I will give as the example the $\sin(2x)$ for $x$ in the third quadrant with $\sin x = -\frac35$. Without using the formula for the sine of double angles, this becomes a geometry problem of medium complexity. (It will look more complex here because I am not appending diagrams. I have drawing hints in brackets.).
Consider a line $OX$ [we can draw it as a horizontal line with $X$ to the right of $O$] and a line $YY'$ perpendicular to $OX$ at $O$ [drawn with $Y$ above $O$ and Y' below $O$] with $\overline{OY'}=3$. Add a line $OP$ perpendicular to $YOY'$ meeting at $Y'$ with $\overline{PY'}=4$ such that angle $XOP$ is obtuse
[$P$ is drawn to the left of and lower than $O$]. Then by the Pythagorean theorem $\overline{OP} = 5$, and the angle between $XO$ and $OP$ taken in the
"long" [counterclockwise] direction is in the third quadrant. In fact, by this construction, this "improper" angle is $x$.
Having set up the diagram, we now prepare an angle of $2x$:
Extend line $PO$ in the direction of $O$, to point $Q$, such that $\overline{OQ}=4$. Construct angle $QOR'$ equal to angle $Y'OP$. Then angle $XOR'$, a first-quadrant angle, is $2x$. Mark point $R$ on $OR'$ such that
$\overline{OR} = 5$; angle $XOR$ is $2x$.
Now we go about finding $\sin(2x)$. Drop a perpendicular from $R$ to $OX$ meeting $OX$ at $U$. In right triangle $RUO$, angle $UOR$ is $2x-2\pi$ which, for purposes of trig functions, is the same as $2x$. So $\overline{RU} = 5\sin{2x}$.
To find $\overline{RU}$, draw a line parallel to $XO$ through point $Q$, meeting $OR$ at $S$ and $RU$ at $T$, and drop a perpendicular from $Q$ meeting $OX$ at $V$; then $\overline{RU} = \overline{TR} + \overline{QV}$. In right triangle
$OVQ$, angle $VOQ = x-\pi$ since it is a vertical angle with angle $POX'$, so
$\sin \angle VOQ = \frac35$, which gives
$$
\overline{QV} = \overline{OQ} \sin \angle VOQ = 4\cdot\frac35 = \frac{12}{5}
$$
And in right triangle $QTR$, angle $TQR = x$ since angle $OQT$ forms alternate interior angles with angle $VOQ = x$, and angle $TQR$ is complementary to angle
$OQT$. Thus $\sin \angle TQR = +\sqrt{1-(3/5)^2} = \frac45$. Then since $\overline{RQ}=3$,
$$
\overline{TR} = \overline{RQ} \sin \angle TQR = 3 \cdot\frac45 = \frac{12}{5}.
$$
Thus $$\sin (2x) = \frac15\left( \cdot\frac{12}{5}+\frac{12}{5} \right)= \frac{24}{25}$$
If your teacher is assigning this problem under these conditions, it must be that she is motivating the derivation of the formula for th sine of twice an angle. Because nobody, once in possession of that formula, would ever go through this complexity gratuitously.
