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?

Start with the known Maclaurin series
$$\cos x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$$
and
$$\sin x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}\;.$$
The given series has factorials of even numbers in the denominator, so I began by shifting indices to make them $(2n)!$ to match the cosine series. After that it was mostly a matter of following my nose: at each step in the calculation below there’s really only one thing that suggests itself strongly.
$$\begin{align*}
\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}&=\sum_{n=2}^\infty(-1)^{n-1}\frac{n-1}{(2n)!}\\
&=\sum_{n=2}^\infty(-1)^{n-1}\frac{n}{(2n)!}-\sum_{n=2}^\infty(-1)^{n-1}\frac1{(2n)!}\\
&=\sum_{n=2}^\infty(-1)^{n-1}\frac{n}{2n(2n-1)!}+\sum_{n=2}^\infty\frac{(-1)^n}{(2n)!}\\
&=\frac12\sum_{n=2}^\infty\frac{(-1)^{n-1}}{(2n-1)!}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}-1+\frac12\\
&=\frac12\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)!}+\cos 1-\frac12\\
&=\frac12\left(\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}-1\right)+\cos 1-\frac12\\
&=\frac12\sin 1 + \cos 1 - 1\\
&=\frac12\sin 1+\cos 1-\cos 0
\end{align*}$$
