Given the general equation $f(t) = Y \sin (\omega t + \varphi)$ where $\omega$ is known and two points, $y_1 = f(t_1)$ and $y_2=f(t_2)$ the solution is
$$ Y = \frac{ \sqrt{ y_1^2 + y_2^2 - 2 y_1 y_2 \cos (\omega(t_2-t_1))}}{\sin ( \omega(t_2-t_1))} $$
$$ \varphi = 2\pi - \tan^{-1} \left( \frac{y_2 \sin \omega t_1 - y_1 \sin \omega t_2}{y_2 \cos \omega t_1 - y_1 \cos \omega t_2} \right) $$
Why?
I expanded the sine function into two components
$$ f(t) = A \sin \omega t + B \cos \omega t $$
where $Y=\sqrt{A^2+B^2}$ and $\tan(\varphi) = \frac{B}{A}$. The two points are
$$ y_1 = A \sin \omega t_1 + B \cos \omega t_1 $$
$$ y_2 = A \sin \omega t_2 + B \cos \omega t_2 $$
or in matrix form
$$ \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{pmatrix} \sin \omega t_1 & \cos \omega t_1 \\ \sin \omega t_2 & \cos \omega t_2 \end{pmatrix} \begin{bmatrix} A \\ B \end{bmatrix} $$
with the inverse
$$\begin{pmatrix} \sin \omega t_1 & \cos \omega t_1 \\ \sin \omega t_2 & \cos \omega t_2 \end{pmatrix}^{-1} = \frac{1}{\sin( \omega (t_2-t_1))} \begin{pmatrix} \mbox{-}\cos \omega t_2 & \cos \omega t_1 \\ \sin \omega t_2 & \mbox{-}\sin \omega t_1 \end{pmatrix}$$
or
$$ \begin{bmatrix} A \\ B \end{bmatrix} = \frac{1}{\sin( \omega (t_2-t_1))} \begin{bmatrix} y_2 \cos \omega t_1 - y_1 \cos \omega t_2 \\ y_1 \sin \omega t_2 - y_2 \sin \omega t_1 \end{bmatrix} $$
So
$$ Y = \sqrt{A^2+B^2} = \sqrt{ \left( \frac{y_2 \cos \omega t_1 - y_1 \cos \omega t_2}{\sin( \omega (t_2-t_1))} \right)^2 + \left( \frac{y_1 \sin \omega t_2 - y_2 \sin \omega t_1}{\sin( \omega (t_2-t_1))} \right)^2 } $$
and
$$ \varphi = n \pi + \tan^{-1}\left( \frac{B}{A} \right) = n \pi + \tan^{-1}\left( \frac{y_1 \sin \omega t_2 - y_2 \sin \omega t_1}{y_2 \cos \omega t_1 - y_1 \cos \omega t_2} \right) $$
Yes, of course it is possible. Proceed in the following way:
Let the three values at $-30,0,30$ be $y_1,y_2,y_3$ repec. Then you get three equations in $A,B,C$ \begin{equation}
\begin{split}
A\sin (-30+B)+C=&y_1\\
A\sin (B)+C=&y_2\\
A\sin (30+B)+C=&y_3
\end{split}
\end{equation}
From the first and last equation, after some trigonometric manipulation you get $$2A\sin B \cos 30=y_3-y_1\Rightarrow A\sin B=\frac{y_3-y_1}{\sqrt{3}}$$
Then from second equation you get $$C=y_2-\frac{y_3-y_1}{\sqrt{3}}$$
Now from the last equation you get after using the $\sin(A+B)$ expansion formula $$A\sin B \cos 30+A\cos B \sin 30=y_1\\ \Rightarrow \frac{(y_3-y_1)}{2}+\frac{A\cos B}{2}+y_2-\frac{y_3-y_1}{\sqrt{3}}=y_3$$ So you can find $A\cos B$ from here. So, now you can use $\displaystyle \sin^2B+\cos^2 B=1$ to get $$A=\sqrt{(A\sin B)^2+(A\cos B)^2}$$ Once you get $A$, then you can get $B$ from $$B=\tan^{-1} \left(\frac{A\sin B}{A \cos B}\right)$$ restricting $B$ in $[-\pi/2,\pi/2]$, you can get $B$.
Note: This procedure is valid for arbitrary angles, they need not be necessarily $30^\circ$ apart.
Best Answer
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 $b$:
$$ \begin{aligned} \frac{f(0) - c}{f(\tfrac{\pi}{2}) - c} &= \frac{a\sin b}{a\cos b} = \tan^{-1} b \\ \implies b &= \operatorname{atan2}\left(f(0) - c, f(\tfrac{\pi}{2}) - c\right) \end{aligned} $$