What is the largest circle that fits in $\sin(x)?$ Imagine dropping a circle into the trough of $\sin(x)$. Would it reach the bottom or get wedged between two points on the curve? Depends on the size of the circle. 
So, what is the radius of the largest circle that will reach the bottom of the curve $y=\sin(x)$?
This problem was inspired by this similar one I found in a calculus textbook: Find the radius of the largest circle that will reach the bottom of the curve $y=x^2$ without getting stuck. I was intrigued by the answer of $r=1/2$. 
I have tried attacking this problem from several angles but all have failed. Perhaps an exact numerical solution is not obtainable, I don't know. All help will be appreciated.
Thanks!

Nice question.
If we are dropping the circle so that its center is right above $\frac{3\pi}{2}$ then the coordinates of the center of the dropped circle will be $(x_0,y_0) = (3\pi/2,r-1)$ where $r$ is the radius of the circle. Now, the distance from this point to any point $(x,y)$ on the curve of $\sin$ is
$$
  \sqrt{(x_0 - x)^2 + (y_0 - y)^2}
= \sqrt{ (3\pi/2-x)^2 + (r-1)^2 - 2(r-1)\sin x + \sin^2 x},
$$
and this can be no less than $r$. The resulting inequality is 
$$
     2r(1 + \sin x) 
\leq (3\pi/2-x)^2 + 1 + 2\sin x + \sin^2 x
$$ 
which leads to 
$ r \leq ((3\pi/2-x)^2 + (1+\sin x)^2)/2(1+\sin x)$. Then the answer should be 
\begin{align}
   r
&= \inf \{ ((3\pi/2-x)^2 + (1+\sin x)^2)/2(1+\sin x) \mid x \in (0,\pi) \} \\
&= \frac{1}{2}\inf \{ (3\pi/2-x)^2/(1+\sin x) + (1+\sin x) \mid x \in (0,\pi) \}.
\end{align}
Let the function in the infimum be $f$. Differentiating yields
$$
  \frac{-2(3\pi/2 - x)(1+\sin x) - \cos x(3\pi/2-x)^2}{(1+\sin x)^2} + \cos x
$$
and if you solve for where this is $0$ you get 
$$
 \cos x((1 + \sin x)^2 - (3\pi/2-x)^2)
= 2(3\pi/2-x)(1+\sin x).
$$
One solution is at $x = 3\pi/2$, using L'hopital's rule, 
$$
  \lim_{x \to 3\pi/2} f(x)
= \frac{1}{2} \lim_{x \to 3\pi/2} \frac{-2(3\pi/2-x)}{\cos x}
= \frac{1}{2} \lim_{x \to 3\pi/2} \frac{2}{-\sin x}
= 1.
$$
This is $r$ (it just remains to show that the critical point is the minimum for $f$, and the first derivative test verifies this).
