I have a book that has a brief history of the complex numbers and it covers de Moivre's formula:
$(\cos(x) + i\sin(x))^n = \cos(nx) + i\sin(nx)$.
I am very curious as to how this result was originally found, or derived, BEFORE Euler's formula was around. Also, what was the original proof of this?
Euler was the first to write down de Moirve's formula which I found in The Elementary Mathematical Works of Leonhard Euler starting on page 29. Euler found $$ (\cos(x)\pm i\sin(x))(\cos(y)\pm i\sin(y))(\cos(z)\pm i\sin(z)) = \cos(x \pm y \pm z)\pm i\sin(x\pm y\pm z) $$ from here he deduce what we call de Moirve's theorem $$ (\cos(x)\pm i\sin(x))^n = \cos(nx)\pm i\sin(nx) $$ Then it follows that \begin{align} \cos(nx) &= 1/2[(\cos(x)+ i\sin(x))^n+(\cos(x) - i\sin(x))^n]\tag{1}\\ \sin(nx) &= 1/2[(\cos(x)+ i\sin(x))^n-(\cos(x) - i\sin(x))^n]\tag{2} \end{align} Euler than expanded equations $(1)$ and $(2)$ by the binomial theorem. He then let $x$ be a small angle approximation and $n$ be infinitely large so that $xn$ was finite. Let $xn=v$. Then $\sin(x) = x = v/n$ and we get \begin{align} \cos(v) &= 1 - \frac{v^2}{2!} + \cdots\\ \sin(v) &= v - \frac{v^3}{3!} + \cdots \end{align} Using equations $(1)$ and $(2)$ and letting $j = n$ be the infinitely large number so that $jx=v$ so that $nx = v = v/j$. Now we have $\sin(x) = v/j$ and $\cos(x)=1$. With these substitutions, we get \begin{align} \cos(v) &=\frac{(1+iv/j)^j+(1-iv/j)^j}{2}\\ \sin(v) &=\frac{(1+iv/j)^j-(1-iv/j)^j}{2i} \end{align} In the linked document, the author who compiled the document (Euler) previously has shown that $(1+z/j)^j = e^z$. Letting $z=\pm iv$, we get \begin{align} e^{iv} &= \cos(v) + i\sin(v)\\ e^{-iv} &= \cos(v) - i\sin(v) \end{align}
