Is there a geometric proof of $ \sin(3t) = 3\sin(t) - 4\sin^3(t) $ First, a bit of notation. Let $x(t) = \cos(t)$ and $y(t) = \sin(t)$ . Let $x$ and $y$ be abbreviations for $x(t)$ and $y(t)$ .
Using de Moivre's formula, proving that $ \sin(3t) = 3\sin(t) - 4\sin^3(t) $  is straightforward. I'm curious whether this formula has a geometric proof or one that doesn't use complex numbers and preferably not calculus.
For comparison, the double angle formula has a wonderful geometric proof here.
Here's the straightforward proof:
$$\sin(3t)$$
$$\Im(\exp(i\times 3t))$$
$$\Im(\exp(it)^3) $$
$$\Im((x+iy)^3) $$
$$\Im(x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3) $$
$$\Im(x^3 + 3x^2yi - 3xy^2 - y^3i) $$
$$3x^2y - y^3 $$
$$3 {\times} (1-y^2) {\times} y - y^3 $$
$$3y - 3y^3 - y^3 $$
$$3y - 4y^3 $$
And thus
$$ \sin(3t) = 3\sin(t) - 4\sin^3(t) $$
 A: Since you also ask for possibly a non-geometric proof which doesn't involve calculus & preferably not calculus, note you can show this using several Trigonometric identities.
Note the $2$ basic trigonometric angle sum formulas of
$$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \tag{1}\label{eq1}$$
and
$$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \tag{2}\label{eq2}$$
Also, there's
$$\cos^2(t) = 1 - \sin^2(t) \tag{3}\label{eq3}$$
Now, you can split $3t = 2t + t$, and use the $3$ formulas above, to get
\begin{align}
\sin(3t) & = \sin(2t + t) \\
& = \sin(2t)\cos(t) + \cos(2t)\sin(t) \\
& = (2\sin(t)\cos(t))\cos(t) + (\cos^2(t) - \sin^2(t))\sin(t) \\
& = 2\sin(t)\cos^2(t) + \cos^2(t)\sin(t) - \sin^3(t) \\
& = 3\sin(t)\cos^2(t) - \sin^3(t) \\
& = 3\sin(t)(1 - \sin^2(t)) - \sin^3(t) \\
& = 3\sin(t) - 3\sin^3(t) - \sin^3(t) \\
& = 3\sin(t) - 4\sin^3(t) \tag{4}\label{eq4}
\end{align}
As your question text shows, it's somewhat easier to use de Moivre's formula instead.
