<s>[INST] How to show $\phantom{d}_d C_0+\phantom{d}_d C_4 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) $? I want to show following identities
\begin{align}
  &\phantom{d}_d C_0+\phantom{d}_d C_4 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) \\
  &\phantom{d}_d C_1+\phantom{d}_d C_5 + \cdots = 2^{d-2} + 2^{\frac{d}{2}-1} \sin(\frac{d \pi}{4}) \\
 &  \phantom{d}_d C_2+\phantom{d}_d C_6 + \cdots = 2^{d-2} - 2^{\frac{d}{2}-1} \cos(\frac{d \pi}{4}) \\
 & \phantom{d}_d C_3+\phantom{d}_d C_7 + \cdots = 2^{d-2} - 2^{\frac{d}{2}-1} \sin(\frac{d \pi}{4})
\end{align}
Can you give me any hint or computation for this identities?
 [/INST] HINT:
As the subscript of each subsequent binomial coefficient differ $4,$
If $x^4=1\implies x=\pm1,\pm i$
Set $x=\pm1,\pm i$ one by one in the following identity $$(1+x)^n=\sum_{r=0}^n\binom nr x^r$$
Find $$\dfrac{(1+1)^n\pm(1-1)^n\pm(1+i)^n\pm(1-i)^n}4$$
$1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)$
$(1+i)^n=2^{n/2}\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)^n=2^{n/2}\left(\cos\dfrac{n\pi}4+i\sin\dfrac{n\pi}4\right)$
Similarly, $(1-i)^n=2^{n/2}\left(\cos\dfrac{n\pi}4-i\sin\dfrac{n\pi}4\right)$
Can you take it from here?
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