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I am proving the formula $$ \sin{ \left(x+y \right)} =\sin{x} \cos{y}+\cos{x} \sin{y}$$ by using Euler's formula. This sum formula is needed when proving the derivative of sine.

I am only wondering if then I make a circular argument, since all proofs of Euler's formula which I have seen require knowing the derivatives of sine and cosine.

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  • $\begingroup$ No, it just requires that $e^{i s} e^{it} = e^{i(s+t)}$. $\endgroup$
    – copper.hat
    Aug 30, 2020 at 19:38
  • $\begingroup$ It really depends on your starting point. Rudin ("Real & complex analysis") defines $\exp, \cos, \sin$ in terms of their power series and derives relationships from there. $\endgroup$
    – copper.hat
    Aug 30, 2020 at 19:40

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Ahlfors in his complex analysis book defines $e^z$ to satisfy the equation $f'(z)=f(z)$ for all $z$ in the complex plane, and $f(0)=1$. From this definition he derives the power series expansion of $e^z$ and show it converges everywhere. Also He shows $e^{a+b}=e^ae^b$ using Leibniz rule for derivatives. Then defines $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ and $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$. From this Eulers formula is derivable without the derivative of $\sin$.

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  • $\begingroup$ How does he just define sine and cosine like that? $\endgroup$
    – mathslover
    Aug 31, 2020 at 3:30
  • $\begingroup$ Well, I would guess it has to do with the relationship for derivatives since we defined e'(z)=e(z), You can check sin'(z)=cos(z) and cos'(z)=-sin(z), and sin''''(z)=sin(z). It is very similar to how we define cosh(z)=e^z+e^-z/2. But these differential definitions of sin and cos are logically nice since they are defined on the entire complex plane, and there is no circularity in definitions, however they may feel less geometrically intuitive than cos(theta)=a/h of a circle. $\endgroup$
    – Mars
    Aug 31, 2020 at 15:52

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