Find the sums of the series: $$ \sum_{n=1}^{\infty}\frac{\cos nx}{n!},\ \ \ \sum_{n=1}^{\infty}\frac{\sin nx}{n!} $$
I did something like this: $$ \begin{aligned} &\sum_{n=1}^{\infty}\frac{\cos nx}{n!}+i\sum_{n=1}^{\infty}\frac{\sin nx}{n!}=\sum_{n=1}^\infty\frac{(\cos x+i\sin x)^n}{n!}=e^{\cos x +i\sin x}-1=\\ &=e^{\cos x}\cdot e^{i\sin x}-1=e^{\cos x}(\cos(\sin x)+i\sin(\sin x))-1\Rightarrow\\ &\Rightarrow \sum_{n=1}^{\infty}\frac{\cos nx}{n!}=e^{\cos x}\cos(\sin x)-1,\ \ \sum_{n=1}^{\infty}\frac{\sin nx}{n!}=e^{\cos x}\sin(\sin x) \end{aligned} $$ However, the answer section says that $\sum_{n=1}^{\infty}\frac{\cos nx}{n!}=e^{\cos x}\cos(\sin x)$, and I have no idea where the number $1$ got from there.
