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Suppose we have two graphs of sine function in the form $y= \sin(x+\alpha)$ and $y= \sin(x-\alpha)$. These two graphs determine some closed regions which areas depend on $\alpha$ parameter. Now we want to determine the greatest possible circle which can be inserted into generated region without cutting the graphs.

Some cases of $\alpha$ seem to be easier than other.
For example, intuition tells that for $\alpha = \pi/2$ the searched circle should have a radius equal $1$, see picture below (but how to prove it it's an open issue), and it is the greatest possible circle for all possible sine waves, but ...

enter image description here

  • what would be the solution (radius of the greatest circle) for a general case of $\alpha$?
  • what methodology for the solution should be used for such kind of problem ?

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First off, notice that $\sin(x+\alpha)$ is shifted as much to the left of the origin as $\sin(x-\alpha)$ is to the right. Hence one of the the largest circles will always be such that it is centred at the origin. Since the circle doesn't cut any of the sine curves, the sine curves must touch the circle only once, i.e. the tangent lines to the sines must coincide with the tangent lines to the circle at two diametrically opposite points.

In short, the tangents to the sines at their points of intersection with $y = mx$ must be perpendicular to $y = mx$ for the circle to be greatest. The abscissae of the points of intersection are given by

$$ \sin(x\pm \alpha) = mx. $$

If the solution to this equation is $x = \pm a$, then the points of intersection are $(a, \sin(a +\alpha))$ and $(-a, \sin(a - \alpha)$. The tangents at these points are perpendicular to $y = mx$. Hence,

$$ \cos(a \pm \alpha)\cdot m = -1 \\ \implies m = \frac{-1}{\cos(a \pm \alpha)} $$

Thus,

$$ \sin 2(a\pm\alpha) = -2a $$

Clearly, $-0.5\le a \le 0.5$. And the equation of the circle is given by

$$ x^2 + y^2 = a^2 + \sin^2 (a \pm \alpha). $$

The value $a$ probably needs to be found out using numerical methods.

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    Solution seems to be good, but maybe it is possible somehow further simplify equation $\sin 2(a\pm\alpha) = -2a$ ? – 2017-02-24
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    Evidently also for $\alpha=\pi/2$ $\sin 2a =2a$ then $a=0$ and $r^2=\sin^2(\pi/2)=1$ what gives good result.. – 2017-02-24
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    Yea, maybe it can be further simplified. Thanks for the verification though. – 2017-02-24
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    Thank you for the solution. It would be interesting to see also a function $r=f(\alpha)$ if not with a formula then at least graphically.. – 2017-02-24