In this question, it's said that the path of a cycloid can be given as this parametric equation:
$$\begin{align*}x &= r(t - \sin t)\\ y &= r(1 - \cos t)\end{align*}$$
and is shown here:
With unit radius, we can say
$$p(t)=<t-\sin t, 1- \cos t>.$$
I don't understand why we have minus signs in those terms. Imagine the vector from the origin to the point $p$ on the circumference as (a vector from the origin to the bottom of the circle at $t$) + (a vector from the bottom of the circle to the circle's center) + (a vector from the center of the circle to $p$), the only negatives would appear in the third summand. In particular, they would have both negative components when the point is both below and left of the center -- which should only happen 1/4 of the time, not all the time.
So why is it minus all the time, not just 1/4 of the time?
Okay so we have
So we know that, like you said, we can express the position vector of $P$, as the sum of three other vectors, that is, according to this image, $$\vec{OP}=\vec{OT}+\vec{TC}+\vec{CP}$$ And I'm not going to go through it all because you already know it, but what seems to be giving issues is that $$\vec{CP}=<-r\sin(\theta),-r\cos(\theta)>$$ and you are wondering why it is always $-r$ because it looks like the signs of the components of this vector should be changing every quadrant of the circle. Well in this case it is helpful to refer back to the unit circle.
And we can see that when $\theta \le \frac{\pi}{2}$, both components of the vector should be negative; which lines of with both sine and cosine being positive and then multiplied by the negative coefficient $-r$. Then for $\frac{\pi}{2}<\theta\le \pi$ we know that the vertical component should be positive and the horizontal should be negative, which once again lines up with the fact that sine remains positive but cosine becomes negative.
And so on, you can continue to check these and they all checkout, hence the coefficient $-r$ should always be negative and you end up with a minus in the final position vector.