Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=\binom{n}{k+1}i^k$, where $i=\sqrt{-1}$, to obtain
\begin{align}
\sum_{k\ge 0} \binom{n}{2k+1}(-1)^k &= \sum_{k\ge 0} \binom{n}{2k+1}i^{2k} \\
&= \sum_{k\ge 0} \binom{n}{k+1}i^k \frac{1+(-1)^k}{2} \\
&= \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1}i^k + \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1} (-i)^k \\
&= \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1}i^{k+1} - \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1} (-i)^{k+1} \\
&= \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k}i^k - \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k} (-i)^k \\
&= \frac{1}{2i}\left((1+i)^n-1\right) - \frac{1}{2i}\left((1-i)^n-1\right) \\
&= \frac{(1+i)^n - (1-i)^n}{2i} \\
&= \frac{\left(\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4)\right)^n - \left(\sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4)\right)^n}{2i} \\
&= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(-n\pi/4)+i\sin(-n\pi/4))}{2i} \\
&= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(n\pi/4)-i\sin(n\pi/4))}{2i} \\
&= \sqrt{2}^n \sin(n\pi/4)
\end{align}