I am asking if it's possible to find the derivatives without geometry, circular thinking, and definitions that don't make sense to introduce based on the intuitive meaning of sine and cosine (i.e that assume you already know the answer.)
Note: I take the Pythagorean identity, the zeroth and first derivatives at zero, and that sine is odd and cosine is even to be used as the definitions for sine and cosine.
Edit: I will also assume they are bounded by $1$ and $-1$ and never constant on a continuous interval.
I tried using the Pythagorean theorem and that sine is odd and cosine is even and reached that $$\cos\theta+i\sin\theta=e^{g(\theta)}$$ where $g$ is an odd function. However, I couldn't prove that $g''=0$ (equivilant to $g''$ is even) which we can use (along with $\sin'(0)=1$ and $\cos'(0)=0$) to deduce $g=i\theta$.
So, I thought maybe it's much easier. If we differentiate the Pythagorean identity, we get: $$\sin\theta\sin'\theta=-\cos\theta\cos'\theta\tag{$\star$}\label{1}$$ Squaring both sides to avoid multiple cases for the signs, we get: $$(\sin\theta\sin'\theta)^2=(\cos\theta\cos'\theta)^2$$ What are the possibilities here?
We certainly know $\sin^2\theta≠\cos^2\theta$ and $(\sin'\theta)^2≠(\cos\theta\cos'\theta)^2$, so it seems that the only possibility is: $$(\sin'\theta)^2=(\cos\theta)^2\\ (\cos'\theta)^2=(\sin\theta)^2$$ My question is: Is this logic true, can we prove such a thing? If so, we can proceed as follows, differentiate the equations: $$\sin'\theta\sin''\theta=\cos\theta\cos'\theta\\ \cos'\theta\cos''\theta=\sin\theta\sin'\theta$$ Now, substitute from $\eqref{1}$: $$\sin''\theta=-\sin\theta\\ \cos''\theta=-\cos\theta$$ Afterwards, we can use ansatz $e^{\lambda\theta}$ and the zeroth and first derivatives at $0$ to solve for the exponential forms of sine and cosine. As a result, we can deduce the derivatives and the angle-sum formulas too.