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

1.	If \[i=\sqrt{-1}\], then \[\frac{{{e}^{xi}}+{{e}^{-xi}}}{2}=\]
A.	\[1+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{4}}}{4\,!}+.....\infty \]
B.	\[1-\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{4}}}{4\,!}-.....\infty \]
C.	\[x+\frac{{{x}^{3}}}{3\,!}+\frac{{{x}^{5}}}{5\,!}+....\infty \]
D.	\[i\,\left[ x-\frac{{{x}^{3}}}{3\,!}+\frac{{{x}^{5}}}{5\,!}-.....\infty \right]\]
Answer» C. \[x+\frac{{{x}^{3}}}{3\,!}+\frac{{{x}^{5}}}{5\,!}+....\infty \]

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

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