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I realize that the equality from the title is wrong, but I do not understand why. A correct equality would be $\frac{\cos(\alpha) + i\sin(\alpha)}{\cos(\alpha)-i\sin(\alpha)} = \cos(2\alpha)+i\sin(2\alpha)$ by multiplying with the conjugate of the denominator (i.e. the numerator). However, I wanted to try another way too: $$\frac{\cos(\alpha) + i\sin(\alpha)}{\cos(\alpha)-i\sin(\alpha)} = \frac{\cos(\alpha) + i\sin(\alpha)}{\cos(\alpha) + i\sin(-\alpha)} = (\cos(\alpha) + i\sin(\alpha))(\cos(\alpha) + i\sin(-\alpha))^{-1} \\[4ex] \overset{\text{Applying de Moivre's formula}}{============} (\cos(\alpha) + i\sin(\alpha))(\cos(-\alpha) + i\sin(\alpha))= \cos(\alpha - \alpha) + i\sin(\alpha + \alpha) = 1 + i\sin(2\alpha)$$

I don't think I made any mistakes or at least I'm not able at all to spot them, so what exactly is wrong with my attempt? Any help is greatly appreciated!

J__n
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    No (it works for exponents in $\Bbb Z$) but $(\cos(\alpha) + i\sin(\alpha))(\cos(-\alpha) + i\sin(\alpha))\ne \cos(\alpha - \alpha) + i\sin(\alpha + \alpha).$ – Anne Bauval Apr 06 '23 at 18:06
  • @AnneBauval Yes, second to last step. – Sasikuttan Apr 06 '23 at 18:09
  • In class, we've learnt that for $z_1 = r_1(\cos(t_1) + i \sin(t_1))$ and $z_2 = r_2(\cos(t_2) + i \sin(t_2)),\ r_1,r_2 > 0$, $z_1 \cdot z_2 = r_1 \cdot r_2(\cos(t_1 + t_2) + i\sin(t_1 + t_2))$. Was the professor wrong? I'd say the mistake is that $\cos$ is an even function, so the minus can be removed, now leading to the correct equality, but, if what I just said is correct, then my question becomes in what circumstances do I need to be careful at the sign of the cosinus. – J__n Apr 06 '23 at 18:11
  • So basically, OP should check the way they applied the sum of angles formulas for cosine. – Elliot Yu Apr 06 '23 at 18:12
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    @J__n Well, I guess my now deleted comment is relevant in a different way... Notice how your teacher's formula has the same argument for the sines and cosines in each of the complex numbers, $z_1 = r_1 (\cos \color{blue}{t_1} +i\sin \color{blue}{t_1})$. This formula does not apply to a number of the form $z_2' = r_2 (\cos(\color{red}{-t_2})+i\sin\color{blue}{t_2})$ – Elliot Yu Apr 06 '23 at 18:14
  • @ElliotYu ah, that makes sense! I don't think I would've ever thought about that. Thank you for pointing that out, and thanks to all for the time and help on such a basic mistake! – J__n Apr 06 '23 at 18:17
  • Though it does apply if you remember that $\cos(-t_2)=\cos(t_2).$ So imo your main "fault" was not to try to apply it, but to write different arguments for $\cos$ and $\sin$ in your rhs ($\alpha-\alpha$ and $\alpha+\alpha$). When the formula applies, the arguments are the same, as you wrote in your first comment. – Anne Bauval Apr 06 '23 at 20:34

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Since De Moivre’s formula can be applied only for those complex number in standard polar form, therefore the wrong step should be correct to $$ \begin{aligned} {[\cos \alpha+i \sin (-\alpha)]^{-1} } & =[\cos (-\alpha)+i \sin (-\alpha)]^{-1} \\ & =\cos \alpha+i \sin \alpha \end{aligned} $$ Hence we have $$ \begin{aligned} \frac{\cos \alpha+i \sin \alpha}{\cos \alpha+i \sin (-\alpha)} & =(\cos \alpha+i \sin \alpha)(\cos \alpha+i \sin \alpha)\\& =\cos 2 \alpha+i \sin 2 \alpha \end{aligned} $$

Lai
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