<s>[INST] Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of 
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$
Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \frac{\sin(1)}{2} + \cos(1) - \cos(0)$$
So I wrote down the partial sums of the given series and $\sin(1)$ and $\cos(1)$:
$$
\qquad\qquad\quad\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \quad - \frac{1}{4!} + \frac{2}{6!} - \frac{3}{8!} \cdots
$$
$$
\;\;\sin(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n+1}}{(2n+1)!} = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots
$$
$$
\cos(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n}}{(2n \qquad)!} = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} \cdots
$$
Looking at the numbers I can see that Wolfram|Alpha's result is correct: $\frac{1}{2}1 -  \frac{1}{2!} = 0$ and $\frac{1}{2}\frac{-1}{3!} + \frac{1}{4!} = \frac{1}{4!}$, so the $\cos(1)$-series is shifted by $1$ since there is no $1$ at the beginning of the given series, so it needs to be subtracted from $\cos(1)$: $-cos(0)=-1$. But how do I get here without Wolfram|Alpha?
 [/INST] let $m=n+1$, so your series reduces to $$\sum_{m=2}^\infty(-1)^{m-1}\frac{m-1}{(2m)!}=-\sum_{m=2}^\infty(-1)^{m}\frac{m}{(2m)!}+\sum_{m=2}^\infty(-1)^{m} \frac{1}{(2m)!}$$
$$=-\sum_{m=2}^\infty(-1)^{m}\frac{1}{2(2m-1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$
Note that $\frac{m}{(2m)!}=\frac{m}{(2m)(2m-1)!}=\frac{1}{(2)(2m-1)!}$, and also we evaluated the second series at $m=0$ and $m=1$
Replacing $m$ by $k+1$ in the first series and simplifying, we get,
$$\sum_{k=1}^\infty(-1)^{k}\frac{1}{2(2k+1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$
$$\sum_{k=0}^\infty(-1)^{k}\frac{1}{2(2k+1)!}-\frac{1}{2!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$
$$\sum_{k=0}^\infty(-1)^{k}\frac{1}{2(2k+1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1$$
$$=\frac{\sin (1)}{2}+\cos(1)-\cos(0)$$
</s>