I assume you're looking for series expansions of $\cos(ix)$ and $\sin(ix)$. Given Euler's formula, $e^{ix}=\cos(x)+i\sin x,$ and the fact that $\cos,\sin$ are real for real $x$, we may equate the real and imaginary parts of both sides of the above equation in order to find series for $\cos x$ and $\sin x$. Using the series expansion of the exponential function, and the fact that powers of $i$ cycle, we arrive at ($\color{Red}{\text{can you tell why?}}$): $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots;\qquad \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$ (One issue here is circularity: the above formulas are typically used in deriving Euler's formula in the first place! But I will assume this issue is of no concern for the purposes of this question.)
From here, we may replace $x$ with $ix$ in the expansions of cosine and sine to obtain the series expansions of $\cos(ix)$ and $\sin(ix)$, in which case we get ($\color{Red}{\text{can you check this for yourself?}}$): $\cos(ix)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots;\qquad \sin(ix)=i\left(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots\right).$
We see from the above that $\cos(ix)+i\sin(ix)$ is purely real; indeed it is exactly the series expansion of $e^{-x}$ we would expect, and we know that is also purely real for real $x$. This means that taking the real and imaginary parts of $e^{-x}$ will not give us $\cos(ix)$ and $\sin(ix)$: the only way taking the real and imaginary parts of an expression of the form $a+bi$ will gives us back $a$ and $b$ respectively is if both $a$ and $b$ are real numbers, which is not the case here.