Find the coordinates of a point on a circle

From the picture, it seems that your circle has centre the origin, and radius $r$. The rotation appears to be clockwise. And the question appears to be about where the point $(0,r)$ at the top of the circle ends up.

The point $(0,r)$ ends up at $x=r\sin\theta$, $y=r\cos\theta$.

In general, suppose that you are rotating about the origin clockwise through an angle $\theta$. Then the point $(s,t)$ ends up at $(u,v)$ where $$u=s\cos\theta+t\sin\theta\qquad\text{and} \qquad v=-s\sin\theta+t\cos\theta.$$

With an angle of 115° in a clockwise direction, you can find your point (x,y) as shown in your diagram with the following math:

Any point $(x,y)$ on the path of the circle is $x = r*sin(θ), y = r*cos(θ)$

thus: $(x,y) = (12*sin(115), 12*cos(115))$

So your point will roughly be $(10.876, -5.071)$ (assuming the top right quadrant is x+, y+)

The standard circle is drawn with the 0 degree starting point at the intersection of the circle and the x-axis with a positive angle going in the counter-clockwise direction.

Thus, the standard textbook parameterization is: x=cos t y=sin t

In your drawing you have a different scenario. The way it is drawn, the starting point is at the top and increasing degrees is in the clockwise direction. Thus, the standard parameterization must be modified to accommodate your situation.

Take a look at what the value of x is in your picture at the starting point and then what happens as t increases. x starts at 0 and then increases to a maximum of 1 and then returns to 0 when t = Pi.

Now you want to compare that behavior to a standard graph of sin and cos to decide which one matches that need. X=sin t behaves that way, so now you have the parameterization of x. Note, it is not x=cos t as a standard math book teaches you because in trigonometry class they typically have 0 degrees at the intersection of the x-axis and the unit circle.

Now, y in your drawing starts out at 1 and then decreases until you hit 0 and then -1 at PI. A look at the a graph of either sin or cos shows that cos behaves that way.

So... Y=cos t.

I'm posting this because it can aid someone that knows about sin and cos, but has a problem in which the 0 degree starting point is in a non-standard position and the direction of positive degrees is in the clockwise direction, not the ccw direction.