Yes, you can do it in 3 points, considering $f(0)$, $f(\pi)$, and $f(\frac{\pi}{2})$.
$$
\begin{aligned}
f(0) &= a\sin b + c & \\
f(\tfrac{\pi}{2}) &= a\sin(b + \tfrac{\pi}{2}) + c & &= a\cos b + c \\
f(\pi) &= a\sin(b + \pi) + c & &= - a\sin b + c
\end{aligned}
$$
To find $c$:
$$
\begin{aligned}
f(0) + f(\pi) &= 2c \\
\implies c &= \frac{f(0) + f(\pi)}{2}
\end{aligned}
$$
To find $a$:
$$
\begin{aligned}
(f(0) - c)^2 + (f(\tfrac{\pi}{2}) - c)^2 &= a^2\sin^2 b + a^2\cos^2 b= a^2 \\
\implies a &= \sqrt{(f(0) - c)^2 + (f(\tfrac{\pi}{2}) - c)^2}
\end{aligned}
$$
And finding $b$:
$$
\begin{aligned}
\frac{f(0) - c}{f(\tfrac{\pi}{2}) - c} &= \frac{a\sin b}{a\cos b} = \tan^{-1} b \\
\implies b &= \operatorname{atan2}\left(f(0) - c, f(\tfrac{\pi}{2}) - c\right)
\end{aligned}
$$