# Is there a geometric proof of $\sin(3t) = 3\sin(t) - 4\sin^3(t)$ [duplicate]

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

$$\sin(3t) = 3\sin(t) - 4\sin^3(t)$$

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.