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.
Best Answer
Your work is correct. The problem is probably a misprint.