Evaluate a binomial-like sum

Question

Let $x \in \mathbb{R}$ and let $n \in \mathbb{N}$ then evaluate: $$\sum_{k=0}^n{n \choose k}\sin\left(x+\frac{k \pi }{2}\right)$$

I could only go up to breaking this sum in two parts of $\sin x$ and $\cos x$ in the following way: $$\sum_{k=0}^n{n \choose k}\sin\left(x+\frac{k \pi }{2}\right)=\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k{n \choose 2k}\sin x+\sum_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k{n \choose 2k+1}\cos x$$ I have the intuition that $\sin(x+y)=\sin x\cos y+\sin y\cos x$ should be used at some point. Can someone please help me out?

Answer 1

$\sin\left(x+\dfrac{k\pi}2\right)$ is the imaginary part of $$e^{i\left(x+\dfrac{k\pi}2\right)}=e^{ix}(e^{i\pi/2})^k=e^{ix}i^k$$

using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

$$\implies\sum_{k=0}^n\sin\left(x+\dfrac{k\pi}2\right)$$ imaginary part of

$$=\sum_{k=0}^ne^{ix}(e^{i\pi/2})^k=e^{ix}(1+i)^n$$

Finally, $1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)$

$\implies(1+i)^n=2^{n/2}\left(\cos\dfrac{n\pi}4+i\sin\dfrac{n\pi}4\right)$ by Proof for de Moivre's Formula

Answer 2

$\sum_{k=0}^{n} {n\choose k} sin (x+\frac {k\pi} 2)$ is the imaginary part of $\sum_{k=0}^{n} {n\choose k} e^{i (x+\frac {k\pi} 2)}=e^{ix} \sum_{k=0}^{n} {n\choose k} e^{i \frac {k\pi} 2}$ and $\sum_{k=0}^{n} {n\choose k} e^{ i\frac {k\pi} 2}$ is the binomial expansion of $(1+e^{ i\frac {\pi} 2})^{n}=(1+i)^{n}$.