What is locus of a fixed point on a circle of radius $r$ rolling over the curve $y=\sin x$? What is locus of a fixed point on a circle of radius $r$ rolling over the curve $y=\sin x$? I have been struggling on the problem for many days, but I could not solve it. Geometrically, the locus seems to be a series of cycloid-looking figures mounted on the exterior of the sine curve, but analytically it seems very difficult to find the equation of the locus.

It involves an elliptic integral of the second kind.
Let $M = M(\theta) = (\theta, \sin \theta)$ be the contact point. The tangent at this point is  $T = (1, -\cos \theta)$ so the center $I$ of the circle is such that $\vec{MI} = (\frac {r \cos \theta} {1+\cos ^2 \theta}, \frac {r} {1+\cos ^2 \theta})$.
The fixed point $P$ on the rolling circle is obtained for $M$ by rotation around $I$ of an angle $s$ equal to the length of $OM$ along the curves, that is $s = \int _0^\theta \sqrt{1 + \cos^2 \theta} d\theta$. Explicitly, $IP= (\frac {r \cos \theta} {1+\cos ^2 \theta} \cos s + \frac {r} {1+\cos ^2 \theta} \sin s , \frac {-r \cos \theta} {1+\cos ^2 \theta} \sin s + \frac {r} {1+\cos ^2 \theta} \cos s) $.
So the only step that makes the curve non-rational is the integral, which indeed is an elliptic function, impossible to express with elementary functions alone. 
