Vtyjkyuk

The parametric formula for an epicycloid is:
$$x=(R+r)·cos(θ)-r·cosBiggl(θbiggl(1+fracRrbiggr)Biggr)\
y=(R+r)·sin(θ)-r·sinBiggl(θbiggl(1+fracRrbiggr)Biggr)$$

The angle θ is the angle between the positive x-axis and a line running from the origin through the center of the generating circle (with radius r).

If instead of the angle $theta$, you are given the angle between the positive x-axis and the end of the curve (we will call this angle $beta$), how do I find $theta$?

$$β = arctanfrac(R+r)·sin(θ)-r·sinBiggl(θbiggl(1+fracRrbiggr)Biggr)(R+r)·cos(θ)-r·cosBiggl(θbiggl(1+fracRrbiggr)Biggr)$$

Define $t=1+frac R r$ to make the equation
$$beta=tan ^-1left(fract sin (theta )-sin (t,theta )t cos (theta )-cos (t,theta)right)$$

If $beta$ is small, composing Taylor series, you would have
$$beta=frac t (t+1)6 theta ^3+Oleft(theta ^5right)$$ and then an estimate of $theta$.

Otherwise, I suppose that a numerical method (such as Newton) would be required.

Let us try with $R=2r$, that is to say $t=3$ and suppose $beta=1$; this would give $theta_0=frac1sqrt[3]2$ and the iterates would be

$$left(
beginarraycc
n & theta_n \
0 & 0.793700526 \
1 & 1.229465478 \
2 & 1.185508561 \
3 & 1.185309641 \
4 & 1.185309636
endarray
right)$$

Edit

We could have a better approximation using the simplest Padé approximant; this would give
$$beta=frac10t (t+1), theta ^3 60+3(t^2+10t+1), theta ^2 $$ and then a cubic equation. For the worked example, this would give $theta_0 approx 1.29716$.