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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.

Kamal Saleh
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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 $~x.$

user2661923
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  • Amazing rep coincidence! Thanks for your answer! – Kamal Saleh Apr 01 '23 at 21:59
  • @KamalSaleh, you accepted the answer so give it a +1. – user25406 Apr 02 '23 at 12:16
  • @user25406 I did :) I am assuming that the first two points aren't valid since the Taylor series was originally not defined for real numbers. Is that right? – Kamal Saleh Apr 02 '23 at 21:52
  • @KamalSaleh You have it backwards. The Taylor series was originally defined for real numbers rather than Complex numbers. This applies to the Sine function, the Cosine function, and the function $~f(x) = e^x.$ – user2661923 Apr 02 '23 at 22:04
  • @KamalSaleh Notice that in my answer, I just found and corrected an analytical error. Euler took the entire Taylor series of the Sine function and multiplied it by the scalar $~i = \sqrt{-1}.$ – user2661923 Apr 02 '23 at 22:07
  • @user2661923 I meant complex numbers. Oops. – Kamal Saleh Apr 03 '23 at 17:04