If i = sqrt{-1}, then frac{e^{xi} + e^{-xi}}{ | vedclass

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$(A)$ We know that the expansion of $e^z$ is $1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \infty$.Substituting $z = ix$, we get $e^{ix}

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$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \infty$

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$(A)$ We know that the expansion of $e^z$ is $1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \infty$.Substituting $z = ix$, we get $e^{ix} = 1 + \frac{ix}{1!} + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \dots \infty = 1 + \frac{ix}{1!} - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \infty$.Similarly, $e^{-ix} = 1 - \frac{ix}{1!} + \frac{(-ix)^2}{2!} + \frac{(-ix)^3}{3!} + \frac{(-ix)^4}{4!} + \dots \infty = 1 - \frac{ix}{1!} - \frac{x^2}{2!} + \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \infty$.Adding these two expressions:$e^{ix} + e^{-ix} = 2 \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \infty \right)$.Therefore, $\frac{e^{ix} + e^{-ix}}{2} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \infty$.

If $i = \sqrt{-1}$, then $\frac{e^{xi} + e^{-xi}}{2} = $

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