Eliminating the parameters of a cycloid

Question

I was given two parametric equations and I need to eliminate the parameters, but I have no idea how. That pesky $\theta$ is hard to get rid of. Any suggestions? I tried many trig identities

$x$ = $\theta$ - sin $\theta$

$y$ = 1 - cos $\theta$

Answer 1

From the second equation, you get that $\theta=\arccos(1-y)$. Now, from the first equation you get that\begin{align}x&=\theta-\sin\theta\\&=\arccos(1-y)-\sin\bigl(\arccos(1-y)\bigr)\\&=\arccos(1-y)-\sqrt{1-(1-y)^2}\\&=\arccos(1-y)-\sqrt{2y-y^2}.\end{align}

Answer 2

From $y= 1- cos(\theta)$, $cos(\theta)= 1- y$ so $sin(\theta)$$= \sqrt{1- cos^2(\theta)}= \sqrt{1- (1- y)^2}=$$ \sqrt{2y- y^2}$ and, or course, $\theta= arccos(1- y)$. $x= \theta- sin(\theta)= arccos(1- y)- \sqrt{2y- y^2}$.