What does $e^{2ix}$ and $e^{-2ix}$equal to?

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What does $e^{2ix}$ and $e^{-2ix}$equal to?

Does it equal to $2\cos(x)+2i\sin(x)$ and $2\cos(x)-2i\sin(x)$?

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2 Answers

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Euler's formula states:

$$ e^{ix} = \cos x + i \sin x $$

so \begin{align*} e^{2ix} &= e^{i(2x)} \\ &= \cos 2x + i \sin 2x \end{align*} and \begin{align*} e^{-2ix} &= e^{i(-2x)} \\ &= \cos (-2x) + i \sin (-2x) \\ &= \cos 2x - i \sin 2x \end{align*}

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No, but $$ e^{\pm 2ix} = \cos 2x \pm i \sin 2x $$

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