Minimal distance between points on two graphs What is the minimal distance $d$ between the graphs of $y_1 = \sin x$ and $y_2 = 2 + \sin x$? The trivial observation is that $d \le 2$ (just set $x = 0$ in both equations), but numerical computations show that for $x_1 \approx 0.4787$ and $x_2 \approx -0.4787$ (it is surprising that $x_1+x_2 = 0$, isn't it?) we have $d((x_1, y_1(x_1)), (x_2,y_2(x_2)) \approx 1.4423$. Can we find the exact value? I already tried using RIES but it gave no results.
 A: As yohBS commented, the problem is to minimize the square of the distance, that is to say $$F(x,y)=(x-y)^2+(\sin(x)+2-\sin(y))^2$$ Computing the partial derivatives, $$F'_x(x,y)=2 (x-y)+2 \cos (x) (\sin (x)-\sin (y)+2)$$ $$F'_y(x,y)=-2 (x-y)-2 \cos (y) (\sin (x)-\sin (y)+2)$$ What is "clear" is that $$F'_x(x,-x)=4 x+2 (2 \sin (x)+2) \cos (x)$$ $$F'_y(x,-x)=-4 x-2 (2 \sin (x)+2) \cos (x)=-F'_x(x,-x)$$ which make both partial derivatives equal to $0$ if one of them is $0$.
So, the problem reduces to the minimization of $$G(x)=4 x^2+(2 \sin (x)+2)^2$$ that is to say to the solution of $$G'(x)=8 (x+(\sin (x)+1) \cos (x))=0$$ The plot of this function shows a root very close to $-\frac 12$. So, using Newton with $x_0=-\frac 12$, the first iteration gives $$x_1=-\frac{1}{2}-\frac{\left(1-\sin \left(\frac{1}{2}\right)\right) \cos
   \left(\frac{1}{2}\right)-\frac{1}{2}}{1+\left(1-\sin
   \left(\frac{1}{2}\right)\right) \sin \left(\frac{1}{2}\right)+\cos
   ^2\left(\frac{1}{2}\right)}\approx -0.478634$$ while the solution is $\approx -0.478722$ (it is worth to mention that one iteration of Halley method leads to $x_1 \approx -0.478725$) 
Using Newton with $x_0=-\frac {\pi} 6$, the first iteration gives $$x_1=-\frac{1}{24} \left(3 \sqrt{3}+2 \pi \right)\approx -0.478306$$ and using Halley would give $$x_1=-\frac{603 \sqrt{3}+4 \pi  \left(87+\sqrt{3} \pi \right)}{4608} \approx -0.478750$$
Using $x_1$ as an approximation of the solution, the minimum distance is then given by an ugly formula (too large to fit on the page); in any manner, at this point $F(x_1,-x_1)\approx 2.08031$ corresponding to a distance $\approx 1.44233$ you already found.
You could notice that $F(-\frac {\pi} 6,\frac {\pi} 6)=1+\frac{\pi ^2}{9}$ to which corresponds a distance $\approx 1.44797$ quite close to the rigorous minimum. Developing $F(x,-x)$ as a third order Taylor series built at $x=-\frac {\pi} 6$ gives a minumum for $$x=-\frac{1}{6} \left(8 \sqrt{3}+\pi -2 \sqrt{39+2 \sqrt{3} \pi }\right)\approx -0.478741$$
All of this can be confirmed by an analysis on the contour plots.
A: It is antisymmetric about the point $(1,1)$.
For minimal distance the common normal should be normal to either graph.
The minimal distance = $d$,  $ (d/2)^2 = x^2 + ( 1 - \sin x)^2 $ which simplifies to:
$ ( x- ( 1 - \sin x) \cos x) =0, $  which has a numerical solution $ x = \pm 0.478722$
So the two points for minimum distance are between the antisymmetric points:
$$ (0.478722,.460646), (-0.478722,1.53935 ) $$

