The idea is that you start at time #t=0# at the base of the Ferris wheel, and for practical reasons say that the radius is #r=1#, so you start at the height #y=-1#. Then, since you're taking trig, I think you suppose that the wheel spins with the same angular speed, so #theta(t)=omega t# is the function of the angle, where #omega# is constant. Without loss of generality and to simplify, we can suppose that #omega=1#. Then we need to project see the point where we are at the time #t# on the height of the wheel, so
#y(t)=cos(t+c)#
where #c# is a constant depending on the starting point. In order to set #c#, since we want #y(0)=-1#, #cos(c)=-1#, hence #c=pi#
So the function you need to draw is basically #y=cos(t+pi )#
graph{y=cos(x+pi) [-10, 10, -5, 5]}
in order to understand what happens see this video
The height realtive to the ground is basically the same except that you have to add the height of the center: so it is #y=cos(t+pi)+1#.
