Suppose that you are given
$f(x_1) = m$ and $f(x_2) = n$.
for a certain value of x where $x\epsilon[0,2\pi]$. Also, you are given that their respective derivatives in same points:
$f'(x_1) = p$ and $f'(x_2) = q$
Given just these 4 parameters, is it possible (and how so) to find a, b and c for given function. ($a, b, c = const$)
$f(x) = a sin(b+cx)$
Note that $f(x)=a\sin(b+cx)$ implies $f'(x)=ac\cos(b+cx)$, so $c^2f(x)^2+f'(x)^2=a^2$ is constant. In particular, we need $c^2m^2+p^2=c^2n^2+q^2$. Thus (provided $m\ne\pm n$), we find $$c=\pm\sqrt{\frac{q^2-p^2}{m^2-n^2}}$$ and with this $$a=\pm\sqrt{c^2m^2+p^2}=\sqrt{\frac{q^2m^2-p^2n^2}{m^2-n^2}}.$$ Finding matching $b$ now is not hard any more.
The special case $m=\pm n$ is left as an exercise :)