We know that $$e^{ix}=\cos x+i\sin x$$Substituting $x\mapsto-x$: $$e^{-ix}=\cos x-i\sin x$$Multiplying the two equations: $$\cos^2x+\sin^2x=1$$Is this a new proof of the Pythagorean identity that doesn't have any circular reasoning? I believe so since Euler's formula can be proved using the Taylor series, which I don't think has any relation with the Pythagorean theorem. But I want to make sure.
I got this proof when I was trying to find $\cos(i)$ on my own.
No, it isn't really a proof at all.
This is because Euler deliberately choose to arbitrarily (and artificially) define $~e^{i\theta}~$ as $~\cos(\theta) + i\sin(\theta).~$
That is, $~e^{i\theta}~$ was intentionally defined to provide syntactic sugar for the expression $~\cos(\theta) + i\sin(\theta).~$
Euler relied on the fact that then $~e^{i\alpha} \times e^{i\beta} = e^{i(\alpha + \beta)}.$
Further, Euler noticed that (it could be argued) the definition makes sense because:
You can start with the Taylor series of the Sine function, and multiply the entire series by the scalar $~i = \sqrt{-1}.$
You can also use the Taylor series for Cosine of $x.$
Combining the two, you can compare the combination to the Taylor Series for $~e^x,~$ where you substitute $~ix~$ for $~x.$