Proof of Euler's formula that doesn't use differentiation?

So I saw a 'proof' of the sine and cosine angle addition formulae, i.e. $\sin(x+y)=\sin x\cos y+\cos x \sin y$, using Euler's formula, $e^{ix}=\cos x+i\sin x$. By multiplying by $e^{iy}$, you can get the desired result.

However, this 'proof' appears to be circular reasoning, as all proofs I have seen of Euler's formula involve finding the derivative of the sine and cosine functions. But to find the derivative of sine and cosine from first principles requires the use of the sine and cosine angle addition formulae.

So is there any proof of Euler's formula that doesn't involve finding the derivative of sine or cosine? I know you can prove the trigonometric formulas geometrically, but it is more laborious to do.

For all real $z$, we have: $$e^z=\lim_{N\to\infty}\left(1+\frac zN\right)^N$$ so it seems like a good definition for complex $z$ as well. Letting $z=ix$ and using De Moivre and L'Hôpital gets you $\cos x+i\sin x$.

I will write down one of Eulers own proofs.

Euler starts with writing down De Moivres Formula (can be prooven by simple induction using some basic trig identities).

$$\cos(nx)+i\sin(nx)=\left( \cos(x)+i\sin(x)\right)^n$$

He says that $n$ is very large ($n \to \infty$) and $x$ is very small ($x\to 0$). The product of both will be a finite number called $\omega =nx$. Then he applies this as substitution for De Moivres Formula: $$\cos(\omega)+i\sin(\omega)=\left( \cos(\frac{\omega}{n})+i\sin(\frac{\omega}{n})\right)^n$$

Euler now applies the limit $n\to \infty$: $$\cos(\omega)+i\sin(\omega)=\lim_{n\to \infty}\left( \cos(\frac{\omega}{n})+i\sin(\frac{\omega}{n})\right)^n$$

using small angle aproximations $\cos(x)\approx 1$ and $\sin(x)\approx x$: $$\cos(\omega)+i\sin(\omega)=\lim_{n\to \infty}\left( 1+\frac{i\omega}{n}\right)^n=e^{i\omega}.$$

In the last line he applied the limit representation $e^x=\lim_{n\to \infty}(1+\frac{x}{n})^n$.

The proof might not be rigourus but it captures the main ideas in a very simple and beautiful way.

There is a way to prove Euler's formula without using power series. Try integrating $\frac {1}{1+x^2}$ using partial fractions to get a formula for the complex logarithm. You then have to use polar conversion formulas.

Another possible definition of sine and cosine is as solutions to the equation $$ u''+u=0, $$ with $\sin$ the one with $u(0)=0$, $u'(0)=1$, and $\sin$ defined as $-\cos'$ (uniqueness theorems implies this gives a workable definition). It is also apparent that if $u$ satisfies the equation, so does $au'$: $$ 0=a(u''+u)' = (au')''+(au'). $$ So we have the differential equation $$ \frac{d}{dx}(\cos{x}+i\sin{x}) = i(\cos{x}+i\sin{x}), $$ ($(\cos+i\cos')' = \cos'+i\cos''=\cos'+i\cos=i(\cos-i\cos')$) so $\cos{x}+i\sin{x}$ is a solution of $u'=iu$ with $u(0)=1$. But we know (from properties of the exponential, wherever you want to get them from) that $e^{ix}$ is also such a solution. Hence the uniqueness theorem gives $$ e^{ix}=\cos{x}+i\sin{x}. $$

Now, the addition formulae can also be proved by differentiation: we have that $$ u(x) = \sin{(x+y)} $$ is a solution to $u''+u=0$ with $u(0)=\sin{y}$, $u'(0)=\cos{y}$. Then uniqueness for differential equations implies that $$ u(x) = u(0)\cos{x}+u'(0)\sin{x} = \sin{y}\cos{x}+\cos{y}\sin{x}, $$ because $\cos$ and $\sin$ are linearly independent. You can do $\cos{(x+y)}$ in the same way.