Problem - 3156
Show that
\begin{align*}
C_n^0-C_n^2+C_n^4-C_n^6+\cdots &=2^{\frac{n}{2}}\cos\frac{n\pi}{4}\\
C_n^1-C_n^3+C_n^5-C_n^7+\cdots &=2^{\frac{n}{2}}\sin\frac{n\pi}{4}
\end{align*}
Let $x=i$ in the expanded form of $(1+x)^n$:
$$(1+x)^n=(C_n^0-C_n^2+C_n^4-C_n^6+\cdots)+i(C_n^1-C_n^3+C_n^5-C_n^7+\cdots)$$
On the other hand,
$$(1+i)^n = \Big(\sqrt{2}\big(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\big)\Big)^n=2^\frac{n}{2}\Big(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\Big)$$
Setting the real and imaginary parts of these two relations equal gives the desired results.