Let the sinusoid be
$$y=a\sin x$$ and the point $(u,v)$.
I have omitted the phase paremeter because by translation it can be absorbed in $u$. And I have omitted the frequency parameter, because it can be absorbed in $u,v,a$ by scaling.
Now the squared distance is
$$(x-u)^2+(y-v)^2=(x-u)^2+(a\sin x-v)^2$$ and you can obtain the extrema by canceling the derivative
$$x-u+a\cos x(a\sin x-v)=0.$$
This is a nasty nonlinear equation that you have to solve by numerical methods. Then direct minimization is also possible. But the shape of the curve is a little scary.
Notice the inequality
$$(x-u)^2+(|a|-|v|)^2\le(x-u)^2+(a\sin x-v)^2\le (x-u)^2+(|a|+|v|)^2.$$
Considering the minimum of the upper bound, you get an $x$ interval where the searched minimum is guranteed to lie by solving
$$(x-u)^2+(|a|-|v|)^2=(|a|+|v|)^2.$$
From the second derivative,
$$1+a\sin x(a\sin x-v)+a^2\cos^2x=0.$$
This is a quadratic equation in $\sin x$, which you can solve analytically. It will give you the locations of the extrema of the first derivative, hence it separates the roots. By detecting the intervals with a change of sign, you can apply regula falsi.
Combining with the previous criterion, you can be sure to find the minimum with a finite number of attempts.
