + 441/2500) = sqrt(5625/2500) = sqrt(225/100) = 15/10 = 3/2

|PQ| = sqrt(2^2 + 0^2) = 2

Therefore, cos(angle RPQ) = (72/25) / ( (3/2)*2 ) = (72/25) / 3 = 72/75 = 24/25. Which matches the given value. So that's correct.
</execution_4.1>

<execution_4.2>
Then angle RPS is 180° - angle RPQ, so cos(angle RPS) = -cos(angle RPQ) = -24/25. That's a quicker way, but does that hold geometrically?

Wait, if angle RPS is 180° - angle RPQ, then their cosines would be negatives. But is that the case here?

Wait, if angle RPQ is the angle between PR and PQ (to the right), and angle RPS is the angle between PR and PS (to the left), then these two angles are supplementary. Because if you have a point R above the x-axis, the angles between PR and the positive x-axis (angle RPQ) and between PR and the negative x-axis (angle RPS) would add up to 180°, since they are on opposite sides of PR. Wait, no. Wait, actually, if you consider the angle between PR and PQ (which is along the positive x-axis) and the angle between PR and PS (which is along the negative x-axis), then those two angles are actually supplementary. Because if you go from PR to PQ, that's angle RPQ, and from PR to PS, that's angle RPS. Since PQ and PS are in opposite directions, the angles RPQ and RPS would add up to 180°, making them supplementary. Therefore, angle RPS = 180° - angle RPQ. Therefore, cos(angle RPS) = cos(180° - angle RPQ) = -cos(angle RPQ). Since we found cos(angle RPQ) = 24/25, then cos(angle RPS) = -24/25.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But the question is asking for cos(angle RPS). Let's compute that.

First, angle RPS is the angle at point P between points R, P, and S. So, we need to find the cosine of the angle between vectors PR and PS.

<prompt_5.1> Calculate the cosine of angle RPS using the dot product formula. </prompt_5.1>
-----

Based on execution_4.2:
But let me just check again if the angles are indeed supplementary. Let's imagine PR is in the first quadrant, making angle RPQ with the positive x-axis (PQ). Then angle RPS is the angle between PR and the negative x-axis (PS). Since the positive and negative x-axes are 180° apart, the angles RPQ and RPS should add up to 180°, hence supplementary.

<prompt_5.2> Verify the geometric relationship between the angles RPQ and RPS. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
To find this, we can use the dot product formula:

cos(theta) = (PR · PS) / (|PR| |PS|)

First, let's find vectors PR and PS. Since PR is from P to R, which is (36/25, 21/50), and PS is from P to S, which is (-2, 0). So:

PR · PS = (36/25)(-2) + (21/50)(0) = -72/25 + 0 = -72/25

|PR| is the magnitude of vector PR, which we already calculated earlier as 3/2. Let me verify that:

|PR| = sqrt( (36/25)^2 + (21/50)^2 ) = sqrt( (1296/625) + (441/2500) ) = sqrt( (5184/2500) + (441/2500) ) = sqrt(5625/2500) = sqrt(225/100) = 15/10 = 3/2. Yes, that's correct.

|PS| is the magnitude of vector PS, which is sqrt((-2)^2 + 0^2) = sqrt(4) = 2.

Therefore, cos(theta) = (-72/25) / ( (3/2)(2) ) = (-72/25) / 3 = (-72/25) * (1/3) = -24/25.

So, cos(angle RPS) is -24/25.
</execution_5.1>

<execution_5.2>
Therefore, angle RPS = 180° - angle RPQ. Therefore, cos(angle RPS) = cos(180° - angle RPQ) = -cos(angle RPQ) = -24/25.

That seems straightforward. So, the answer is -24/25. But wait, cosine can be negative, which would imply that angle RPS is obtuse, which makes sense if R is above the x-axis and S is to the left. The angle between PR (which is in the first quadrant) and PS (which is along the negative x-axis) would indeed be greater than 90 degrees, hence negative cosine.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.1:
But let me make sure there isn't a simpler way to see this without coordinates.

<prompt_6.1> Reflect on the simplicity of the solution and consider if there is an alternative method using trigonometric identities. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Given that we know sin(angle RPQ) = 7/25, so angle RPQ has sin(theta) = 7/25, so cos(theta) = 24/25. Then angle RPS is 180° - theta, so cos(180° - theta) = -cos(theta) = -24/25. Therefore, the answer is -24/25. That's much simpler. So, perhaps there's a way to see that angle RPS is supplementary to angle RPQ, hence the negative cosine. But I need to confirm if angle RPS is indeed supplementary to angle RPQ.

Wait, angle RPQ is between PR and PQ (right side), angle RPS is between PR and PS (left side). Since PQ and PS are in opposite directions, the angles RPQ and RPS add up to the angle between PQ and PS, which is 180°, so yes, angle RPQ + angle RPS = 180°, so they are supplementary. Therefore, their cosines are negatives. Therefore, cos(angle RPS) = -cos(angle RPQ) = -24/25.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
**Final Answer**
\boxed{-\dfrac{24}{25}}
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the final answer is $\cos \angle RPS = \boxed{-\frac{24}{25}}$.
</Final_answer>
</think>